EMI & Other Chapters in 1 Shot | Physics | KCET Complete Revision | 120+ Fixed Marks 🔥

hello everyone good morning welcome to PW Canada YouTube channel so Kion okay so last few parts so hopefully so next four part this is third part so questions first right so first don't worry okay okay okay yes yes hello hello hello good morning good morning good morning everyone yes I can see lot of uh numbers and names yes Namaste Namaste good morning yes thank you sari thank you so much good morning good morning everyone good morning yes six chapter King one sir six chapters don't worry okay so last last two parts now first okay be hon okay so lat okay most important questions and most important concept okay not to worry okay so six chapters right unless okay Sox hours ma okay maximum one and half hour to two hours that's all okay yes yes question right okay yes so yes Shan Shan yes yes Kaa let's start okay so one minute wait okay one one minute wait so okay so this is for cheing okay yes com okay yes yes yes right yes okay so right right NES hello hello good morning right so sure just normal is it clear okay now no yes 100% ready okay so let's get started without wasting our time because we have a six chapters to complete okay so so physics subjects the first one electromagnetic induction actually electromagnetic induction alternating current previous class okay okay thank you so much pleas because yes okay every after explanation not to worry okay so sometimes okay right yes yes it is a revision class for you already Okay so definitely help right yes okay thank you thank you yes alternating current okay now next one dual nature of radiation and matter after that after that we are moving with atom atoms and nucle actually okay now last one semiconductor Electronics semiconduct semiconductor Electronics so simple Concepts okay P Junction and doing okay okay right yes exam okay exam electromagnetic induction in the three marks okay electromagnetic induction in three questions or three Mar okay of radiation and matter in the three questions and atoms in the last two last to last to four okay to four so let's consider three only and in nuclei it is a fixed that is the three questions okay three questions and three questions total questions next one semiconductor Electronics three questions 18 so total 18 18 questions from this session 18 questions okay good morning everyone good morning so welcome back to the PW YouTube channel and revision okay right but numb problem right so right maximum okay so let's get started and notes after 12 hours okay after 12 hours not right so after 12 hours don't worry okay so after 12 hours sometimes okay yes next one so that's all extra extra extra now it's time to start our lesson so first one electromagnetic induction electromagnetic induction are you ready ready yes yes download any downlo okay yes Mah very good very good very good okay yes okay ready ready ready ready yes 100% yes the first one magnetic flux question multiple times so first of definition magnetic flux magne flux the number of magnetic field lines passing perpendicular to the surface area Okay simple definition the number of mag magnetic field lines passing perpendicular to the surface area if I consider this is a b magnetic magnetic field and this is the area vector and this is a perpendicular to the surface area right perpendicular to the surface area and angle between these two that is a Theta okay so mathematical formula magnetic flux that is equal to B vector and a vector B do a vector dot product of a two Vector it is given by b a COS Theta is it clear b a COS Theta right now so this five depends on the strength of the magnetic field area of a crosssection area and Theta that is the angle between b and a right so now suppose unifor Magi uni B is constant a is constant a area of crosssection this is area Okay area of the surface area fix suppose B is a uniform same a is same and it is depends on only Ona between b and a so finally what is theit me square or wber okay Tesla met square and wber W capital W small letter B clear and quantity this magnetic flux is a scalar quantity sir magnetic field is a vector quantity and area is also vector quantity how this magnetic flux is a scalar quantity you know right first year the dot product of two vectors is a scalar so H it is a scalar quantity yes that's all question magnetic flux minimum magnetic flux minimum so formula magnetic flux that is given by ba a COS Theta right ba a COS Theta suppose suppose suppose consider this is surface area okay this is surface this is surface and this is surface andic to the surface area Vector is always perpendicular to the surface maybe upward vertically up maybe vertically up or vertically down okay maybe upward or downward vertically up or vertically down are vector a field lines magne fi Lin can you identify can you identify is minim for example for example okay this is the surface vector vector this is area Vector okay vector vector is zero point when Theta is 90° sir 90° Now define the angle between A and B what is the angle 90° right so 90° what about the flux ba a COS 0° so flux is b a sorry 90° cos 90 it is a zero so flux finally it is a zero wording instead of 90° what is the magnetic flux okay what is the magnetic flux if magnetic field lines okay so magnetic field lines magnetic field lines magnetic field lines passing magnetic field lines passing parallel to the plane parallel to the plane parallel to the plane or perpendicular perpendicular perpendicular to the area Vector area Vector area Vector is it clear is itar yes flux is zero When A and B angle between A and B 90° Daga it is a zero or else the magnetic field lines passing parall to the plane yes very good very good very good yes yes and next one okay next one maximum when Theta is 0° when Theta is 0° or 180° 180° 180° how this 180° for example dially let's consider this is surface area this is surface area okay let's consider this is surface area and area Vector area Vector area VOR M right area Vector M and Magi Magi along the area Vector so this is a b and this is this is area Vector okay and what is the angle between them Theta equal to0 am I right Theta is equal to0 so what about 180° what about 180 so let's consider this is also surface okay and area Vector area Vector okay and magnetic field okay let's consider this as a magnetic field and area Vector now okay so let's consider this is area vector and this is the magnetic field okay this is a magnetic field and what is the angle between them opposite anti parall to each other 180° right 180 magnetic flux magnetic flux it is given by ba cos Theta cos Theta first Z it is a one cos0 is a one so therefore B plus ba 18us ba so finally I can write plus r minus B A plus r minus B A and plus indicates directed outward from the surface flux negative directed into the plane magne Fields simple okay like electric field only clear what is the magnetic flux okay what is the magnetic flux when magnetic field when magnetic field lines okay when magnetic field lines when magnetic field lines passing passing anti parallel okay anti parallel magnetic field passing passing perpendicular to the plane okay perpendicular to the plane perpendicular to the plane or parallel parallel ATA anti parallel anti parallel to area Vector clear magnetic field lines passing parall to the plane parall yes clear clear answer clear yes yes ah K first P yes perpendicular to the plane perpendicular to the plane yes for Maxima very good very good very good okay so let's move on to the next one farad Los okay farad Los okay there will be induced EMF so induced current on concept so first one the phenomenon in which an EMF and ends a current is induced in a circuit by varying magnetic field so called electromagnetic induction this is a definition definition so next one first law first law inste that the magnitude of induced EMF in a circuit is equal to the time rate of change of a magnetic flux linked with the circuit mag flux okay that is a ba right ba cos Theta maximum because Vector same magn D ID DT that is the rate of change of a magnetic flux okay so firstus D the direction of induced EMF hence hence the direction of induced current in a closed loop opposite to the change in a magnetic flux what is the meaning of this one lens law okay lens explain for example due to this change in magne flux okay due to change in magnetic flux conservation is that clear right okay now for this is for a loop okay this is for a loop number ofs one number ofs greater than one it can be two it can be three it can be four n number minus n D5 / DT okay D5 divid DT so it's what is the inded lens ex okay because many times they have asked what is the direction of induced current that poity of induced is such that it tends to produce a current which is opposed change in a magnetic flux that produced it and for example mple examples try to solve as many as problems okay presentation okay for example okay this one okay bar magnet what is the magnetic field Direction north to south right north to south okay but north to south mag flux because I'll be considering this one as a moving towards the coil moving towards the Magi anlock okay North Pole towards anticlockwise a anticlockwise m anticlockwise Direction okay according to okay so next one ex right yes so start answering this question yes pole okay maget so according to Loop okay Loop okay South Pole right South Pole that is a clockwise right so clockwise clockwise induced but magne maget clockwise suppose suppose okay correct let's consider this is a loop okay this is a loop and this is a bar magnet okay this is downward consider this as a South Pole okay already induced EMF produ due to rate of change of magnetic flux direction if it is a South Direction only clockwise right clockwise butse for example right inded EMF sorry current EMF clockwise [Music] but opposite indf anticlockwise for example what is that is 9.8 9.8 m/s acceleration but due to this opposes the motion induced coil okay yes so therefore right answer which is the right answer opposite direction that is option C that is option C yes 50% okay now next one huh this one I know it is a numerical right so question is the magnetic flux linked with a coil varies as 5 is = 3 t² + 40 + 9 the magnitude of in EMF induced at A T is equal to 2 second and flux FL according to First Law according to the first magude of therefore keep that as it is okay magnitude first so D divid DT according to the F first law so d as it is and what is the value of five that is the 3 t² + 40 + 9 okay so further further differentiation three constant keep it as it is what is the differentiation of T square that is a 2T plus four is a constant what is the differentiation of a t that is a one and it is a constant differentiation of a constant that is the zero so e is equal to e and EMF okay 3 into 2 okay 3 into 2 that is a 6t + 4 so this is the EMF but also they have given time at two at 2 second so EMF that is equal to 6 t + 4 e is equal to 6 into 2 into 4+ 4 that is 16 Vol right option b option b very good option b option b is the right answer very good so lead board a gamer PES God G V Rocky and S creation rames H Deepak oh okay now let's move on to the next question cases of induction direction direction to be okay try to understand the concept okay type ofic okay right yes yes it is enough okay now first try to understand PL key okay consider this is a PLU key and this is a battery and this is a positive terminal and this is a negative terminal current FL this is I this is also I and this is also I so previous chap so direction of Magi that prodi around it then what is the magnetic field at Point P point a point a and this is point B this right side point B and this is a point a okay because there is let's consider right hand side what is the direction of a magnetic field what is the direction of a magnetic field use right hand thumb rule Max right hand thumb rule okay so this thumb indate direction of a current and Cur finger IND direction of magnetic field what is the direction of a current downward okay and E side inward right outward so outward that is a plus K that is a z AIS so magnetic field Direction plus K that is z Direction plus Z direction right so plus Z Direction outward magnetic field do the direction right now okay nowe to this caring conduct magne field prod what is the direction of magnetic field outward right outward outward using Loop law Loop law direction of a magne field outward then what is the current produced in the coil what is the current produced in the coil what is the EMF produced in the coil straight you can use this thumb as a current direction and a finger as a direction of a magneet field not only direction of a magnetic field outward direction of magne field outward indicate direction of Mag field outward what about this curve curve finger indicates that is the direction of current due to this current so current direction anticlockwise right anticlockwise so Direction y anticlockwise so first case you know initially the switch is open the current in the circuit is zero hence induced EMF okay EMF is zero therefore induced current is also zero magne fi very good okay next one when switch is closed the current flows in a circuit circuit flow circuit flow EMF will not be equal to zero am I right EMF will not be equal to zero directionally current produ that is anticlockwise anticlockwise okay now current direction current direction in Loop is clockwise what is the reason direction of a magnetic field and this is the direction of a current anticlockwise direction it opposes a change opposes a change constant according to lens law it opposes a change OPP this current induced will be current that is induced current will be anticlockwise clockwise so opposite direction induced current so induced current this is induced current so what about this one this is due to battery this is due to battery this is due to battery is it clear right is it clear so first so clockwise induced current okay induced current clockwise clockwise Direction first case that is opposite to the anticlockwise anticlockwise opposite clockwise direction right yes okay next two okay right now next dire find the direction of induced current in the coil due to external circuital CC direction of current okay key closed current FL okay so first direction of downward right downward what is the direction of this one okay what is the direction of this one outward right outward but first initially switch is closed already switch CL okay oppos what if if I consider this is a positive and this side is a negative so positive and this is a negative and current ection flow Direction ection FL now direction of a current upward and what is the direction of a magnetic field into the plane so e Cas magnetic field Direction minus K magnetic field Direction minus K okay so yeah minus Magi okay magne field Direction into the what is the direction of a current what is the direction of a current due to this current okay what is the direction of a current due to this current for so due to this into the plane into Theus K so into the IND now into the use right hand rule only but instead of that so into thee thumb into the plane curl clockwise Cur in clockwise so this is current due to this current okay because mag due to due to a lens law opposite opposite directionally okay opposite direction induced current induced current is that clear when a switch is open the current flows in the zero okay in the circuit is zero so first one initially switch is closed in the CC directionally anticlockwise right anticlockwise so first done second done because Swit is not equal Z but what is the direction anticlockwise clear yes EMF anticlockwise current clockwise ah current in the sense external circuit in the clockwise but induced current opposite directionally getting my point okay right clear first confir first one second one and third one all yes okay yes very good very good EMF tends to produce a current in opposite to in opposite to incre opposite okay very good very good okay fine no problem okay now yes analysis of a flowing data okay following dat data okay following data so first one velocity of a loop or ring move away from the current carrying wire that is along plus I first two slides okay so again I'm considering loop again I'm considering Loop current switch CL so current FL okay current FL current FL yes current FL so what is the direction of current downward okay what is the direction downward this side what is the magne field out of the plane out of the plan dot in the present dot in the present mod this is a magnetic field that is outward in Direction okay outward in Direction now what is the direction of a current in the coil not induced EMF induced current what is the direction of a current in this coil due to this current not induced not induced okay so first one direction of magnee to this current anticlockwise direction anticlockwise right anticlockwise is it clear now okay let's consider first two let's consider this is a constant okay I is constant okay is Conant thatf is zero first condition next one suppose this is a magne fieldon okay okay move Direction plus xais okay plus X AIS now from the wi distance from the increase distance distance distance is increasing then what what happens to the magnetic field degre why previous concep moving chares magm Bal mu I 4i IED by a straight wire bracket of sin Theta 1 plus sin Theta 2 what is that a what is that R indicates that is the distance okay Conant but inded current direction CH if primary coil current increases that is for in case of a transformer not in this case okay that is a for a Transformer Transformer what you are saying that is for Transformer Ro correct invers proportional FSE anticlockwise because how parameters are related five is dependent on magnetic field area is constant area of the loop is constant dependent on B and E is dependent on D5 divid DT and 5 dependent on B so that is nothing but it depends on B itself right what happens to the e e also degrees if B de am I right supp very good EMF inded in anticlockwise in the same direction because Direction inded induced induced current prod right anticlockwise very good very good wonderful okay so first due to this current anticlockwise inded EMF induced EMF this is current due to this one current due to this one yes very good very good yes Paul welcome back to the session right yes one second yes so first case clear first case clear done next one again Loop same okay so current is is constant okay current due to thisi that is out of the plane that is a plus K cap okay plus K cap now due to this direction right anticlockwise this is anticlockwise anticlockwise this is a current okay now velocity of a loop or ring move closer to the current carrying wire as a distance decreases what happens to the magnetic field as a distance as a distance increases magnetic field right IND therefore anticlockwise indf prod is it clear clockwise wonderful wonderful step byep learning okay clockwise anticlockwise right now next one yes same current direction same amount of current is Flowing okay same amount of current is Flowing okay so direction am I right but it is not induced current okay it is not yeah it is inded current but lens law okay now velocity of a loop or ring Move Along plus J or minus J Direction the magnetic field Rems the same or constant now tell me tell me what is the direction everybody Direction everybody Direction everybody what is induced current what is induced current equal to zero or not not equal to zero yes equal to zero or not equal to zero yes zero very good b a right B is conation of con dation of a Conant Z only EMF right simple this is zero clear clear very good very good very good okay now let's move on to the last case if the both coil and magnet are moving now here also tell me what is the induced DF Z or not equal to zero yes yes thumb rule right hand maxw right hand thumb rule you can use that that is the only way right that is the only way you can find out what is the direction of magnetic field two parameter current magnetic field equal to zero if both coil and magnet are moving with the same velocity same velocity CH mag is there any change in magnetic flux no Whenever there is no change in magnetic flux em zero EMF change Zero what happens to the induced current that is also zero simple there is no change in magnetic flux so magnetic flux constant if five is constant Pi is constant e is equal to D / DT that is equal to Z differentiation of a constant that is a zero very good okay now let's move on to the this part okay so this is the last case of induction what is the direction of induced current what is the direction of EMF so first one Circle me Loop it can be square it can be any shape but it should be a closed loop it can be any shape but it should be a closed loop is it clear okay CES should be yes now first entering the magnetic field magnetic field Direction inward inward direction that is B is B is along minus K cap okay along minus K cap and what is outside outside the magnetic field this this square box Ina that is the magnetic field it is out of the this box Magi out of the so Zer am I right out of the magnetic field indent is zero area area of the loop increasing increasing increasing okay so inside the loop inside the loop what is the direction of induced current what is the direction of induced current this is the direction of a current okay this is the direction of a current this is the anticlockwise right anticlockwise direction right yes or no this is the anticlockwise direction am I anticlockwise Direction but anticlockwise anticlockwise anticlockwise what is the direction of induced current according to lens no you should not change induced current induced current is anticlockwise direction am I right anticlockwise yes this is oh right right right sorry right right yes so first into the plane what is the direction of a magnetic field into the plane clockwise right clockwise at Center okay at the center clockwise at the center magnetic field is constant right magnetic magnetic field is constant area of the loop is constant area is constant inside the field okay inside the field B is constant then what is the flux flux is also constant you flux constant what about e zero and what about induced current that is also zero but clockwise clockwise okay that is the current produced okay but increase Conant support am right support yes right so clockwise IND Direction IND current direction anticlockwise right anticlockwise already due to this Magi current is in this direction so first one first one okay first one anticlockwise second one clockwise this is a in this case Okay this middle one it is a zero IND current is zero the last one IND this current is clockwise clear yes into the yes clockwise last case is a clockwise first case it is a anticlockwise clear yes very good very good okay now let's move on to the questioning part induced parameter induced parameter and flux if it is a function of a time if a flux is a function of a time how area decreases the loop entering while enter is it clear yes Paul after this session after 12 hours you'll be getting okay okay right fine now let's move on to the next class I mean next slide yes if F five is the function of a Time e is generally okay e is generally D5 divid by DT am I right EMF definition and for instantaneous EMF this is with a negative sign okay D5 ID DT D5 divid DT and for average EMF that is 5 2 minus 51 divided by T2 minus T1 this is for instantaneous okay instantaneous at a particular time at a what is the EMF what is the average EMF another got it okay yes Paul concentrate here pram please concentrate right yes now let's move on to the induced current and induced charge induced current and IND charge generally okay generally that is D divid DT I'm considering magnitude okay current cannot be negative right cannot be negative negative cannot be negative right IND generally IAL V from from OHS right froms law potential difference EMF EMF is nothing but potential difference only potential only getting a point okay now current conduction current DQ divided by DT because already we have seen no conduction current that is a DQ by DT and induced current that is I that is due to change in a magnetic flux right so so I that equal I IND current okay DT R next one DQ DT that is equal to e d/ DT into 1 by R right so DT DT get cancel DQ that is equal to D / R this is a very important relation okay equation okay equ Capal letter okay 2 minus 51 divid R do same do same okay clear okay yes VB gaming okay right okay right okay now questioning part your favorite 30 30 seconds first question first magnetic flux linked with a closed circuit of a resistance 10 ohm VAR is with a Time VAR is with a time okay that means flux is the function of a Time flux is that function of a Time instantaneous flux I mean instantaneous EMF as well as average EMF stud right the induced EMF in the circuit at A T is equal to 0.2 is induced EMF out induced EMF therefore induced at instant time right 0.2 second so e equal to D5 ID DT along with the negative sign EMF can be negative right at T is equal to T1 e is equal to minus D5 D5 value 5 value 5 t² - 4 t + 1 okay now further simplification further simplification e is equal to bracket 5 is a constant differentiation of a t t² that is a 2T minus 4 is a constant T differentiation of a t that is a one and plus differentiation of a one that is a constant therefore it is a zero it is a zero further simplification 5 into 2 that is a 10t minus 4 okay so at T is equal to at T is equal to 0.2 seconds right at 0.2 seconds eal minus 10 into 0.2 - 4 0.2 into 10 that is e equal to minus that is a 2 - 4us ofus 2 2 - 4us 2us minus ofus 2 that will be plus 2 andr IND e r but IND current but what if they asked so plus 2 divid R value 10 ohm so it will be 0.2 aamp that is a induced current okay so that is option induced EMF I the option D option D is the right answer very good 5% only 5% right so clear yes do you have any questions no right correct right okay now let's move on to the next question try yes try this question 45 seconds 45 seconds 45 seconds yes so question is a magn field of a flux density magnetic field flux magnetic field of a flux density one wer per M Square acts normally to 80 tons coil of 0.01 M square area if this coil is removed from the feed of field in a 0.2 second then the EMF induced so condition magnetic field of a flux density magnetic field of a flux density D B okay B that is a one W one wer per M Square okay per met Square per met Square okay and number of tons 80 80 tons and area that is 0.01 M square and the time they have given that is 0.2 second okay now e is equal to okay e equal to minus plus ignore that D5 ID DT and we know that 5 is B do a and five b value is to one right okay first so that D ID DT B do a right so area of the loop remains constant area change so e can be written as with respect to n number of terms n into a DB / DT okay so values values Subs okay what are the values e is equal to any value there 8 area there 0 01 into DB 1 divided by time change in time that is 0.2 so 80 into 0.1 10 0.01 0.1 divid 2 0.2 into 10 okay now e equal 80 into 0.1 2 e will be 4 Vol that is a plus 4 Vol okay minus and plus in the directly you can consider it as option D 7% again 7% only 7% okay 57 57% okay yes option D right next one yes correct so generally question they have given okay a square Loop of a side 2 cm enters a magnetic field with a constant speed of 2 cm/ second as shown the front edge enters the field at a tal Z 0 which of the following graph correctly depicts the induced EMF in the following graph correctly okay following the graph okay take a clockwise Direction positive clockwise Direction positive okay what is the direction of a magnetic field it is into the plane right into the plane as the Magi Direction inward current produces clockwise rightwise as enterree IND oppos Direction opposite IND am right so first anlock anlock clockwise posi anticlockwise negative netive eliminate negative wait okay negative Direction but starting from zero it is wrong it is zero starting from zero now positive there it is also wrong directly you can go ahead with option D without worry okay starting yes anticlockwise negative Niara correct very good very good okay so fine very good very good so option is cular loop okay oh you proved me wrong 100% correct now let's move on to the next question okay basic logic lo lo okay anlock because current already in the loop it is a clockwise Direction IND anticlockwise at the center of the loop I mean center of the magnetic field it is a constant zero but magnetic field in the outse clockwise anticlockwise so which is the right answer variable clockwise anticlockwise zero which one which one near a circular Loop of a conducting wire as shown in the figure an electron moves along a straight line the direction of induced current if any Loop if in sorry if any in theop Loop is so Direction ofr okay sorry direction of electron okay direction of electron so already into the plane that is Magi right so okay this is the loop so this is the loop okay already first anticlockwise EMF first anticlockwise EMF as it enters the Mido right as it is moving okay this is okay so this is clockwise Direction so this is this is anticlockwise Direction e side in the entri anticlockwise clockwise clockwise anticlockwise first anticlockwise next clockwise so which one is right option a variable we cannot predict first anlock as it is moving away moving away Magi clockwise clockwise directionally inded DF okay yes thank you am I right option A is the right answer yes 8.82 salute okay now let's move on to the next question direction of a current okay direction of a current direction of a current induced in a wire moving in a magnetic field is found using direction of a current induced 10th class throughout the year which one is right which one is right yes which one is right option A okay induced current current induced option b option b option b okay let me explain so first right hand clle hand rule right hand rule okay right hand rule to find out direction of a magnetic field am I right when current flowing through it fine next one flemix left hand rule to find out direction of a force acting on a current carrying conductor charge in a magnetic field so Force discuss Fleming's right hand rule used to find out induced current or induced EMF induced EMF EMF okay next one ampers Loop Direction okay direction of direction of you know DL okay along the okay along DB not DB that is a b magnetic field Amper circuit law now tell me which is the right answer which is the right answer yes which is the right answer answer right so that is as you expected that is a Flemings right hand rule Flemings right hand rule that is option C so option C is the right answer yes 15% 15% 68% that is for magnetic field that is for magnetic field okay okay right hand Palm rule Maxwell's Flemings right hand Palm rule K that is for easy purpose okay right now let's move on to the next question try to answer this question okay yes first one magnetic field changes at a rate of 04 Tesla and DB ID DT and in a square of a side 4 cm kept perpendicular to the magne field okayi I mean indent so side side length mag field if the resistance in the coil is 2 into 10^ minus 3 ohm then induced current induced current induced Cur D along with the you have to substitute simple trick okay next one that is Bid a sub here DB do a divided DT okay now e is equal to a is a constant take it out dbid DT clear now so first EMF Cate e that is equal e equal EMF what is the area L therefore can I write it as a l Square okay because yes both same now DB value is to DB value that is a DB divid DT now so so first e is equal to e is equal to l into L that is a 4 cm 4 into 10- 2 4 into 10^ minus 2 done next one DP divid by DT that is a 0.4 clear so further simplification 4 into 4 16 okay so 4 into 4 16 so EX okay so 4 into 4 into 4 10us 2 10us therefore 10us okay so e is equal okay so current current IND current that is Eid by R what is the value of e 64 into 10^ minus 5 and what is the value of resistance that is 2 into 10 power minus 3 okay you not 2 1's 2 3 are 2 2 that is IND current that equal 32 into yes yes is nothing but 100 only so further simp 0.32 that is I is equal to 0.32 amp clear so that is option which is the right answer 0.32 option b option b is the right answer very good 60% very good good comeback okay very good yes fine now let's move to the next concept motional so let's consider this is X distance from the rod this is a rod okay this is a rod and moving with a speed V and this is a magnetic field in a minus k K Direction okay and this is the length of the rod fine and Y will be change in a magne flux then what is the mag of EMF what is the formula for this this is a Theta is equal to0 why Theta is equal to0 for example this is the magne flux okay this is mag flux length I mean velocity and magnetic field velocity what is the values D bracket of B do a divided by DT a is a constant constant B is a constant okay uniform magnetic field B is a constant D ID DT into a area question sir why a is a is not Conant length is a fixed okay length is fixed I'm okay with that but what about this x this going to vary of the loop this is the loop okay en part this is a loop loop are because of motion of this rod in a magne field that is L so B is equal to D / DT a into sorry L into x l into X okay L into x l constant therefore take it out B DX DT first first concept Motion in a straight line velocity Distance divid by time velocity blv BL V so finally EMF is given by b l v where V is a velocity velocity and L is the length of the conductor length of the rod length of the rod okay okay now first of all due to magnetic force and electrons they move from one end to another end what is the high potential and what is the low potential right what is the high potential and what is the low potential okay right question next question okay lead no no no okay cation sinet value becomes 0 sorry so right okay yes now what is the high potential what is the low poti right different different fing this fingers along the direction of motion of a rod Char so this fingers the direction of a motion of a rod velocity of a rod and yes cross cross now direction of use this right hand okay direction of velocity along x-axis direction of motion of a rod and what is the direction of Mag field inward therefore poal so in this particular case high potential and this is a low potential is it clear for a Direction I mean you can use cross product cross product Screw rle Product by screw rule screw rule okay you can use a screw rule fine yes Paul is ncrt reading required sir for physics or coaching material the stud so it just depends on our mindset of the student okay fine very good fine write thumb rule yesan correct okay that is of B Omega I square that is for a rotational motion this is horizontal motion okay this is horizontal motion high potential this one and low potential this one here magnetic field Inward and the motion of a rod in this direction V cross V cross to find out high potential low potential so direction of a magnetic field this is direction of a magne field okay direction of a motion and direction of a magnetic field inward therefore curl it and open this thumb and this is the high potential so this becomes high potential and this becomes a low potential and high potential HP and low potential LP clear how to find out a potential of a rod very good R now next one what if this Rod has a resistance okay resistance resistance resistance resistance velocity and this is the length Okay magnetic field inward in Direction same motion of a conductor that is a velocity and direction of magnetic field Inward and open this thumb and this one becomes high potential and this one becomes low potential fine now IA this is high potential okay this is a high potential this is a plus and this is a minus and what about this one clear this is R so this is equalent to this so e BV it is a velocity of a rod in a magnetic field clear yes yes moan next one pal yes yes definitely BM definitely useful thank you thank you thank you so much let's move on to the next concept generally if you want I'll explain or else I'll move on to the next concept okay if you want this concept I'll explain or else I'll move on to the next concept yes money 30 minutes okay 30 minutes break okay yes no okay fine fine fine fine got it got it got it got it yes lead so first place srti second sit s creation third one pesh and Rama God gamer rames V niharika yes there 113 Bo great good okay okay now so I'll remove this question I mean this concept because as you don't want okay yes fine next concept next induced EMF induced EMF already calculate same diagam so e equal to blv am I right blv next one current flowing through the circuit that is Eid r blvid r clear next one retarding our opposing force exerted by action of induced current with respect to this Frame magnet field inward okay magnet field inward therefore C and open this finger so thumb finger indicate high potential and low potential so potential high potential low to low poal right current direction high potential to low potential high potential to low potential this is a low potential so current upward direction right high potential to low potential current already this conductor is in a magnetic field what is the direction of a force due to magnetic field what is the direction of a magnetic F force due to magnetic field which law you have to use same same this one what is what you have to do this now magnetic force Direction use this fingers indicates the direction of a magnetic field okay and this thumb indicates a direction of a current and this one indicates direction of a force acting on a conductor now direction of a magnetic field inward direction of a current open this thumb which side upward and this side direction of a force so this one direction of a induced Force okay direction of a IND I mean Direction due to direction of force that is due to magnetic field okay so this is actually force acting in a conductor p i l sin Theta where s this is the L okay this is a l length of the conductor small letter L or capital letter L let me consider small of the cond so this of length that means a current along the direction of current that that will be the length of the conductor and magnetic field Inward and magnetic field minus kis and length of the conductor plus J and what is the angle between L and this one B that is Theta is 90° right Theta is 90° so this one becomes b i l sin Theta sin 90 becomes one sin 90 becomes a one b i l okay b i l now here you need to understand here you need to understand this F is equal to okay this magnetic force equals induced force and magnetic force will be equal induced force and magnetic force will be equal okay so now b i l okay B I L current already how much it is b i value B LV divid R into l so b² L Square V ID R so finally induced Force okay the force due to current that is b² l s v / R so finally next one next last slide okay previous this magnetic force that is equal to that is due to induced current that is equal to b² l s and V divided R now we are exting a force on so F external what is the value of f external if if this is right rod okay this Rod move in a magnetic field okay Conant okay Conant velocity con external Force should be matching with this induced Force right that is a magne force so Fus I same equ F that is equal F thatal that is the magnetic force that is b² l s v divided R but you have Direction the external force that is a plus I okay but I want just magnitude just I want magnitude therefore b² l s v ID R so finally mechanical Power so rate of doing work rate of doing work rate of doing work inste that okay velocity motion of the what is the angle between this F and v motion of cond then what is the angle so angle between V and f that is zero okay now do remember from the basic this work done okay Power Force into velocity into cos Theta same way here I'm writing okay this formula we are not going to use this formula we are not going to use instead of that we are using this F external V cos Theta now cos Theta sorry cos Theta so cos0 cos0 that is 1 so make this to P is equal to F external into V so F external lare v r b² l sare v² ID r b² l sare v Square another e equal to b l v e sare b² l Square V square right so replace so power it is given by e s/ r this is a required Theory to solve a question okay ready now let's start with this question yes start yes yes question is a rectangular Loop sides 10 cm and 3 cm okay 10 cm okay so 10 cm length length and this will be 3 cm okay fine fine moving out of the region of a uniform magnetic field magnetic field B directed normal to the loop directed normal to the loop inward Loop director Inward and into the plane next one if you want to move loop with a constant velocity this is the velocity 1 1 cm 1us 2 because all the units are in then required mechanical force required mechanical forcea yes so I hope you remember the formula that is a b sare l sare v divided by V divided by divid by R yes r divided by R so external Force what is the external Force external force is nothing but mechanical force and the mechanical force why we are calculating external Force we are applying we are doing a work right so substitute the values substitute the values f is equal to what is the B value 0.5 whole squ into length length 10 cm yes 10 cm Loop side 10 cm which one 3 cm right 3 cm because it is because it is to the velocity okay this is a par to is zero because V is a parallel okay V is a parallel okay length is parall to the V now cim 10^ - 2 into 3 into 10^ - 2 into V velocity 10 power -2 divided by 1 ohm 1 mohm that is 10 power - 3 ohm so further simplification this is 0.5 1 1 1 by 1 by and here okay that is multiplied by- 2 - 2- 2 that is to to - 6us so finally that is a 9 by4 okay 9 by 4 into 10 powerus 3 okay 10 powerus 3 further simplify 4 into 9 Sorry 4 9 divid 4 so simplification 2.25 2.25 into 10 power minus 3 Newton yes which is the right answer which is the right answer so this is option A that is 2.25 into 10^ minus 3 Newton got it okay very good now let's move on to the next concept homework yes homk okay yes correct okay fine lead so five to five very good average time for each second she taking almost 32 second 32 and 33 very good Sid creation around 30 seconds paves 39 second very good God gamer ramama rames VES very good Niti n okay R and okay fine you and S creation right consider that is a homework next one yes a rod of a length 2 m slides with a speed 5 m per second so speed speed speed okay this is the speed and this is the length on a rectangular conducting frame as shown in a figure there exist a uniform magnetic field 0.04 Tesla perpend to the plane of the figure if the resistance of the rod 3 ohm the current through the rod start M yes start what they are asking current through the EMF as current through the rod so length of the rod is two length of the rod length of the rod and this is the speed okay and this is the magnetic field inward in Direction okay so according to this okay CR screw motion of a that means this indicates thumb thumb indicates high potential so this will be high potential and this will be low potential now our question is mainly they're asking about current current is given by e divid r right what about e that is a b LV right so e find bs2 0.04 into LS2 2 m and V to so 04 right 0.4 0.4 now next one next R sorry I I that is equal to 0 0.4 this is a three rightly and divide by 10 4id 3 10- 1 so 4 by 3 I equal 4 by 3 into 10^ minus1 amp option therefore further simplification 3 1's are 3 and 1 and your point and 3 3 are 9 and again one so 1 1.33 okay 1.33 into 10^ -1 amp so options 1.33 into 10us one next procedure convert that into therefore 1.33 into 10us divide by multiply and divide by 100 right so 100 multi by 1.3 133 into 10us 1 10 + 133 into so IAL 133 so right answer for this question 133 mamp so option b right option b very good 8.77 practice is main okay practice is a key for your success okay now next one self inductance okay self induct what about self-inductance it opposes a change in a current okay okay con okay hi B hi Kanan yes timings fine now this is a positive and this is negative so current current circular Loop okay soent okay flowing through a conductor so direction of a current this is the direction of a current use this right hand rule only right hand Max right hand rule only Direction ofr so direction of a current direction of magnetic field outward so magnetic field flux change in a magnetic field right change in sorry flux B do a area same but what about the from the last fourth chapter right BDL according to B sub as current increasing magneet field is also increasing in as current is increasing okay as current increases flux changes so 5 is equal to l i this is I okay so flux is equal to l into I but this is self inductance okay 5id I what is the unit unit is given by what is the unit of a flux that is a VOR per W per amp is the unit of a self inductance Mutual inductance getting Point okay now according to thew ealon eal minus D DT iute minus D / DT braet of l i self inductance constant proportional constant Take It Outside minus l d i ID DT so finally we have with respect to current okay with respect to current we have a EMF that is D5 ID DT that is with respect to flux change but current change EMF produ that is opposite Right Time Break sir last self inductance self induct got it okay yes fine now let's move on to the questioning part yes calculate calculate calculate bega bega yes answer answer this question okay question the current in a coil 122 this is important right life okay right right next one current in a coil changes fromr amp to zero that is a d divided DT okay fine time second 0.1 second and the EMF induced 100 Vol so e is equal to minus l d i divid d right and self inductance right netive therefore I'll consider just magnitude okay so e value is to 100 that is equal to l into change in current 4us 0 divid time 0.1 so cross multiply 0.1 into 100 that is equal to 4 into l so 10 into 4 L L is equal 10 10id 4 okay so 4 10id 4 and 2 so 2.5 Henry 2.5 Henry that is option C or or Henry Henry that is a h okay now so so 48% answer correct very good next question a current in a coil changes from 2 Amp to 5 ampere in 0.3 second the magnitude of EMF induced in the coil is 1 Vol the value of self-inductance take it as a homework same as a previous question okay same as the previous question but same question this is e value and this is di value this is the DT value okay so find M right next one indu MF 49 49 36 49 13 slides yes no complete the chapter okay chapter complete okay okay okay so okay yes okay now induced DF in the coil in the okay induced this is very important okay very important now whenever there is a change in current through the coil already it opposes change by developing EMF so current current keep that in mind okay if a current is increasing through the coil it opposes by developing EMF and if the current is decreasing it supports by developing EMF okay so first one according to the formula we know that e equal D DT and with respect to indu this inductor so first question if a current is increasing current increase but what is the condition they have given increasing for example same same one second sorry that is opposing [Music] so this okay of current INR is increasing is opposite therefore this is potential this is low potential this is high potential this side and this is low potential so here e is equal to BV oh sorry blv okay that is minus l d i ID DT yes current is [Music] increasing this L that is inductor okay repl Direction high potential to low potential current flow high potential to low potential right also called as in yes okay yes current is constant loop again there is a loop current is constant current change that's all will current is Conant okay so now IND current condition current is decreasing okay current is decreasing current is decreasing current Cent support support supp deges right very good okay this is is high potential this is a low potential okay okay so this is low potential and this is high potential current high potential to low potential right High po this ISR this is this is so negative positive negative positive support batteries are connecting one after one one after one very good previous question option current increase right current is increasing current inre right clear yes very good very good now support same polarity Okay negative connecting to positive negative connecting to positive negative connecting to positive OPP negative connected to negative positive connected to positive yes it's a simple okay now let's move on to the sorry okay no problem now next one okay so let's consider this is 5 m Henry sorry five Henry now this is potential means point a and this is point B Tye okay okay start yes yes so resistor resistor battery inductor so first of all condition current in the circuit 10 Amper and it is decreasing de first of all support support very good support soent so right so this is point this is resistance 10 ohm okay and this is 5 Vol and current flow the I and this is a low potential this is high potential and this is a 15 Vol and support support support this is a negative potential and this is a positive potential and this is a b and this is EMF e deop i right now Val find Val what is e value what is e value just to find out e value Val okay equal okay L is equal l l d ID DT l d IID by DT l s there 5 I there 10 s to 50 old okay 50 Vol 2 ampere per second okay 2 ampere per second I okay current in the circuit decreasing decreasing decreasing look decreasing decreasing decreasing at 2 ampere per second that is a 10 Vol right e place 10 10 okay support positive oppose negative simple okay now apply open loop apply open loop apply open loop apply open loop open loop apply VA let's consider this only path okay open loop VA same direction of current direction uh Direction across minus negative sign I 10 ohm okay I next high low potential to high potential plus 15 low potential to high potential plus 10 Vol that is equal to VB clear step clear step clear open loop now let's substitute the values VA that is five I 10 R 10 + 15 + 20 that is a 25 VB VB S 2 find VB so 5 into 100 + 25 that is equal to VB do 30 minus 100 that is equal to VB so VB is equal to 70 that is a minus 70 Vol -70 Vol so okay so this is the final answer okay so that is 70 Vol clear yesus 70 not plus okay- 70 that means this is low potential this is high potential okay this is a low potential clear yes LS to 5ry D ID DT rate of change of current 2 ampere per second 2 ampere per second two total 10 right two 2 into five that is a 10 pram I hope it is clear for you yes okay very good current 10 Amper yes d by DT 10 okay okay no problem okay no problem mention current in a Circ 10 yes okay now let's Okay Okay correct question first starting okay now a current in a coil a current in a coil of inductance inductance and L changes from this is a Di and this is a DT magnitude of average induced EMF magn of average induced EMF e is equal to minus plus therefore direct magnitude consider therefore l d i / DT and E is equal to LS to 0.2 30 second that that's all 30 second Matra okay and d i yesterday 5 - 2 divid D to the 0.5 so further simplification e is equal to 0.2 into 5 - 2 to 3 divid 0.5 multiply divide okay to eliminate that decimal points okay multiply divide by 10 so e is equal to 2 into 3 / 5 so e is equal so 5 6 so 0 so 1.2 1.2 that is option b option b very good 80% answer wonderful PES second and S creation God gamer good keep it up guys R Rocky Kik ak ak top a gamer May and anaa Kupa patil H danash mul manula niharika it's meiru rames very good UD okay top 20 okay now let's move on to the next question next question it is graphical based question okay the current flowing okay start yes the current flowing through the through an inductance coil of a self inductance 6 M 6 M Henry at a different time in time instance is as shown the MF induced between T is equal to 20 and 40 indf okay okay because no problem okay okay now S2 EMF so according to the condition EMF inded EMF that is minus plus therefore directly I can l d IID DT D DT along yis I along x-axis that is T So directly here you can see eal l l 6 M that is a 3 minus 3 into di change in y axis okay find D divided DT that is change in y- axis divided by change in x-axis okay now change in y axis four so final final answer is to minus two okay minus 2 and final one two okay okay three okay so 3 - 4 3 - 4 that is -1 okay d i divided by okay divid by not divided by so final minus initial first that is my 3 - 4 initial sorry final minus initial okay this is a final and this is initial number and divided by final minus initial that is a 40 minus 20 so that is minus1 / 20 so negative therefore what I'll do d by DT okay so what is the value here it is -1 / 20 so finally here you can substitute okay minus of minus yes 1 divid 20 okay now e is equal to so minus and minus plus therefore 6 ID 2 into 10us eal that is 10 -4 Vol okay 10 power -4 vol clear H that is option b option b very good very good very good option b okay very good very good very good okay now okay yes sorry so refresh okay yesid by Dila but instead of values they have given graph D divided by D along yis it is a current along xaxis that is a time in current because they have given right 40 so 20 and 40 so final minus initial so fin 40 and initial current 202 here that is a four so 3 - 4 divided by final time initial time 40 minus 20 that is a 20 so D minus 1 by 20 right so lead okay lead okay okay Mano I'll try I'll try okay I'll try my best I'll try my best okay now let's move on to the next question homework okay homework next one yes magnetic energy just formula not magnetic energy so work done is nothing but magnetic energy only okay this is energy also you can call it as energy energy M energy density so energy divided energy density that is B squid 2 Omega sorry omeg m b sare divid Eli capacitor electri simple okay yes F sorry L induct ient 2 B Sid 2 M Electric fi e s 2 b s 2 m clear yes okay potential energy right questioning questioning part let's start your answering questions the energy stored in an inductor of a self inductance L Henry carrying a current I direct answer 10 second 10 second only just 10 second which one l s i s l sare iare l sare i sare l is Conant con yes right okay so that is of L squ that is option b l i Square option b fine so no option b 85% very good next one series and parallel combination of inductors okay series series okay L1 and L2 so series effective inductors and series L inductance in series that is L1 plus L2 so this is L1 and this is L2 and inductance in parallel we can get it as a L1 L2 divid L1 plus L2 like resistors only okay like resistors only okay no next one okay next topic okay calculate self induced EMF of a coil is 25 Vol when the current in it it's changed at a uniform rate from 10 ampere to 25 aamp in 1 second the change in energy change in energy generally w e that is equal to l i square right change in energy that is of L final I Square minus initial I sare simple okay now equal sorry e equal to okay start yes answer this question e is equal to minus lus so I'll ignore it l d i / DT so e is equal to e 25 that is equal to l into D IID DT so 25 - 10 / 1 okay so 25 into 15 l so L is equal to 25 / 15 soal by 3 okay 5 by3 Henry 5 by3 Henry Val okay so half into 5 by3 into l i² h sorry l f sare LF Square 25 s minus 10 s simplification so 5 / 2 into 3 into 25 s that is a 6 25 minus 100 so further simplification 5 divid 6 2 into 3 6 into 525 okay so 525 less than 25 this one this ratio let's consider it as a 1 by one only one so 525 the value simplification trick no so 5id 6 less than one right so 525 525 724 it is greater than 525 wrong 625 this is greater than 525 wrong 540 it is than 525 it is also wrong so this is the right answer so directly simplif simplification that is a 4 37.5 J 437 4 37.5 J okay so that is option A that is option A 12% right that is the main intention of my live sessions okay okay but question okay now yes Mutual inductance direct induct so generally flux in case of a self inductance self-inductance same formula in place of l m Sub in place of L M sub this is a mutual inductance so mutual inductance Henry okay culations okay yes right now so coaxial line I mean coaxial solenoids so if formul mu n s v mu permeability of a free space n sare number of turns per un l number of T per unit length so n number of TS per unit length Okay so this is for L1 smaller one and this is for L2 bigger one okay so finally mut indu n v but two number ofs bigger bigger smaller Co N1 N2 next v s so mu n² V so mu N1 N2 V / L right okay now so last question of this chapter yes okay yes when when the number of a tons and length of a solenoid okay okay number of TS and length of the Sol both are doubled you know doubled try to answer this question almost 60 to 70% to% self inductance okay self inductance L okay L is equal to so generally so L is given by mu n v okay L is equal to Mu n² / okay l s are into yes because number of number of vol that iser so 1 l 1 L get cancel L is equal to M so let's consider this as equation one okay okay now m n n sare there number ofs double number of turns double and length also double and this is eight okay okay this is considered this is equation one not malindu okay now L is equal to Mu 2 S 4 into n² 8 ID 2 L 2 2 are 4 yes l l Das l l Das this is L okay now mu n² a / l e value that is l two times two times double that is self inductance becomes a double self inductance becomes a double yes 51% great 51% correct approach no no no no area will become of no no no no area become right okay yes okay now homor mut inductance right electromagnetic induction principle Emi electromagnetic induction right so finally uh according generally number NBA Theta cos Theta omeg okay angular velocity so is equal to okay so 5 is equal to NBA cos Omega T by using farad law D / DT okay so e becomes N B A Omega sin Omega T sin Omega T So NB a Omega e that is the S Omega T this is the final relation e e eal v v sin Omega so form this is very important formula in understanding AC alternating current yes okay alternating current start in half an hour okay so half an hour time to yes for e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e hello hello hello hello hello good afternoon everyone welcome back to the session KCT revision series I hope yes come on boys and girls hello hello hello yes was Muhammad hi sahil hello C gaming AB Movie Club mang okay yes UK same okay ready ready yes one again coming back so hello ready ready yes yes okay ready good good so one andet AC okay after that we'll we'll take a short break after that we can start with a dual nature of matter and radiation okay yes very tough chapter after this session lakshmikant easy easy okay yes ready lead board first ready yeah 100% ready good so again in the first place k r and ak bgm ak BM fourth position Al pades next one NAA AG gamer Mayu Kik and Rocky Top 10 good congratulations and okay yes dcod okay brand Bangalore always yes mother alternating current in short okay and from this chapter you can expect three questions okay from this chapter you can expect fine fine fine yes okay now so first V right mu V is equal to V sin Omega T right okay yes shal okay Mayu okay from kbur hello hello yes abijit okay yes so first difference what is the difference between DC and AC Al dcurrent F simple okay I V versus V means potential right from zero okay right from zero time also do not change got it okay simp and the sources of these are the example for sources of a DC okay sources of a DC sir what are the sources of sources of DC example battery okayy these are the sources of a DC direct current yes why do we use AC so DC sorry why only AC alter so first of all cell and Battery what is the potential difference maximum 12 maximum maximum 50 but electric device fridge ugly TV ugly laptop ugly okay AC ugly work require lot of voltage difference potential difference so differ that is 220 so 220 volage 220 Vol that is the main reason for a small small am so that is the main reason we are using ACR and reasons there but it is not the right time to discuss about it but it is a revision series let's concentrate on our topics okay now exactly exactly lot of resist resistance grade area prism correct okay not only now coming to AC AC and alter alter alternating current alternating magude ande keep on changes with respect to time time AC alternating Okay so now AC alting so so first one Gra yes so these are the form of alternating current okay these are the form of alternating current majority last two okay is it clear right sir increasing straight yes B pram is it clear okay now are the okay because short short of okay thank you thank you so much okay weing current was corre okay so free okay so if you have any concern with respect to content right now so let's concentrate on these two okay now next concept what is this already discussed definition go ahead and what are the condition to be current as a AC y condition now alternating not DC now first one amplitude must be constant ude must be Conant it is not AC amplitud should be constant same now next one alternating off cycle is positive and off cycle negative First CR and continue same positive negative positive negative positive negative positive and negative got it okay now next one next important point the AC continuously in a magnitude and periodically reverse the direction is what is the current zero at Point at some point the magnitude of a current is a maximum highest okay again becomes zero Z maximum value again zero in at this point okay at this current will be maximum amplitude will be maximum so this one we call it as a I not this is also I not maximum amplitude of a current we call it as I Peak values Peak values Peak value Peak value of current okay next one last point that is frequency biscuit questions okay biscuit questions so in India what is the frequency of alternating it is a 50z example USA okay United State of America 60s 10 10 60s get point and voltage in India in India that is a 220 voltage that this is a vrms okay vrms VMS root mean Square velocity okay root mean Square okay not velocity root mean square root mean Square voltage but question sir what is this root mean Square RMS current RMS value of a current VMS V RMS 220 vrms not Peak value amplitude that is the vrms value why is0 in USA okay okay fine now next one what is AC already discuss under that iag e is equal to e sin Omega T sin Omega T do you remember v v Omega t v i directly proportional directly proportional by ohms law okay Define the potential difference applied across the conductor is directly proportional to the magnitude of a conduct I mean magnitude of a current flowing through it so V equal V om IAL I sin Oma T these two formulas are very important okay especially that V is equal to V sin Omega T got it okay Peak current yes correct root2 multiply that will be Peak value amplitude of that particular parameter current current voltage voltage right now angular frequency in a radians per second next one I not Peak value or maximum value of AC clear clear now next one only only 10 second okay first shows the variation of a current with a time in a four devices P Q R and S the device in which alternating current flows is alternating current is graphical presentation just to now not right sir this formula relation to HS La yes yes yes okay thank you as thank you so much explan explanation first one p positive of cycle negative of cycle positive of cycle negative cycle positive cycle negative off cycle okay straight this is DC this is DC first one last one that is a zero okay okay I mean that should be equal okay which is the right one so but approximately okay let's consider approximately which one is the right option A okayi off cclee negative off cycle amplitude should be same should be same but I'll consider that is a P that is a ption correct okay now let's move on to the next concept that is the concept of average or mean current generally mathematics mean current mean average for example okay for example what is the average current what is the average current current defined as a rate of flow of charges with respect to time okay total charge in a total time right total charge in a total time total charge as a q divided by total time for example initial time T1 and final time t for example I'm starting from 0 second T1 T2 T2 is a final time minus initial time T1 am I right okay total charge that is equal to I average T2 minus T1 okay graphically present area under area under area under it area under get Point okay now same this is I versus T only but instead DC AC current AC current AC current period of time okay E Period of a time only let me consider this is a T1 and this is a T2 and this is the current so of current will be varied total sum of all the current yes that is also correct okay now here T1 in the t2 because alternating okay let me consider average for example so total charge total charge so total charge that is equal to integration current T1 to T2 okay T1 to T2 I into DT so total charge this okay yes negative part okay now with respect to a AC with respect to AC e formula with respect to DC e formula right now next one concept of average and mean current average and mean current correct it is defined as study current under DC let's consider as a DC which transport the same amount same amount of of charge in a given circuit in a same time as the alternating current and the same as alternating current AC does in the same circuit in a same equal interval of time okay that is a off of its period now it is also defined as mean of the instantaneous value of voltage or current taken over off cycle so try to understand first this is okay this is the I current this is a DC current okay resistance resistance so T1 okay t22 and resistance DC curent total charge that is equal to okay total charge is equal Toc bracket of T2 minus T1 why it is simple meanc equal to Total charge divid total charge divid T2 minus T1 justr okay now now next one next one same but do AC current okay AC current AC this is AC current okay I AC this is also resistance okay this is also resistance R okay clear now integration okay that is a t T12 T2 according to the definition according to the definition according to the definition definition equal total charge that means total charge by DC that is equal to Total charge by AC correct total charge of DC I DC T2 minus T1 that is equal to integration T1 to T2 i d correct so F simp T2 T1 right denominator so I DC that is equal to integration T1 to T2 I DT I ACR okay T2 minus T1 this is the this only we can write it as a I mean okay I mean I average I average that is equal to integration T1 to T2 I into DT divided T2 minus T1 okay is to Mega and uh okay try to understand the situation okay got ital I mean that is equals to that is equals to instead of same meaning got it yes very good now next one for any function these formulas are very very important okay okay for I average just before okay just before I average greater symb lbb okay T1 to the T2 limits ID T2 minus T1 next one mean square mean Square already mean right mean square mean sare already simp very good major major abishek yes okay yes question yes sir so comp okay comp so total 45,000 seats free seats okay 45 seats 45,000 seats ranking okay on the day of examin almost to on the 20 to 25% okay serious candidate 10 to 15% okay okay now never okay so right okay I'll explain it okay yes okay now are never now root mean Square RMS RMS first RMS first Square first last okay Square first Square first Square okay root okay square root mean Square okay root mean square root square okay Square first first two square I sare next one mean next one root next that will be the answer okay so T1 2 T2 limits I sare DT ID T2 minus T1 that's all okay abijit yes luck also matters luck sometimes matter okay but hard work always okay workation okay don't depend on luck depend on your hard work and dedication okay like for example okay for example study first to 10th next to 12th next to degree okay so that inspires you lot motivation just time but inspiration always works got it okay yes okay fine yes complete okay next so root mean square e formula fine question formula this is very important this is very important very important okay right next one so has what is your doubt tell me yes Vish sir directly proportional to KCT God okay okay yes yes Guru Prasad thank you thank [Music] youas uh yes suas what is your doubt you can ask me okay okay now let's get into get into the topic okay mean okay mean of sin Omega T integration mean of cos Omega t mean of sin 2 Omega T cos 2 Oma is equal to Z for full cycle what about half cycle C okay you have not used this trick in for a time being okay okay so okay what is your doubt okay now next one mean of sin Square Omega t mean of cos Square Oma T that is equal to this one going to help you a lot in AC chapter circuit questions are important Circ questions C questions okay CC question don't worry okay why sir off cycle not important okay Transformer yes okay okay B Movie Club n ranking next one I mean and I average for off cycle is B okay 2 I not divided by pi 0.637 into okay now type okay start question is find the I uh sorry find the RMS value of alternating current I is equal to I not sin Omega t for off cycle for off cycle I RMS irms and irms square first Square okay square mean square mean root this is the actual scene next under square root okay I I sin Omega T sin Omega T whole Square okay mean next one I Square next one sin Square Omega T sin Square Omega t i s yes sare under square root mean of sin sare Omega T Square Oma sin square of om that is half okay next one that is equal to I under square root 1 by 2 I not divided by root yes okay so finally irms that is equal I ID < tk2 V because I and V are directly proportional yes not not Ganesh not okay now V RMS VMS is equal to V divid root2 clear simple so 0.1 1 okay 0.63 0.63 0. 063 all right double huh okay 0.63 okay 077 yes yes yes correct correct correct [Music] okay okay right next one next one an electric bulb okay second stop stop stop stop integer type 45 seconds yes so an electric bulb is designed for a 20 V DC okay old DC current if this connected to AC source and gives same brightness find the peak voltage okay so according to the condition V RMS that is equal to V ID < tk2 am I right root2 Peak voltage Peak voltage V rearrange this equation as < tk2 into vrms VMS 20 so V is equal toare root 1.414 into 20 2828 Vol this is V not okay 2828 28 28 huh one student answer very good very good okay 2828 very good very good very good now let's move on to the next question yes direct direct second 10 second answer 10 second just 10 second 10 second 10 second yes so a voltage signal is described by V equal to V for 0 less than or equal to T less than or equal to T by2 is equal to 0 for T by2 less than or equal T less than oral T IGN soal V for off cycle it's a mean value off cycle it's a mean value direct answer that is option A V not divided by V not divided by < tk2 very good 86% very good very good very good okay now let's move on to the next question ah inductive reactants already first one the effective resistance are this is a effective resistance inductive react indac opposing in nature okay so resistance resistance opposition offered by the inductor to the flow of current is called inductive reactance and is donated by XL okay XL omal 2 T tal 2 T and andly related right so 2 right Omega that is equal 2 2 pi f as we know that frequency is inversely proportional to the time period got it okay 2 is a constant Pi is a 3.14 f is a frequency and this one L is a inductor inductance of a inductor yes yes yes wait some for time okay now next one graphical representation this is XL and this is frequency how frequency and XEL are related frequency and XEL are related frequency and Xcel are related Excel is directly proportional to the frequency frequency incre Okay so if you practice only pyq for the KD exam how many questions we can able to attend at least 25 to 30 okay 25 to 30 questions ining incling questions okay now next one directly propor so that is the reason it is a straight line that is the reason it is a straight line now next one capacitive reactant opposing nature of a capacitor to flow of alternating current is called called capacitive reactants and formula XC that is equal to 1 / Omega C but the last one Omega L but it is in a numerator but it is a denominator okay it is in a denominator and further you can write it as a 1 / 2 pi FC 2 pi FC and how this XC and F are related XC is inversely related to fre qux actually Z so z z so initial Point lower value High capacitance frequency this one Dees frequenc right know yes next one sir with 90% confident in second year syllabus and 75 in first year syllabus Now 50 plus score yes definitely confidence is the key okay confidence yes Paul delete eics culture right platform learning platform right yes sorry yes okay last you answering my questions sometimes uh yes Amad Khan namaskara namaskara okay marshmallow leave it care including location okay if you're doing good no problem if you're doing something bad you should be very careful technology improve okay right yes yes akbm correct okay right so as it is inversely proportional you can see curve so first One X Next One X X is given by Omega L that is also given by 2 pi okay XC it is inversely x c but 1 2 f c this is L okay clear meaning of the symbols right next one phas diagram both are scalar quantities right scalar quantities but Direction right direction but why but why why we need to mention that mention that into Vector quantities another question mark So e okay rting vect okay phas diagram help so resistance is same as reactance so same as resistance react indu IND react okay now yes now okay yes spark gaming welcome back okay now so just you know a definition just to represent just to represent current volage in a vector rotating vector okay so now so first firstam this is okay nothing buta okay now this rough diagrams okay proper sometimes diagam excuse okay now so let's consider this is voltage this is a voltage okay now next one yes current yes doed line perpendicular to the time AIS xaxis yes K good afternoon okay now hello hello everyone conate on the lecture okay now Peak value both for voltage as well as current that is the leading okay that will be you know Peak can you identify which one is reaching maximum Peak value that will be leading trick okay yes uh ISU sir sorry for disturbing gos law vector quantity now Vector it is related to vector quty magnetic field for example electric field electric field is a vector quantity yes it is related to vector quantity magnetic field magne field is also Vector quty yes it is related to Vector okay yes we lacks we leads I la IAD [Music] okay I so leads V leads I and voltage leads current I LS I lacks voltage I lacks voltage both are correct okay but always present with I because while expressing your answer but be careful but vector rotating for examp okay this is a current this is current Direction let me call it as a amplitude of current amplitude of current and for [Music] example AIS yes AC AC okay now tell me this one or this one positive or negative exactly positive positive ah huh positive H positive positive positive okay positive only okay so we leads we leads this one so V and okay so first V next one next one so this is I and this is V now tell me which one is leading and which one is like lagging which one is leading and which one is lagging which one is leading and which one is lagging uh pram Yu yaa device yaa device ah I lead I leading because Peak value okay compared to this okay so hence in this case okay I leads V by 5 by by phas difference phase difference okay next one V lacks I by 5 is that clear so let's consider still it is I only means amplitude of current this is I and tell me and tell me positive XX negative sorry posi yis negative y yes positive y negative y yes wasanta okay sorry yes yes negative negative negative negative after that means I is leading V lacks I so V will be in this direction okay V will be in this direction so let's call it as a V and V and resistance R okay R in the AC circuit C in the AC circuit l in the AC circuit LCR in the AC okay again one more time okay so mention this is this this curve indicate I mean this is sinal wave indicates a voltage time T2 T1 T2 T1 is less than T2 1 2 3 4 5 6 7 and 8 9 10 T1 T1 will be less than T2 so max right V is leading I is lagging I is lagging lagging v i lags V I have mentioned this is a current okay this is a current and this is a voltage right from zero Max Max I leads V but how many degrees but I don't know what is the particular phase difference difference in right 95% clear very good very good Rya Ana AK pades Kupa PA a Mayu Karthik and Rocky so top 10 top 10 now okay so okay very good okay now next one next concept register in series sorry register in AC Circ right volage according TOS law V is equal to I am I right this is AC Source okay V okay current direction change but consider this is a AC source and resistor is connected to a AC Source okay now what is v v sin Omega T that is equal to I but I because we are considering relationship between V and I we R sin Oma Sina now IAL T and what about v v is equal to V sin Omega t i v i v sign Omega Oma sorry omaa is there any phase difference is there any phase difference tell me yes yes is there any phase difference is there any phase difference is there any phase difference phe difference example okay yal to a sin Omega sin Omega T example okay Y is equal to a sin Omega plusus 5 therefore they are in Phase right so okay this is I and this is V maximum minim therefore they are in they are in Phase there here you need to remember these formulas okay there is no phas difference here I can write NO phase difference phase difference is zero that is no phase difference no phase difference no phase difference clear okay now if this is the current okay if this is a current if this is a current amplitude of current what about v i not represent what about V in the same direction in the same direction in in phas in Phase that is V clear there is no phase difference in Phase means zero difference yes zero phase difference okay yes clear it clear yes okay now V thank you thank you so much once more just V equal what about it is the potential that that is nothing butage V equal V om V sin om equal I okay pH difference same line got it yes next one suppose pure inductor suppose an inductor is connected to ac voltage with instantaneous value V is given by again indor without dering without what inductor inductor yes sadik Shake laka K will be starting from 8th April okay 8th April in the start just imagine long longm okay that two for around 4,200 I guess 4,200 right yes so inductor support oppose so I equal to I sin Omega T [Applause] sameus by2 actually okay minus 5us 5 if five value five value I is equal to I sin Omega T minus < by 2 and what about v v equal V sin Omega T which one is lagging and which one is leading which one is lagging and which one is leading which one is lagging and which one is leading yes begera yes C gaming correct answer very good very good okay yes V yes 4,200 I guess okay current lags very good sign negative IND lag lags okay so current I lacks we by how many degrees Pi by2 we leads leads I by < by 2 okay this is graphical presentation okay maximum height this is a maximum Peak height and this one here no V is leading because it is reaching first and UD maximum maximum this one V leading I clear yes very good very good and Graal presentation that is I that is I and what about v v v leads but how many degrees 90° 90° right this is V not so Pi by2 clear Pi by2 yes okay so the current elect I mean a it right next one so in phas I mean pH di current lags by pi by2 now in this case capacitor capacitor current lead I mean support yes yes yes okay yes supports so supports I I is equal to I not can I write directly sin Omega t + 5 but what about 5 value it is again 90° by 2 sin Omega t + 5 and v v is equal to V sin Omega sin Oma Oma yes which one is leading and which one is lagging I leads by 90° very good very good very positive right positive positive positive so this one indicates this one indicates I leads I leads V by Pi by2 okay I leads Pi by2 this one this one should be PI by2 right this one should be Pi by2 so this is pi by2 clear yes clear okay now so which one is reaching maximum height current and here voltage so current leading I sorry current leading V so next one it is a I not okay which is if it is a I not what about V not what about V not positive y negative or or negative or negative V with angle 90 Pi by2 yes very good jages and mov Club mov Club wrong posi positive okay because we leads we I leads okay so this is V not and this is pi by2 clear clear yes negative very good very good very good okay clear now let's move on so along xaxis X this is xaxis across capacitor and register okay we this should be this should be resist okay now so all these are current I not I not and I not the first one first one with respect to inductor inductor defitely this is aular that is is plus y AIS and on the same straight line okay same straight line same straight line okay so so this is V not and this is V and this is okay diagram so this says I not in all the cases I not with respect to registor okay with respect to registor first to White okay now so we V so first XIs and V not along y- axis clear the first one it is clear this ISS voltage across voltage across inductor voltage across inductor and this is current in all the cases same so therefore I'll be considering I not only all the cases I not only now V capacitor V that is a 90° that is is 90° this is 90 so VC VC okay voltage across capacitor and what about resistor on the same axis on the same axis on the same axis so this one is VR clear V LR so direct direct right so okay so first one form first Formula V Net v net under square root V r² plus v l minus VC whole squ clear next one in terms of I and Z impedence resistance formally I that is equal to Z total impedence total resistance of the circuital resistance of the CC imp okay next one so under square root v s i sare r r sare plus under bracket v l v l i s XL s minus I sare X c² clear okay one second under square so I into XL minus I into XC Square IAL zal I r² + XL minus XC Square EI i e get cancel so finally you will be getting you will be getting that Z is equal to UND root XL - XC s + r² okay whole Square under square root xl- XC square + R square are the tan 5K this one and this one okay and the power factor this is very important R minus r ID z z total impedence shini H aara sir suggest some DPP for KCT on the okay and power factor this one we call it as a power factor power factor okay where five is the phase difference between current and voltage okay now VMS cos power factor formula this is very important okay already form so I hope yes 30 second 30 secondy LCR Circ straight line meaning XIs straight line net okay like force resultant vectors resultant Vector okay yes dasan what is your doubt tell me I'll clarify your doubt and also akm yes yes yes is sh that's enough okay so 77% while soling the questions okay I'll explain it okay now 20 old AC AC okay let's consider 60 second because exp Vol AC is applied to a a circuit of a resistance and coil with a negligible resistance if the voltage across the resistance is a 12 volt and the voltage across coil is simple simple question this is a net this is a net voltage that is a 20 Vol and resistance across uh sorry voltage across resistance is to volt voltage across resistance that is the 12 Vol and what is the voltage across inductor okay okay across coil indu by the formula V net is equal to under square root V r² + v l minus v c square right so capacitor therefore you can eliminate the C we can eliminate that's C this one you can eliminate so V net there V net that is a 20 under square root V R square 12 squ plus v l is that we need to find out v l minus 0 whole Square sing both okaying both thatal 144 simp v² that is equal to 400 minus 144 v² that is equal 256 right so 200 and 100 300 56 and 44 44 that is the 400 yes and v l is equal to 16 Vol right so voltage across this inductor of the coil is to the 16 Vol right so that is option a option A is the right answer hey very good 72% wonderful wonderful okay yes yes Som extra work DPP questions Sol okay fine okay now next one the average power dissipated in a pure inductor pure inductor pure inductor what is the angle between I and V first of all power power decip is it is given by I RMS V RMS and cos 5 in a pure inductor L pure inductor what is the angle between I and V 90° very good 90° so five value in a pure inductor okay pure inductor 90° so cos 90° that is equal to 0 so power 0 into anything is zero so therefore right answer option C right answer option C right answer option C very good wonderful wonderful now next one not the average power disip n okay option C last one last one option C okay good 90% answer let's try to answer this question yes question is the average power dissipated in AC circuit is 2 wat if the current flowing through the AC circuit is 2 ampere and impedence is 1 ohm what is the power factor power factor power factor that is a cos 5 that is equal to R / Z okay power factor right so Factor okay cos 5 p is equal to irms VMS and cos Factor andms VMS irms and VMS okay the average power deip two okay two fine if a current flowing through the circuit is 2 ampere and impedence they have given 1 amp okay fine simple p is equal to V cos 5 right what is the power 2 I 2 2 I into R in place of R I can write it as a z because V is given by I into R but it is a AC circuit therefore I into Z right so I into Z cos 5 so you not two two get cancel one left hand side mik I is equal to 2 Amp Z is equal to 1 amp into cos 5 and 1 2 into 1 2 only cross multiply cos 5 that is equal to 1 by 2 that is equal to 0.5 0.5 that is option A that is option a option A is the right answer only 20% answer right so these classes will be there till maybe eight n okay next dual nature of radiation just photoelectric emission equations that chap is also done but numerical okay now next question 45 seconds okay 45 seconds let's try to answer this one a multimeter reads voltage at a certain AC source as 100 volt okay AC Source v v VMS VMS okay what is the peak value of voltage Peak value V already we know that relation between V RMS that is equal to V divid root2 and V not V required therefore root2 V RMS and V not that is equal to root 1.414 into V 100 V that is equal to 1 141.5 Vol clear what is the right answer 141 okay option C option C is the right answer yes 84 very good very good okay 140 1.4 voltage okay 14 1.4 voltage very good option C no answer okay next one try to answer this question so 10 second answer just a 10 second an ideal inductor is connected across AC Source voltage that current in the Circ ideal inductor that is L current leading current leading and lagging okay current because inductor this is L inductor L therefore current lacks therefore current is ahead of the voltage no lacks voltage in a phase by no Pi by2 is ahead no lags voltage by pi by2 very good very good that is option D option D is the right answer option D is the right answer only 12% answer only 12% inductor inductor Capac okay now next question an alternating voltage source of a variable angular frequency angular frequency variable angular frequency Omega and fixed amplitude B is connected in a series with capacitance C electric bulb of the resistance R inductance is zero when Omega is increased so current formula so first of all relation power of the Buri okay okay okay okay fine not okay Power brightness brightness brightness depends brightness depends on power okay brightness depends on power p i r okay but in place of R what we have to use i s z Now p is directly proportional to I sare okay p is directly proportional to I Square now I resistance relation V ID Z right Z in the sense XC that is equal to 1 Omega C increase Omega is increased Omega increase therefore the increase sorry formula actually X so okay also increases therefore brightness is also increases brightness also increases therefore which one is the right answer the bulb glows di the bulb glows brighter net impedence of the circuit remains unchanged total impedence of the circuit in increases so this is direct answer that is brighter option b option B net imp so this is this is wrong this is wrong next one total impedence of the circuit increases IND right so right answer is that is bulb glows brighter bulb glows brighter okay sir means in the concept repeat definitely you score very good ranking from YouTube classes okay that guarant I'll give next one in a circuit in a circuit the phase difference between the alternating current and Source voltage is Pi by2 Pi by2 difference phase difference phase difference phase difference Pi by2 following cannot be element ising by current is lagging by by2 but differ right pH difference therefore option C resistance option C option C phase difference zero phase difference Pi by2 phase difference Pi by 2 phase difference Pi by2 clear yes option C option C is the right answer yes 96% okay now next one the voltage across the resistor okay take it as homework next one in a circuit yes in a circuit V and R given by V and I are given by 150 this is V equal V sin Omega T right our there and I equal I equal I sin Omega t + 5 value value by 3 and v v Val 150 I not value 150 right 150 150 power disip yes so power disip that is equal to irms V RMS V RMS and cos 5 okay with respect to Peak values that is I ID < tk2 V ID < tk2 cos Pi by 3 next one I V divided by root root cancel 2 into cos 5 60 that is a 1x2 COS 60 1x2 so I not value is to 150 into 150 divid 4 okay so 106 it is more than it is wrong 150 more than 150 more than 15 wrong Z therefore option C so finally 5625 wat is to power dis that is option C option C very good 40% answer okay questions right yes very good sua very good and V Basics H yes very easy moderate easy super moderate Next Level easy so mixed opinion okay fine Kupa noted next one ah in the series LCR circuit shown in the impedence okay sorry take 120 seconds okay take 120 seconds yes LC circuit shown the impedence is imp imp is given by F F okay that is r² + XL minus x c Square XL value XL value that is Omega L omeg 2 pi and substitute the value that is equal to two Pi what about frequency 50 divid Pi into l l to there L1 okay so Pi Pi get cancel XL is equal to 100 2 into 50 100 100 ohm okay now it's time to calculate x c that is 1 divid Omega C again same 1 divid 2 pi f c now x c that is equal to 1id 2 pi into what about f f is a 50 divid pi conf sir where is that F this one okay now Pi Pi get cancel and here it is 2 pi 100 100 c c Subs 2 pi into f is a 50 ID Pi okay into c c that is a 20 micro micro 10 minus 6 okay now so Pi Pi get cancel and here okay huh XC that is equal to okay XC that is equal to [Music] okay 1id so 50 into 20 so 100 into 20 100 into 20 10 power - 6 10^ + 6 and here x thatal 2 XC that is equal to 0.5 into 10^ 3 XC is equal to 500 right equal r s r 300 square that is a 300 squ plus XL value XL that is a 100 100 minus 500 whole Square okay now Z is equal to 3 square that is a 9 and followed by four zeros plus difference of 400 400- 400- 400 squ soal root 25 and so zero 500 that is option D option D 41% answer okay yes very good H okay now let's move on to the next question okay the circuit is connected to an AC Source try to answer this question 30 seconds variable frequency as frequency of a source is increased the current first increases and then decreases which of the following combination of element is likely to comprise the circuit L C and R L and C L and r r and C which one which one first increase nextre frequency yes option b yes direct question option b because it is in okay very good option b option b very good 60% now next one homework next one resonance resonance just one condition resonance X must be equal to XC v l must be equal to VC Z is equal to R okay Z equal r that means imp I mean got it yes spark gaming okay okay akm oh nice akm e place in the very good good keep it up dri grupa P AG gamer uh karik VES it's meiru Ana next one pades R okay next clear okay series resonance okay 1 root LC clear yes okay V good good keep it up okay now next one for a series Circ to be resonance okay resonance I hope you remember okay now 2 Pi root by LC that is for a time period okay frequency okay okay frequency frequency okay yes G slides left okay now for a series LCR circuit first one already X L equal to XC next one Z equal to R next one phase difference between V and I is zero okay V and I is zero right Z is equal to R it is lying on x axis next resonant angular frequency 1 divid under root LC e formula okay next one uh this is a okay ignore next one average power consumption PV becomes a maximum and current becomes maximum okay powerr maximum next one yes question question question question yes at resonance the impedence of a series LCR series LCR Circ isance resonance imp right very good just conditions conditions right so that is option a very good option a option A for this question 71% correct now next one one clear yes clear next so on and ideal resistance R ideal inductance L ideal capacitance C and AC volmers V1 and V2 V1 V3 and V4 are connected across connected to a AC sources as shown in the figure at a resonance at a resonance okay resonance conditions XL is equal to XC VL is equal to VC and Z is equal to R which one R V1 L AC V2 C ACR V3 at a resonance at resonance condition okay at resonance condition X is equal to XC v l is equal to VC that means across L V2 across C V3 which one is right reading in V2 reading in V3 no V2 V3 that is option A is the right one okay option A is the right one yes option A is the right answer option A is the right answer okay now very good very good 57 no okay devil King gaming welcome back okay sh you know okay don't worry now next question in LC circuit at a resonance in a LCR circuit at a resonance the current is minimum maximum power current maximum the current and voltage are in Phase yes current and voltage are in Phase because five value no 0 de and current leads NO phase difference Z impedence is maximum no minimum Z is equal to R right so therefore option b option b option b right yes yes yes okay yes so right answer option b stop the poll yes 100% now next question Transformer Transformer is a device okay used to increase or decrease the voltage okay in acle okay Shalini achara best teacher deserves everything thank you so much okay op feeling devil King gaming yes I'm feeling the same okay right yes okay s you know the 10 minutes okay 10 minutes break fine so Transformer is a device which used to increase or decrease the potential difference the old AG yes it is crazy feeling danash yes all gaming yes yes yes thank you thank you so much okay now decrease and increase and decrease the old therefore two types one do Step Down Transformer Step Down Transformer Step Down Transformer Transformer Step up Transformer Step up Transformer Step Up Transformer okay so step down deing decreasing voltage used to decreasing voltage input volage is greater than output voltage down step step up step up and input voltage is less than output voltage simple step down and step up okay sir capacitor will support current but why capacitor capacitor Rea opposes the current right yes we could not did any question based this which one which one suala which one okay now step down and step up transform Transformer yes ah equation okay H transform yes yes okay now and what principle conf AC generator AC electromagnetic induction Transformer Mutual inductance conf okay now next one okay so Transformer transform yes okay okay now you got right so this is the Transformer now this one input this is input primary last topic okay primary coil and this one output output output secondary coil secondary coil okay now try to understand number of turns in a primary coil and these are number of turns in a secondary coil is it clear and voltage across the primary coil and this one voltage across a secondary coil clear yes now according toal D5 ID DT with respect to n number of turns negative sign indicate negative ignore okay now just okaying okay now so across primary number ofs across primary call and D5 by DT and vs that is a potential voltage across secondary coil that is number ofs secondary coil and D5 divid by DT okay now VP NP DT okay and DT D5 so okay now vs ID VP that is equal to NS ID NP clear turn ratio turns ratio turns rati okay yes any question very good yes after completing the second portions first okay 45 marks second in the soas yes what is your doubt gaming yes ideal Transformer n is equal to one yes for ideal Transformer okay ideal transformer for ideal transformer for IDE ideal Transformer NS ID NP that is equal to 1 efficiency is also one efficiency is also one 100% clear yes very good Transformer yes yes yeses capacitor capacitor is a device Electronic Component which stores the energy right okay now next for for an ideal for an ideal transformer for an ideal transformer for ideal Transformer let's say power at a primary coil power at a secondary Co power input that is equal to power output okay so power power I into V so input current okay current at primary coil into voltage at primary coil in the same way current at a secondary coil and voltage at secondary coil right IP is divided IP is I is IP I VP divided by vs okay VP divided by vs with respect to turns ratio turns number ofs in the primary coil and number ofs in a secondary Co that is equal to IP that is equal to NP divided NS clear yes clear yes VP VP v s VP okay yes okay okay for an ideal Transformer power at a primary coil power at a secondary coil power power input that is equal to power output okay very good very good yes okay now let's get into the questions okay differ right sir LC Del okay LC LCRA okay now let's move on to the questioning part start your answers start your answers the first One Step Down Transformer has 50 tons okay Step Down Transformer has 50 50 tons in the secondary call secondary call number of tons in a secondary call and th000 tons in a primary call okay fine and if a Transformer is connected to a 220 Vol that means it is a input input and primary call and current that is IP primary current at a primary call AC Source what is the output current Andre this is current at a secondary qu just now we have a seen relation right is divided by IP that is equal to VP divided by vs okay that is equal to npid NS C I'll keep it as it is okay next one IP 1 and VP okay next one NPS the number of TS in our secondary primary call that is the th000 divided by this is a 50 okay cross multiply cross multiplication that is that is equal to 1 Z 1 0 get cancel and 100 divid 5 so 20 so 20 amp option b very good very good very good so that is option b very good 100% answer okay now okay fine easy okay now question this is a 10c not more than 10 second okay not more than 10 second the Transformer Transformer works on the principle y principle okay y principle colle princip principle Transformer works on the principle self induction electrical inertia Mutual induction magnetic effect of electrical current easy the option that is option C option C is the right answer option C is the right answer that is option C is the right answer stop yes that is 96% answer now let's move on to the next question okay so dual nature of radiation and matter yes timing break break yes lead board akbm first and Kik it's meu Kupa anaa pades jagadesh and pram hey pram ranked up H 30 minutes okay yes 30 minutes fine 30 minutes start okay okay yes M tea coffee snacks okay okay see you in 30 minutes thank you e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e hello hello Namaste Namaste good evening tea coffe yes s water [Music] sorry hello hello hello begaa start Mor thank you yes let's start yes ready yes so ready yes thank you thank you yes ready one the thumbs up spr yes ah Namaste Namaste yes yes come on come on thank you marshmallow thank you thank you so much yes okay okay so and matter you can expect three questions okay you can expect three questions without fa righto Elric first one photo electrically like threshold frequency cut off frequency cut off first one threshold frequency I hope thresh frequenc threshold frequency for example metal surface okay metal surface certain frequency okay okay new okay new incident frequency new incident frequency new condition new is less than new new is greater than new I mean equal to new and new is greater than new Pho electric emission the third one that is a photo electric emission photo electric emiss what about the first two Pho electric emiss soqu frequenc threshold frequenc yes minimum frequency okay this is the minimum frequency minimum minimum frequency okay this is the minimum frequency required okay this is the minimum frequency that required to emit electrons from the metal surface right yes already explain last two chapters this is third one okay next to fourth one atoms and nuclear last one semiconductor Electronics okay semiconductor Electronics maximum on 1 hour max to Max on 7 Minutes 1 hour 10 minutes okay lowest frequency below which photo Electric does not occur okay threshold frequency that is the minimum frequency below which the photoelectric emission do not occur next one work function minimum frequency minimum energy minimum energy minimum energy that is required okay metalface elim okay so the minimum energy required by the electrons to emit from the metal surface electrons electr okay El so say total energy work function work function El from the surface from the surface so that minimum energy we should the same everyone okay now next one minimum energy required by the electron not Metal by the electron to emit from the metal surface is called work function function in the measure that is H new not where new not is a threshold frequency clear thresold frequency next one cut off wavelength minim maximum maximum wavelength maximum wavelength maximum wavelength up to which okay up to which photo electric emission takes place up to which photo electric emission occurs okay what is the difference between the first statement and last statement minimum frequency for example 100 100 is the threshold frequency elect therefore no photo electric emission 100 is the limit metal surface before that okay 09 but what about maximum wavelength is the cut off with respect to okay right okay so maximum minimum completed okay so now frequency next one work function next one cut off W clear yes sha Chand okay let okay AK okay okay top ranking step okay right next one observation Made In a experiment first one intensity versus photocurrent intensity versus intensity intensity is directly proportional to the number of electrons emitted number of emitted electrons emitted electrons electron from the metal surface for example int say 10 number of electrons emitting from the metal surface also increases directly proportional okay that is a directly proportional number of emed electrons the electrons emitting from the surface okay em from the metal surface okay this is directly proportional to the current I means current Okay small letter I indicates current in a dual nature of radiation and Mally small letter I indicates current capital letter I indicates intensity keep that in mind okay now this one directly proportional straight l this is a straight line so this is intensity and this sorry this is a photocurrent okay this is a photo current this is a photo current and this is intensity intensity clear next one photo current versus frequency now just discuss new is greater than new condition photo electric emission right photo emission so that means initially let me consider this is a photocurrent and this is a frequency okay this is a frequency El greater okay greater certain because this frequency does not depend current Pho okayi cut off maximum length up to which photo current photo emission takes place and say cut off cut for example say 50 say 100 say 15050 why okay clear yes uh intensity can be change by moving sources yes definitely there are multiple ways not only moving okay not only moving you can see that opening of a quartz glass right I hope you remember what you have learned okay what meant by photo uh photo electric effect photo right three types of you know current emission first one field emission that is electric field next one will be collected all these charges of char consider we can Define it as a electric current so due to what photo that is a light so current due to light now photo electric clear yes alone gaming off yes intensity can yes saturation current yes what mean by photo already explained yes just explain the current due to light okay the current due to light is called photocurrent right so exp photocurrent okay so metal metal surface suable frequenc frequenc okayo El the flow of okay the flow of photo electrons due to flow flow of a photo electrons we value is 10 power minus 9 okay this value is 10 power - 9 seconds instantaneous process okay work function means energy required to photo emission yes PRM correct okay what is the difference between electron Vol and work function electron it is the energy associated with electron energy associated with electron the energy gained by the electron when it is applied means potential difference one Vol across the ends of a cond very good yes 10 power minus 8 always sir no it will vary 10us - 7 in Theus 9 instantaneous instant particular time immed unit of energy when electron is moving under potential difference yes pram that is also right okay now so photocurrent versus voltage so photocurrent let's say this is a saturation current okay this is saturation current let's say this is saturation current okay so butal okay electri field colle to em collector to em collect to em electri field electr force opposite charge always the force experienced by negative charge it is opposite to the electric field but positive charge along the direction of electric field right first chapter there is no current okay netive po finally it is here in potential vs stopping potential so this says a potential at a collector potential at collector clear yes clear yes fine now next one Einstein explanation of a photoelectric effect radiation observed by the metal yes not not yes okay okay very good very good thank you thank you so Point okay very important points okay note note the first one photocurrent photocurrent is directly proportional to the number of electrons right number of electrons emitted electrons right number of emitted electrons number of emitted electrons number of emitted electrons so number of electr but number it is depends on intensity intensity okay intensity next important Point kinetic energy of electrons kinetic energy of electrons is directly proportional to the frequency directly proportional to the frequency and this frequency is depends on stopping potential stopping potential it is a negative potential applied across a collector okay fine question okay yes okay now different let me consider fine so first one photo current this is also for photo current and voltage now different intensity different inity let's consider this is i1 okay and this one I2 this is I2 compare i1 is greater than I2 therefore i1 is greater than I2 and more the intensity more the current more the intensity more the current okay now this common stopping potential common stopping potential okay common stopping saturation current first of all you have to know what is saturation current saturation current that is the maximum current okay after that there is no increase in a current that is the maximum current okay now next one stopping potential and it is a negative potential applied across a collector negative potential applied across a collector now in the same way in the same way y different frequencies okay different frequencies different frequencies there let me consider this is intensity single intensity okay single intensity now same intensity constant frequency frequency this is stopping potential one this is stopping potential two that means frequency let's say new one okay new one is less than new2 that means stopping potential 1 is less than stopping potential of a two so stopping potential stopping potential is directly proportional to the frequency and it is not affecting intensity okay this is intensity this is current is it clear okay clear yes first channel LC mega marathon okay upd wait for the update okay mean stopping potential is independent of intensity and depends on frequency very good exactly right okay now next one Einstein explanation of a photoelectric effect okay so this is the surface let's consider this is incident energy okay let's consider this is incident energy so I'll consider it as a e total energy and here electrons electr okay electr that means kinetic energy with maximum kinetic energy you know with a maximum energy and that energy we call it as a work function okay work function work function the minimum energy required for the electron to emit from the metal surface okay emit function ener required by the electrons to emit from the metal surface we call as a work function maxtic energy okay work function therefore total energy could right total energy inci on metal surface energy workun yes work yes El with maximum that is a maximum kinetic energy this is a maximum kinetic energy right this is the actual General actual as well as general relation of this entire energy but nma exam different I mean in terms of a frequency in terms of a wavelength okay in terms of a frequency and in terms of a wavelength okay yes huh concentrate money concentrate money yes in terms of a frequency H new yes H new that is equal to H new work function plustic energy kinetic energy maximum okay and know wave for wav represent HC ID Lambda HC divid Lambda plus maximum energy okay H new that is equal to H new plus maximum kinetic energy so kinetic energy now okay kinetic energy now in terms of a kinetic energy H new minus H new h Comm kinetic energy maximum kinetic energy newus new okay newus new relation so ktic energy relation with respect to stopping potential stopping potential it is given by E into v s where is the charge on electron charge on electron that is given by E vs clear right and these are the formulas with respect to Einstein equations okay yes question right next one questions as you asked questions here questions ready now yes ready thumbs up negative potential yes but mag okay stopping potential Einstein equations right Yesa ready Nowa ready yes ready sir okay now yes okay fourth and fifth chapters fourth and fifth chapter Sam okay yes okay so ready ready yes first 45 seconds question is the figure shows the variation of a photocurrent with anod potential for a photosensitive surface for the three different radiations let a i a ib IC be the intensities of f a FB and FC be the frequencies for the curves a b c respectively so relation okay Rel one first one okay do a okay and this is B and this is C okay let's consider this is a meeting at a single point okay and okay okay I I okay IB that is equal to IC next one this is I do I a clear [Music] possibility F A is equal to FB I is I a is equal to IB I a is equal I not possible you not possible this is also not possible this is also not possible next one FB is equal to FC FB is equal to FB FC and FB and F A so that means this is also not possible therefore right answer option a option A 177% just 17% just 17% leader board H okay leaderboard leader board AKB oh very good STI dri KK it's meiru PES K Ana PR ranked up okay okay nice very good okay right refresh next one next question is I mean next question and here they asked about know okay so particle nature of a light partic nature of a light partical nature particle nature particle nature of a light particle nature of a light that is a photon okay first point what is the energy energy of a photon H new HC divided by Lambda okay HC divided by Lambda worry now next phon travels with the speed of light next one rest mass of the photon is zero and mass of a moving photon is m is equal to H H new divid C Square H divid Lambda C and this one the momentum of a photon momentum of a that is with spe in place of a moment MV MC and M as it is and C we know that it is a h h new divided by C Square okay H c² H Lambda H Lambda e energy divided C that is the speed of light okay right now next one in a photon particle collision okay last one is very important the photons are not deflected by electric and magnetic field they are electrically neutral they are electrically neutral fine okay yes uh okay fine one sppy okay now so okay these formulas are very important for your okay clear okay right next one radiation force and pressure radiation force and pressure number of phons em number of phons okay now this is completely reflected this is this is coefficient of reflection that is coefficient of reflection is one okay so this is incident okay this is incident photons and this is reflected photons reflected photons reflected photons absorb means coefficient of absorption is one and2 incident photons okay incident photons as to reflecta it is completely zero reflection is zero reflection of a photons zero reflection of photons zero yes C okay this is for 2 e divid c this one e by C because only absorption right H okay now next one normal incidences cases either perfectly block body and perfectly reflecting body so all these things okay number of photons are emitted number of protons incident number of protons are emitted from the metal surface right okay clear right now number of photons em P Lambda divid HC this one for formula it is a Master Formula this one formula is a Master Formula based on this normal incident Mill okay this one formula number of photons P power not a momentum it is the power and already Lambda it is a wavelength H is a planks constant C is the speed of light got it okay now next one yeah questioning part ready questioning part questioning part yes start party yes start H okay H okay okay uhuh option D okay no frequency right frequency and here stopping potential stopping poal that is vs okay sir side okay not yes b d c a huh okay now question in an experiment to study photoelectric effect the observations sorry The observed variations of a stopping potential with a frequency of incident radiation is shown in the figure the slope and Y intercept okay yis stopping potential along xais frequency therefore Einstein equation E is equal to 5 + kinetic energy and kinetic energy that e into vs right e into es nowc convert this one to frequency H new is equal to H new Plus e vs okay e vs because so V because they have right they haveed it so e e v not that is equal to H new minus H new not so V not that is equal to H divided e new minus H new divided e okay yal MX C comp in place of Y that is v in place of M that is a slope right slope that is equal to H ID e and intercept intercept minus H new / e clear which is the right answer which is the right answer yes so right answer option a option A right so slope is to h divid e and negative sign clear clear yes option A now next question a 60 VT a source emit monochromatic yes ah a 60 V Source emits a monochromatic light of a wavelength 6625 nanometer the number of photons emitted per second number of photons emitted per second okay so number of questions I mean number of photons that is n is equal to P not Lambda divided by HC and substitute the values P not Power 60 V into lamb 6625 nanom that is 10^ minus 9 HC value that is 6.625 into 10 power - 34 JC into C value that is the speed of light that is 3 into 10^ 8 m/s okay now 3 1's 3 2 are that means 20 so 20 into 662 6.625 that means 10^ 2 into 10us 9 10 2 100 under multip still I'm getting 6625 only divided by 6.625 into 10 power - 34 8 26 right- 26 okay minus 26 am right 26- 8 20 34 yeah now here you can see 6.625 6.625 M this to 20 into 10^ 2 10 power- 10- 7 26 so finally you are having it is a 10 20 into 10 power 19 okay 10 power + 19 plus + 19 therefore scientific scientifically scientific method mention 2 into 10 power 19 + 1 20 okay this is the number that is option b option b option b is the right answer that is 57.14% answer clear yes clear okay now clear yes uh leaderboard uhuh leader board fourth place right next H the energy gap of an LED is 2.4 e electron old when the LED is switched on the momentum of the emitted photons yeah yes so energy that is equal to 2.4 into electron okay into 1.6 into 10^ minus 19 J okay and and here we have a relation with with respect to energy that is Eid c e divid c previous SL explain okay different forms now p is equal to what is the E value that is a 2.4 into 1.6 into 10^ - 19 divided by 3 into 10 Power 8 3 into 10^ 8 3 into 10 Power 8 0.8 okay and 16 into 16 into 0. 16 into 8 okay okay 128 okay 128 decim so around 1.2 right 1. so okay which is the right answer so - 19 and- 8- 27 right- 27- 11 so this is wrong so right answer this one okay momentum that is equal to 1.28 into 10 powerus 27 okay 27 kg m per second clear option a 15 15ra answer clear okay now try to solve this question H try to solve this question so clue Lambda that is equal to H ID P okay and they have given in terms of a kinetic energy right the de brly wavelength of a particle of a kinetic energy a is a Lambda the wavelength of a particle if it's kinetic energy K by4 sotic energy ktic energy can be written as 2 MK under square root okay and this is this is equation one okay now second equation Lambda dashal K by therefore Lambda Das that is equal to H ID under root 2 m k Das right substitute the values hun 2 m k by 4 K by 4 Remains the Same rights okay now M K that is 1 by2 okay 1 that is 2 under root 2 MK okay so what is the value of this already that is Lambda so finally Lambda Dash that is equal to 2 * of Lambda that is option a option A is the right answer very good very good very good okay yes clear right so leader CL is Krishna okay yes okay now try to answer this question try to answer this question 10 second is enough 10 second not on okay not in a photoelectric experiment if both intensity and frequency of the incident light are doubled then the saturation photo electric current Pho electric photoelectric current is it depends on intensity right it is depends on intensity directly dependent frequenc dep therefore which is the right answer which is the right answer really intensity and frequency of the incident light are doubled therefore current also will be doubled that is option b that is option b 95% no simple okay now answer this question homework okay consider this as a homework next one yeah try to answer okay yes which is the right answer explain explain graph already explain which one which one which one uh Kik kbur okay sir okay yes option option D and option next which one is right conf current intensity are directly related current and directly Rel this is the right answer okay option D ktic energy and frequency that is new is less than new not new is greater than new not clear okay so this is simple simple okay now next one come on come on previous one option D yes option D 42% answer good yes again 45 seconds the following graph represents the variation of photocurrent with anode potential for a metal surface okay i1 I2 I3 i1 I2 I3 represent intensities and GMA 1 GMA 2 GMA 3 represents frequency of a curves 1 2 3 respectively first question based on this chapter seconds answer okay so let's try to understand the question okay here I2 that is equal to I3 clear this is i1 frequencies now as we know that frequency frequency is directly proportional to the kinetic energy right so frequency and this one directly proportional to the stopping potential and according to that according that frequenc this is new three okay GMA 3 gamma 3 H gamma 3 and this one gamma 2 Gamma 1 H yes answer yes answer answer yes correct GMA 1 is equal to gamma 2 yes correct i1 not equal to I2 yes i1 not equal to I2 Gamma 1 is equal to GMA 2 correct but i1 is equal to I2 wrong i1 equal I2 wrong clear next one Gamma 1 is equal to gamma 3 i1 not equal to I3 this is right but this is wrong okay this is right but this is wrong next one gamma 2 equal gamma 3 wrong i1 equal i3 wrong so therefore right answer the right answer for this and this one okay option a option A is the right answer okay now next one an electron is moving with an initial velocity V is equal to V not I cap I Axis i j k components I indicates xaxis J indicates y axis and K indicates Z axis okay minim 50 question okay one chapter in the 150 questions 5 okay 5 question Sol okay now ah X AIS fine very good and B is equal to B not j j y AIS Andre electron moving in this direction and magnetic field and entering into the magnetic field so this is V is equal to v i cap and this is B is equal to B J cap right and electron charged particle projected perpendicular to the magnetic field yes any answer last part discuss when charge is projected parall to the magnetic field when charge is projected anti parall to the magnetic field when charge is projected perpendicular to the magnetic field and charge is projected at an angle to the pro magnetic field y yes any answers from your end straight line no par and Par what happens when you throw perpendicular when a charge is projected okay Circle yes circular motion circular and helical at an angle greater than 0 de and less than 90° any angle very good so C yes and okay circular path with a radius with a radius with a radius with a radius R okay with a radius r r symbol I mean formula MV ID QB MV QB and M QB and okay P equal q v QB R okay and wav length H ID P so H equal q b r yes concentrate H is a constant right H is a constant Q is also constant why charge on electron 1.6 into 10us 19 that is also constant and B it is a uniform magnetic field uniform magnetic field and R it is the radius I mean radius of a circular path this R is also constant hence D Broly wavelength Remains the Same that is option A that is option a very good 79 option A okay now next one homework try next one the maximum kinetic energy of a emitted electron emitted photo electron okay energy intensity intensity affects a photocurrent kinetic energy affects a stopping potential as well as frequency right more than 50 no distraction more than 50 okay more than 50 clear option b now option b yes option b is the right answer frequency frequency depends on frequency because kinetic energy is depends on a frequency frequency electron okay right so frequencies also affects stopping potential stopping potential of electrons okay now that is option b IU next next one proton and alpha particle are accelerated through same potential difference accelerated through same potential difference the ratio of their de brogly ratio of their de brl so wavelength it is given by h ided p and let's convert that into kinetic energy 2 m k okay and let's convert this kinetic energy into potential difference okay potential difference that is work done per unit charge and here according to work energy thetic energy divid Q energy sotic energy equals to Q into Delta V right Val H ID under root 2 m Q Delta V now Conant potential differ so is constant two is constant potential difference is also constant and it is depends on mass as well as a charge therefore I can write 1 / under root M and Q okay so Lambda that is a wavelength is dependent on 1 / root mq po start yes po start again yes okay now okay proton okay proton and alpha particle proton and alpha particle charge and mass because Al charge and mass proton charge what is the mass of a proton that is M okay Al particle x24 this is a 2 e and this is a 4M okay now ratio proton and alpha particle right proton and alpha particle proton and alpha particle Lambda PID Lambda Alp so reciproc okay under square root okay under square root mq okay mq Alpha and Q Alpha divided by mass of the proton and charge of the proton okay now substitute the values what is the mass m what is the charge e divided by and what is the mass that is a 4 M what is the charge that is the 2 e 2 e now you no m m e e get cancel so that means 1are root 4 into 2 right 1 1 2 < tk2 1id 2k2 ratio of same potential difference proton to alpha particle proton okay okay so this is for proton divided by alpha particle right so okay so this is alpha particle divided by proton okay now alpha particle divided by proton Buton alpha particle proton divided by alpha particle that is 2 < tk2 22 that is option b option b is the right answer option b is the right answer very good very good very good marshmallow once more okay see this is the formula okay this is constant this is constant this is Con I mean this is constant but variable quanti of the mass as well as a charge because there are two particles proton as well as alpha particle both are charge particle only so charge ofon and mass of the proton Alp particle charge Al partic Mass okay mass of the proton and charge of the proton charge of the AL of the particle model OKAY alone gaming of night poll is not working for you okay okay don't worry right fine lead board lead rank one in the second ranking and three four impr first fourth place and Mula it's me Vu Kik marshmallow and Kupa patil Jagdish patil yeah top 10 right next one yes okay homework that is okay yes 100% answer already Yes ah a particle is dropped okay let's consider 45 seconds for this question yeah a particle is dropped from a height H the de brogly wavelength of a particle depends on the height as again Lambda that is equal H sorry H by P right H divided P so H is equal to H sorry Lambda is equal to H ID MV right particle is dropped for example okay certain initial velocity Z right and what is the velocity at a time okay now for a free particle v² that is equal to u s + 2 a s General equation equation of a motion three okay v² that is equal to this is zero because initial velocity is zero because it is a free fall plus 2 as it is and in place of acceleration acceleration due to gravity B next one in place of a displacement here you can write height H height H okay they have given H now v² V therefore 2 GH okay now substitute Mone h divid mare root 2 G value same so constant constant constant constant but what about this proportional to h / 1 by H power 1 by 1 by H power 1 by2 H power 1 by2 that is option A is the right answer option A is the right answer 177% simple simple question okay H right nextal presentation three questions three questions the variation the variation of a photocurrent with a collector potential for a different frequencies of incident radiation V1 vs2 V3 is a shown is as shown in the graph then saturation current okay and here they have given different frequencies as a V1 V2 V3 okay and retarding potential stopping potential increase in order okay okay which one is right option b V3 is greater than V2 is greater than V1 okay V3 greater V1 is less than V2 is less than V3 V1 greater than wrong V1 is less than V2 less than V3 that is option C that is option C okay option C yes uh how do you approach frequen right so of course it is a negative term okay but for examp 15 is greater than 10 10 is greater than 5 but in of that 5 is greater than 10 10 is greater than 15 okay clear 5 is less than 10 10 is less than 15 simple okay Tes I hope it is clear for you okay now next one next one next one yes D Broly wavelength trick okay trick last one have the answer option C okay now start your question okay the de brogly wavelength of an electron accelerated to potential difference of 400 Vol is approximately 10 second tricks trick okay yes option b option b let's see good good okay now trick trick is Trick trick is Lambda that is equal to 12 29 27 Ang divid root V means potential difference Delta V or in place of a kinetic energy 12.29 divid under square root ktic energy okay that is this is also angstrom clear now potential difference use the same way that formula so 12.29 divid under square root 400 that is equal to 12.29 divided this is okay now 12.29 divided 2 into 10 - 1 into 10us 10 Ang okay angstrom is nothing but 10 powerus 10 okay now next [Music] one 6.4 that is 10 power nanom so 10 power - 11 10 power - 11 so Lambda is equal to 6.14 into 10 11 okay 6.14 into 10us 11 into okay okay yesp and div so 6.14 into 10-9 this is 100 two decim point 0.06 into 10 power - 9 M or Lambda is equal to 0.06 06 nanometer okay so that is option D 11% 11% just 11% okay yes okay yeah okay so we are end of the lecture okay end of the lecture dual nature of radiation and matter end of the lecture SL one 135k homework homework and this is also homework and this is also homework and this one maximum kinetic energy already maximum kinetic energy so depends only on maximum kinetic energy depends only on frequency right directly frequency okay yes tricks so n photons of wavelength Lambda are absorbed by the black body of a mass m the moment the momentum gained by the body is directly NH divided by Lambda NH divided by Lambda that is option A that is option a option A is the right answer option a option A is the right answer got it okay problems are more the in exam problem okay problem okay problem numer application question next one previous question option A H 22% yes try this one a photoelectric threshold wavelength for a silver is Lambda kn and the energy of the electron ejected from the surface of a silver by incident R incident wavelength that is Lambda will be okay so simple question so total energy that is equal to work function plus kinetic energy right work function plus kinetic energy the energy of electron but energy sotic energy okay kinetic energy that is equal to H sorry e minus 5 e minus 5 so in terms of HC Lambda okay H CID Lambda Lambda and energy of the electron okay energy of the electron HC divided Lambda KN HC is common okay H C is common and 1 by Lambda minus 1 by Lambda KN and K is equal to H Lambda minus Lambda divided by Lambda Lambda okay HC so finally kinetic energy that is equal to HC HC Lambda KN minus Lambda divid Lambda minus Lambda this is the right answer what is the right answer H C Lambda minus Lambda Lambda into Lambda KN that is option option Yu option D option okay now let's see okay yes yes sir okay option D okay now let's move on to the next question ah homework next one homework next one yes kinetic energy of electron gets tripled homework Lambda equal h under root 2 m k so that is directly proportional to under square root K okay let consider this is one okay let consider this is one and so K Das K tried three times of a k and Lambda Das much yes huh option C ah option C very good very good very good very good okay Lambda Dash is equal to that is 1 byun3 of Lambda okay that is option C answer option C answer option C okay now let's move to the next question the maximum kinetic energy of electron electrons emitted electrons in a photoelectric effect does not depend upon yes come on come on boys and girls answer maximum kinetic energy of emitted electrons in a photo electric effect does not depend depend Allah does not depend okay so first you know total energy that is equal to 5 plus kinetic energy okay kinetic energy work function intensity frequency wavelength okay so first frequency H new plus kinetic energy okay maximum kinetic energy yes kinetic energy frequency dependent so frequency dependent fine next wavelength HC ID Lambda HC ID Lambda KN plus kinetic energ energy so kinetic energy depends on Lambda also that is the wavelength of light next work function already work function that is a work function so option C is the right answer does not depends on intensity photocurrent depend okay photo current depend photo current depend photocurrent depends on intensity maximum kinetic energy maximum kinetic energy depends does not depends on intensity depend together that okay now let's move on to the next question electron and proton have same electron and proton have same de brogly wavelength then kinetic energy of the electron is same buter of electron okay so what is the relation just before now solve Lambda is inversely proportional to the ener under energy yes formula H idot 2 MK 2 MK now here cross multiplying it okay under < TK 2 m k that is equal to h / Lambda and squaring on both side 2 MK that is equal to h / lamb Lambda and K is equal to K is equal to 1x 2 m under root H divid Lambda yes first know H is constant Lambda is also constant and two is also constant therefore kinetic energy inversely proportional to the mass okay inversely proportional to the mass so IA mass of electron is much lesser than mass of the proton hence kinetic energy of electron much greater than kinetic energy of the proton simple okay so right answer more than that of a proton more than that of a proton more than that of a proton yes 71% answer good next last question of this uh chapter last question of this chap yes which one electron mass is less than proton thus electron has maximum kinetic energy yes correct correct yes option the right okay now so loading loading loading so that is option D yes 86 answer yes atoms and nucleus atoms and nuclei okay atoms and nuclei 3 + 3 total six questions okay six questions okay now lead board yes akbm top anyway next pades dri it's meiru jages patil manula Samar marshmallow grupa karik Shalini radika anaa pritam putas Swami abishek dur okay Shila and VES top 20 okay top 20 yes break okay how much time time break timing timing timing timing yeah you should be happy you should be happy to okay yes 10 minutes 20 minutes 30 minutes yes together mar mar together okay okay 15 minutes is enough okay 15 minutes 20 minutes okay 20 minutes huh start M after 20 minutes okay once rev okay atoms and nucle reion revision e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e hello everyone good evening welcome to the session yes yes begaa answer answer M yes come on come on hello hello hello so one day break okay one day break conductor Electronics complete okay yes good evening good evening good evening everyone good evening good evening good evening yes T is my secret of energy 64th rank so maximum 2 and half hours okay two two and half hours 18 question just imagine 18 question out of 60 okay yes same here yes okay so thumbs up yes yes begera ready ready yes okay maybe maybe possible okay yes okay huh right right right huh okay yes okay so let's get started so chapter name that is atoms atoms in the I mean so ATS andle you can expect at least six questions okay so start so first one you not for alpha particle scattering R for alpha particle scattering okay now first observ still okay minutes EXP so first radio element sorry okay okay almost you can observe here almost partic straight straight line but okay more than 90° more than 90° okay more than 90° know what are the observation we can make out from this diagram okay the first one observation most alpha particles travel through the foil undeflected yes is mostly empty space because Center of atom completely empty the atom is mostly empty space complely nexts some alpha particles are deflected by small angle small Ang the nucleus is positively charged as as alpha particle is positive charge right charge h24 right H e24 h24 so number of protons that is a 2 right so it is a positively charged particle alpha particle is charged part particle and beta particles negative charge particle where gamas are charg less just a race okay now is a positively charged particle I mean positively charged next one occasionally an alpha particle travel back from the file the nucleus carries most of the atom atoms Mass okay the nucleus carries most of the atom M nucleus center of the atom center of the atom right theer okay now next one drawbacks first one you know drawbacks could not explain a Spectra okay Spectra okay balar bracket pund passion next one could not explain stability of an atom so stability which which one is stable and which one is unstable explanation okay yes so next one close the closest distance of approach closest distance of approach and center of the nucle okay okay so small yes okay impact parameter okay so closest distance approach okay do positively charged okay positively charged z e now particle is alsoi charge particle SOI charge right so at this point the velocity becomes zero velocity becomes zero okay n [Music] please okay now right entire energy converted to potential energy right entire kinetic energy converted to potential energy so e distance are so according to conservation of energy initial energy that is equal to final energy right initial energy of M v² of MV s potential K that is 1 4 EP not and z z e into e this is alpha particle divided by R okay so R closest distance approach so r k e div energy clear okay yes next one impact parameter you know this is a nucleus okay this is the nucleus okay this is the reason okay yeah this is a perpendicular distance actually okay this is the perpendicular distance of a line joint okay line joining of a nucleus soers proportional okay proportional the impact parameter is the perpendicular distance between the initial velocity Vector of the alpha particle and the center of the nucleus when the particle is at a finite distance finite distance finite distance okay fin distance okay now next one angle of scattering angle by which alpha particle is deviated from the from it original Direction original actually this is a Theta this is angle of deviation right okay right now lecture okay now so when will be okay Z head on collision head on collision head on collision because okay 180° and rebounds very good rebounds okay yes next one BS model drawbacks okay first model okay now B model of a hydrogen atom the electron in orbit okay in an atom revs around the nucleus in a certain permitted orbits called stationary orbits you know circular Orit okay directly okay yes so p p okay now an electron in an atom revolve around the nucleus in a certain permitted orbits called stationary orbits an electron in a stationary orbit does not radiate energy okay certain certain right C okay so this is the nucleus okay this is the nucleus this is the nucleus okay and Ne so let's consider electron okay electron so elect and what is the radius of this what is the radius according to previous one an electr okay circular next an electron can R around the nucleus only in those orbit for which the angular momentum is integral multiple of H ID 2 pi it is a multiplied integral multiplied okay yes thank you thank you jagesh now is called PL planks value right 6.62 into 10 power 10us 34 second next one an atom can radiate energy only when electron undergo transition from orbit of higher energy level to orbit of lower energy level and R under transition from okay radi energy higher energy level to lower second third fourth lower what if uh the particle atom okay transition from lower energy to higher right clear yes it is a constant okay it is a constant value it is a constant value okay now so this is the energy level Okay so is e EF initial energy level and final energy level CLE force force centripetal force centripetal force and that is equal to Kum Force okay m² ID R and that is equal to K z e and Eid r² am I right from second second MVR that is equal to h / 2 pi H ID 2 pi into n n n is the principal quantum number okay now value so V is equal to NH / 2 piun m R okay 2 pi MH sorry 2 pi m r value okay this is one equation one this is equation two so 2 in one 2 in one m / R into NH / 2 pi Mr R okay whole Square NH right hand side and k z e s divid r² okay so now open the bracket m / R n² h² ID 2 pi m² r² that is equal to k z e s / r² k r s r sare get cancel so further R equation so I want R therefore I'll cross multiply it R is equal to okay R is equal to M n² h² divided 2 pi okay 2 pi m² next one k z e squ okay simplification nowal n² okay h² divided next 2 m k z e sare n sare 1 / 4 Pi Epsilon right 1 4ep r equal h² ID 2 pi M into 1 / 4 Pi Epsilon into Zed into e s right so further simplification so Pi Pi get cancel 2 1 2 2 okay r equal okay 2 h n sare okay sare 2 h² divid pi square and 4 Square four four four okay four so Pi square < square h Pi square and Pi Square so M Epsilon and e s is two constant n² ID Z okay so finally two constant H is also constant that is a 6.6 into 10^ - 34 Epsilon 8.854 into 10^ -2 M mass of the electron 9.1 into 10 - 31 Pi 3.14 e 1.6 into 10^ -9 not Sol we are going to get a not value as 0.59 3 angstrom okay 0 point 593 angstrom okay so finally R is equal to a n² / Z question so R can I say R is directly proportional to the n² Z that is the atomic number okay now next one velocity consider okay velocity MVR divided NH ID 2 pi okay now velocity therefore rearrange the equation 2 pi Mr R and we know that n h ID 2 pi M and what about r r 2 h² Epsilon 2 h² Epsilon [Music] next one M Pi e s m Pi e s n² divid z n² ID Z okay now simplify Pi Pi get cancel okay m m get cancel H can so V is equal to okay okay n okay so Zed by okay e s okay done done so n Epsilon H 4 is okay okay e s divided 4 Epsilon H okay Epson h z divided n okay so V you can say Z ID n directly proportional is it clear only formula is needed yes okay yeah and energy energy energy in N orbit energy in N orbit that is given by 13.6 okay divided by Z Square divided by n² where negative okay elect is bound okay electron is bound state right yes exactly so this value is okay v v so finally V is equal to V Zid n where V is 2.5 2.2 10^ 6 m/ second okay so V that is equal to 2.2 into 10 power 6 m/s 6 m/ second okay and here n is the principal quantum number Z is the atomic number simple okay is next time period so time period yes time period so time time period of a electron around a nucleus in a circular orbit so generally T is equal to distance distance circumference of a circle know 2 i r ID V so 2 pi are the constant therefore R ID n dependent on r divid n so r value any there that is n² right n² divided Z huh n² divided by Z and this one n² divided Z n² ID Z into r value Z ID n okay directly proportional to n² okay n Cube divid Z squ clear yes okay know the 50 50 slides okay now clear next what about current current next current it is given by DQ divided DT right so DQ DT so Q is a charge ignore it and DT time period Ulta reciprocal that is z sare ID n z s divid n so I is directly proportional to the z s ID n is directly okay so draw drawbacks of this one we can go ahead with this could not explain Spectra could not explain stability of an atom closest distance approach already know formula okay next one spectral series spect Spectrum emission spectrum absorption Spectrum okay so important but energy now this formula plays a very important role this form plays very important role okay where Lambda is a wavelength R is the redb constant Z is the atomic number N1 and to which it ised from which it is De exited state in the exited N2 and N1 Z atomic number so exited State lman series Balmer series passion series and bracket series last one p so okay numbers this is N1 this is 1 2 3 4 and five 2 3 4 5 and so on 3 4 5 and so on 4 5 6 and so on 7 5 6 7 and so on 6 7 8 and so on okay so this is the N1 value and this is the N2 value right huh questioning part yes questioning part so try to answer this one try to answer this one try to answer this question yes try to answer this question the first question hydrogen atom initially in the ground state initially in the ground state okay is exited by the absorbing a photon of a wavelength 980 angstrom the radius of the atom in the exited state in terms of a B's radius a will be okay radius so generally radius that is a KN okay a KN n² divid z n Square okay okay ground state and this is the exited State hen atom okay initially in the ground state let's consider this is hydrogen atom so okay with wavelength H2 W length of 980 okay so first initially ground state ground state -3.6 okay n² sorry z s divided by n² hydrogen hyrogen equal to 1 therefore in the ground state I'll consider it as a one okay so -3.6 into 1 squ divided by this is also 1 square so finally you'll be having 13.6 okay 13.6 uh 13.6 energy electron Vol okay now okay ex absorbing energy absorbing Aon of right in of that HC lamb so HC HC Val 1,200 sorry 12500 electron Vol angom divided by wavelength is to there 980 okay 980 Anum so angstrom angstrom get cancel okay electron okay don't mind H because uh okay 980 so 12. 12.75 electron volt okay this is also electron volt right so this is en N2 already 13.6 13.6 right it is absorbed right so 12.75 electron volt 13 13 6 13.6 plus 12.75 26 oh minus ala yes yes minus 13 6 + 12.75 so so this is 0.85 elect okay so e n that is equal to okay e n is equal to so -3.6 Z square that is n Square okay n square and what is the value of n that is equal to minus 0.85 electron Vol so further not minus minus get cancel so n value n value that is 13.6 divid 0.85 you it is 13.6 and it is a 0.85 so it should be greater than 13.6 is less than one right so and it is less than one it is completely 16 so n² that is equal approximation n equal 4 n equal to 4 yes first year physics okay first so a not as it is n Square 4 square that is the 16 divided by Z is a 1 therefore 16 a 16 a 16 a is the answer 16 a is the answer okay so that is option C option C is the right answer very good now try to calculate this one yes in a hydrogen like atom electron makes a transition from an energy level with a quantum number n to another with a quantum number n minus one if n is greater than one okay greater much greater than one the frequency of a radiation the frequency of a radiation emitted as a proportional so frequency so generally say speed speed is equal to F into Lambda so already Lambda 1 by Lambda that is equal to r² 1us Lambda 1 sorry N1 s minus N2 sare am I right yes further 1 divid Lambda okay 1 divid Lambda r that is equal to hydrogen so this becomes a 1 okay 1 square so 1 / n² minus n -1 whole squ okay right huh yes simplification 1 iD Lambda that is equal to r 1 / n² cross multiply it nus1 s okay okay okay to another with a Quantum that is the final so that minus this one okay so interchange interchange so 1id Lambda r as it is so 1 iD N - 1 s - 1 / n whole Square okay so R is equal to n² minus N - s / n² into n -1 s okay simplify this R is equal to n² - n² - minus and this is 1 + 2 N divided by okay divided by n² into n -1 whole squ so simplification plus sare sare and therefor ignore it okay eliminate okay now 1 Lambda that is equal 1id Lambda that is equal to R bracket Off 2 n divid n² n square right so cancel 2 R n^ 3 frequency frequenc so frequency Lambda this is a frequency so FAL C Lambda right so 1 Lambda proportional this is a constant so frequency that is equal proportional to proportion this is this is also constant so 1 divid n CU so option D is the right answer option D is the right answer so po open sorry okay start start yes do you have any questions yes do you have any questions okay nucleus start clear yes hello Niara akbm yeah takes a lot of time okay it takes a lot of time okay morning 11 11: a.m. start okay yes fine now next question yes in the rther fors alpha scattering experiment as impact parameter try try try your best scatter experiment as impact parameter increases okay impact parameter dotted line in the dotted line in the impact parameter B inversely proportional right inversely proportional angle that is B angle that is cattering angle of the alpha particle as well as impact parameter are inversely proportional suppose increase impact parameter incre scattering angle of alpha particle decreases angle degrees direct Theory based question yes yes possibility there okay thank you thank you thank you yes fine now next question so option C option C very good 50% answer now next one three energy levels okay one second yes yes three energy levels of a hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron trans are so first first so first E2 E1 E2 minus E1 E2 minus E1 that is equal to Lambda 1 okay not direct Lambda 1 HC divided by Lambda 1 okay let me consider equation one next one Lambda to yes E3 minus E1 that is equal to H C divid Lambda 1 Lambda 2 so this is equation 2 Lambda 2 Lambda 2 Lambda 3 that is E3 minus E2 that is equal to HC Lambda 3 that is equation equation E3 - E2 + E2 - E1 that is equal to HC / Lambda 3 plus HC ID Lambda 1 okay now get cancel E3 minus E1 that is equal to HC common out side so 1 / Lambda 3 + 1 / Lambda 1 clear yes and E3 minus E1 HC Lambda 2 okay from two from two and four from two and four compare from two and four from two E3 minus E1 HC ID Lambda 2 H C bracket of 1 iD Lambda 3 + 1 by 1id Lambda 1 okay now H C HC get cancel 1id Lambda 2 cross multiply Lambda 1 + Lambda 3 divid Lambda 1 Lambda 3 okay so Lambda 2 formally Lambda 2 formally Lambda 1 Lambda 3 divid Lambda 1 + Lambda 3 so Lambda Lambda 2 in the form of Lambda 2 Lambda 1+ Lambda 2 Lambda 1 Lambda 3 divid Lambda 1+ Lambda 2 so this says option D is the right answer option D is the right answer yes 61% answer good keep it up okay yes uh complete okay now days complete okay only two days two days first portion complete right next next one the radius of hydrogen atom the radius of hydrogen atom in the ground state 0.53 angom after collision with the electron it is found to have a radius of 2.12 angom the principal quantum number n of the final state of the atom is so first ground ground n equal 1 so let's consider 60 90 seconds let's consider 90 seconds okay now observe ground state ground state n is equal to 1 okay now original formul a n Square divid Z right n Square divid z a constant z z is equal to 1 therefore R is directly proportional to the principal quantum number so R1 directly proportional to the N1 Square R2 directly proportional to the N2 square but N2 principal quantum number R1 R2 that is equal to N1 N2 whole Square okay R1 value is to 0.53 huh 0.53 angstrom angstrom get cancel like that no need to convert it okay 1 2 into N1 is2 N1 that is a 1 N2 S2 n whole Square because 1 square one okay n Square okay cross MP 2.12 divided by 0.53 okay 0.53 fine yeah yes so 2.12 divided by 0.53 so 4 so n² that is equal to 4 n is equal to 2 so which is the right answer yes which is the right answer Bea which is the right answer option A don't so option A 76% very good very good okay very good now next uh accordance with a B model of a quantum number that characterizes Earth revolution around the Sun in an orbit of a radius 1.5 into 10^ 11 M with a orbital Speed 3 into 10^ 4 m/s given mass of the Earth simple MVR NH / 2 pi the quantum number okay the quantum number so n value 2 pi MVR ID H so 2 is constant Pi is constant M mass of Earth that is a 6 point okay 6. 6 into 10^ 24 V 3 into 10^ 4 m/s r 1.5 into 10 11 6 into 6.6 into 10us 34 so sub solve for solve for n okay so this is mass and this is speed and this is radius R clear yes so solve so answer option C okay answer option C fine option C clear okay now next one yes it is same as we have solved the previous one right so that is R1 ID R2 N1 ID N2 whole Square same method follow Mone okay consider it as a homework consider it as a homework so answer that is 4761 okay no all 14 lessons all 14 lessons Okay all 14 lessons teach got it okay now so let's move on to the next question yeah uh equivalent current okay so question you know in an in an atom electron re or in the nucleus along the path of radius 0.7 to angstrom making 9.4 into 10^ 14 Revolution per second Andre N ID T per second revolution number of turns per second n okay equivalent current is given by so current is given by Q divid T Q is NE divided by T and group The these number I mean this one NT divided by E so I is equal to what is the number that is 9.4 into 10^ 14 divided t t t e e value e Val e 1.6 into 10 power - 27 okay - 27 so so 1 Point 1 Point so 9.4 into 1.6 so 15.04 15.04 okay 15.04 1.4 1.4 amp I mean current okay 15 option a option A okay yes option A okay option A yes okay so option A is the right answer option A is the right answer okay start okay fine no problem next question homework next question energy of an electron in the second orbit of hydrogen atom is E2 the energy of electron in the third orbit of helium will be so let's consider third orbit of helium okay so generally e is equal to -3.6 z s ID n² first one okay first one hydrogen for hydrogen okay start okay start yes for hydrogen for hydrogen Z is equal to 1 okay and principal quantum number is to second state that is two now e is equal to that is a two Al second exited State 13.6 into 1 square that is still 1 and this is a 2 square okay so- 13.6 divid 4 okay so let it be next one for helium for helium for helium Z is equal to okay helium that is a two right atomic number two okay now and principal quantum number is to third 3 okay now E3 that is equal to -3.6 into z s that is a 2 square divided 3 Square so further simplify E3 that is equal to -3.6 4 divid 9 so E3 / E2 that is equal to -3.6 into 4 / 9 and E2 then any the value E2 E2 value that is a 13.6 okay so -3.6 -3.6 gone and 4 4 16 divided by 9 and E2 so this is E3 value so this is the answer so which is the right answer 16 by 9 E2 option b option b is the right answer option b is the right answer 77% good very good this video will be saved right yes Save okay 12 hours okay 200 300 Pages 200 okay 200es okay leader board akes dri srishti it's me viru Shalini jages Samar manula karik yes okay now let's move on to the next concept okay I okay okay okay now next one okay the period of revolution of electron revolving in nth orbit of hydrogen atom time directoral to n sorry nid z s right 59 n n Cub divided z s n Cub divided Z squ okay NB divided z s n Cub divided Z squ yes so who is the right answer Zila that is nqbe only that is option C okay that is option C very good very good so that is option C oh start mod okay fine fine so next question huh angular momentum angular momentum try yes angular momentum of a kinetic energy so kinetic energy of electron energy of electron find yes ah side okay okay energy in N State okay energy in nth State energy in nth state so e is equal to okay 13 6 z s ID n Square okay so angular moment of angular moment of electron in hydrogen atom is okay 3 H ID 2 pi ktic energy find fine mod that is in terms of okay this is in terms of NH / 2 pi where n is equal to 3 okay so hydrogen atom therefore -3.6 into 1 S ID n² n square is 3 squ so 9-3 6id 9 which is the right answer so mag will get answer okay option b right option b right answer so that is E3 E3 E3 that is 1.51 1.51 electron Vol 1.51 electron volt okay 1.51 electron Vol clear yes now next one H frequency of Revolution so time period n CU divided by Z squ T and frequency are inversely proportional therefore frequency will be Z squ divid NQ so therefore with respect to n Cube that is option D is the right answer right yes calculator we do not use so okay don't mind next ccor use okay yes clear yes now next one take it as a homework next total energy of an electron revolving in the second orbit of hydrogen second orbit n is equal to 2 N is equal to 2 same formula - 13.6 Z Sid n² so use you'll get the answer hydrogen hydrogen Z value becomes one so -3.6 divid n² next the period of revolution of a electron in the ground state of hydrogen atom is T the period of revolution of the electron in the first exited State okay first exited State first exited state right first exited first exited State option C no see ground state and to for a ground state for a ground state for a ground state n is equal to 1 first exited State first exited State first exited state that is n is equal to 2 so time period that is nid z square right so hydren T is directly proportional to the N right so so T directly proportional to n value is to 2 whole Cube so T is directly proportional to 8 8 times so T Das becomes 8 times okay so that is option D that is option D very good very good okay option D okay now next question the energy in electron Vol required to exite an electron from Nal 2 to Nal 4 hyrogen at so change in energy that is given by E2 minus E4 so e generally generally e e is equal to -3.6 z s / n² right so hydrogen at that is okay 13.6 divid N2 s minus okay minus 13.6 divided 4 whole squ okay so now n 2 that is equal to 2 N4 that is equal to 4 substitute and get the answer -3.6 it is a common and 1id 2 - 1 iD 4 1id 2 and 1 divid 4 okay Square yes Square yes simplify correct correct uh previous one that is option D yes answer how much you get take the LCM so 13.6 so 1id 4 1id 16 so four constant -36 4 1 1 by 4 okay so further - 13.6 divid 4 into uh 4 - 1 that is a 3x 4 okay so 3x 4 16 yes plus 2.55 exactly you know what is the reason they already have eliminated negative sign E4 minus E2 E2 M okay now that is the 2.55 that is option D that is option D 88.9 next homework next homework next homework next homework next ah okay so so do if an if an electron okay if an electron if an electron okay if an electron in a hydrogen atom that means Z value one atom jumps from an orbit of level n is equal to 32 and orbit of level n is equal to 2 okay n is equal to 2 the emitted radiation has a frequency so frequency Namaste Namaste namaste yes N1 N1 is two and N2 is three N1 is 2 and N2 is a three okay now frequency already PR that is Cal F Lambda okay frequency that is a z r z s bracket of 1 1id n² - 1 / N2 squ okay [Music] nowc RC this is a z becomes a one okay this Z becomes a one now 1 divided by 2 2 - 1 / 3 S so further simplification RC and 1X 4 1 divid 9 so further simplification 9 - 4 36 and further simplification 5 RC divided by 36 this is the 5 divided by 36 that is option C that is option C 40% answer okay now next question uh so pole pole huh which of the following spectral series of a hydrogen atom is lying in a visible range visible range right Balmer series that is Balmer okay that is infrared okay infrar yes very good yes homework Sol okay the figure shows the energy level of a certain atom when the electron D exited from 3 e to e 3 E2 e okay fine 3 E2 e okay 3 E2 e so H C divided by Lambda okay so let me consider this one 2 e divided byc divided by Lambda next if the what is the wavelength of the electromagnetic wave emitted when the electron de exited from 5 e to e 5 e by 32 e okay 5 e by 3 e e sorry uh is equal to HC 5 5 eus 3 e that is a 2 e divid 3 that is H C divid Lambda 1 okay now 2 e that is equal to 3 HC divided by Lambda 1 and compare equation 1 and two compare equation 1 and two compare 3 HC divided Lambda Dash Lambda Dash Lambda 1 that is equal to H C ID Lambda and here HC HC get cancel cross multiply Lambda 1 is equal to 3 Lambda okay so that is option a option A H 3% answer only 3% yes ah okay okay okay ah okay okay fine fine okay it is not possible okay yes ah leaderboard leader board leader board so next slide start okay yes so next topic akbm paves and danash sishi it's meiru Shalini and Samar jagadesh Samra okay Kik and mura ah okay so let's get started with the nucleus nuclei okay nuclei so you can expect three questions easy question Mass defect binding energy okay short long okay now let's get started with nuclei so first one nucleus and nucleons so first one radius of a nuclei do you have any idea radius of a nucle r is equal to r a power 1x3 do you remember okay 1 by 3 next one volume of a nuclei so 4x3 pi r CU 4x3 R not okay h r not 1 by 3 3 power three power three and a 1 by 3 1 by3 cancel okay Alpha Beta break it is a very small chapter okay semiconduct Electronics okay now first one electron electron Mass so mass number mass number and Z atomic number atomic number atomic number okay so atomic number number of protons IND number of protons number of protons IND number of nucleons okay number of nucleons number of nucleons that is number of protons plus number of neutrons okay combine got it okay now 170 there 174 so 30 30 35 okay density of nucle it is independent independent of a okay nuclear density it isent okay fine now electron so 9.1 into 10 power minus 31 kg kg already mention and proton that is 1.65 okay 1 625 yes 1.625 into 10^ - 27 kg and neutron 1.6 75 into 10 power - 27 okay K 0.55 AMU next one proton proton this to 1.77 and 1 86 what the yes binding energ discuss got it okay now let's move on to the next concept uh range of force range of force and it is a short ranged short range short range in the formally Fe meter feter that is in terms of 10 power minus 15 met in terms of 15 M okay and spin Independence dependence spin dependence and nature of force nature of force attra attractive in nature okay attractive in nature okay attractive in nature attractive in nature attractive but if it is less than 0.8 okay 0.8 FM repulsive non Central force and strength of the force okay strength of the force the position of position dependence independence of a charge and Order of the order of first order of a force okay first order of force independence of a charge and okay what is the strength of force okay okay charge something nucleons proton neutron Neutron proton neutron Neutron proton proton and neutron proton right not only understanding okay yes okay right yes okay now next one so nuclear stability graph okay just go through it stability graph okay now next one and uh Einstein energy Mass relation and mass defect so first of all Einstein mass energy relation eal mc² energy can be converted into mass and mass also can be converted into energy interconvertable by this equation right and mass defect Mass defect Mass defect formula Delta M Small Delta M okay so Z MP that is the number of protons into mass of the proton plus a minus Z into mass of the neutron this is the number of a neutron okay minus mass of the nucleus mass of the nucleus okay this is these two formulas are very important these two formulas are very important okay now next one uh binding energy binding energy for nucleons and its significance binding energy and and okay binding okay yes fine now binding energy of binding energy binding energy binding energy for per nucleon okay per nucleon divided by a this is general formula okay general formula in general in general okay stability binding energy ofon is 4928 me El volt and mm is 1640 Mega electron volt which nucleus is more stable which nucleus is more stable so first one for iron for iron for binding energy per nucleon binding energy for IR that is a 4928 divided by nucleon 56 okay 56 yes yes yeah yes duration okay yes clear 8.8 8.8 very good 8.8 Mega electron volt now next one for bisou for basm okay same binding energy for nucleon yes that is a 164 divided 209 7.8 Mega electron volt okay Mega electron volt okay so this is more stable iron is more stable than bismo clear okay clear right next one so the graph binding energy per nucle gra yes finding binding energy binding energy binding energy that is equal to mass defect into 9315 me Elon Vol and what about this Mass def Delta M so Delta m z MP Plus a minus Z MN minus minus mass of the nucleus so first of all binding energy of alpha particle from the following data mass of the nucleus S2 this m Val so this is M Val and mass of the proton that is a mass of proton and this is a mass of neutron okay mass of the neutron and uh this is the Mega electon Z value and Al partic that is a 24 Z Val two 2 into 17277 plus a is to there that is 4 minus 2 this is a value and this is z value okay into mass of mass of a neutron that is 1. 866 am minus mass of the nucleus 41265 okay so simplify okay simplify and multiply with 931 okay multiply with 931 clear so multiply with whatever the value you'll get it and binding energy multip Delta M into 931 5 480 9 931 El that's all right yes now Q value now next one binding energy of a nitrogen this is a value and this is z value and this is M value okay this is M value they have given okay now same formula you have to use binding energy binding energy that is equal to Delta M okay into 931 Mega electron volt 9315 Mega electron volt get Point okay simp binding energy that is given by Zed mass of the proton plus a minus Zed mass of the neutron minus M okay so B do e z value 7 mass of the proton mass of the proton uh 1727 1727 Plus 14 - 7 14 - 7 mass of the neutron is2 mass of the neutron 86 86 86 okay and what about this 14307 okay is to multiply with 9315 Mega Elon volt that will be your answer okay that will be your answer clear that one will be your answer so finally option option D option D is the right answer got it okay fine next one soor homework that equal work time taken but right worker hour so energy in terms of a mass energy relation MC squid T mc² ID T So p is equal to am of fuel consumed by okay rearr this equation okay Mass big okay mass that is equal to P into T / c² and what is the p p is a 10^ 9 I this one 1 hour 3600 divided c² 3 into 10^ 8 3 into 10^ 8 okay simplify okay you'll be getting 0.04 okay 0.04 answer option D yes op option D right clear yes next one yes which graph in the following diagram correctly represents the potential ener of a pair of nucleons as function of their Sear but potential in the nucle that 15 it is also electron therefore this is the right answer that is the right answer yes okay now next one two protons are kept at a separation of a 10 nanometer let FN and Fe the nuclear force and electromagnetic force between them okay okay 10 power uh 10 nanometer separation FN and Fe the nuclear force and electromagnetic force what is the relation between them so first of all nuclear force short range okay short range right it exist only in terms of femometer femtometer forer okay now electromagne uh force is to large range long range long range therefore Fe is greater than FN much much greater much much greater okay much much greater next the for experiment for a headon collision okay head on collision and what is the angle 180° written rebounds okay rebounds particle with a golden nucleus impact parameter impact parameter angle okay impact meter impact parameter that is cos Theta by 2 cos Theta by 2 okay so B directly proportional to cos 180 by 2 and B directly proportional to C 90 cos 90 is to 0 so hence this is 0o right zero now next one huh semiconductor Electronics so last topic okay last topic so this chapter going to cover uh one hour one hour cover okay 1 hour is maximum yes time time start start okay 20 minutes 30 minutes BM uh first P classes second okay yes YouTube class okay first okay first okay 30 minutes fine so let's take it as 35 minutes okay yes so 35 minutes okay e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e thank you hello hello hello am audible yes AUD hello disant a gamer Wasim Muhammad sisti SSA alron evening evening pral evening already launching classes starting from April 8 okay 8 in the start yes okay maybe after five after five class start okay okay start mwa one second yes Trant okay okay ready now yes ready now so from this uh semiconductor Electronics each questions okay three questions from semiconductor Electronics is it clear okay uh you know the maximum 45 minutes to 1 hour okay maximum 45 minutes to 1 hour chap complete right okay he gameer okay okay okay now lead board AK pades nashri sishi it's meiru Shalini Samar jagadish karik manula top 10 good okay so highest po interact around 300 plus students interact okay 300 plus okay uh welcome Wasim okay fine storm breaker Optics and Ops yes okay maybe ma class live session yes okay yes fine start okay now so a semiconductor Electronics mainly based on cond okay conductivity property semiconductors work semiconductors okay conductivity property called conductivity okay because resistance of a conductor it is depends on temperature also but e semic conduct temp that is the speciality at 0 yes okay right so temperature incre conductivity okay both are increasing But Metal inversely proportional semiconduct directly proportional clear okay so next first Oney valence band and conduction band okay veence band and conduction band so first of all veence band conduction b veence band lowest energy level where all the free will be you know electrons in a valence band move to the conduction band conduction band electrons conduct got it okay now material three materials the first one okay the first one that is the conductors the first one conductors mple times conductors next one semiconductors next one insulators what are the value of energy gap forap so first of all next one second I'll check okay so next important okay so conduction ah so for example the valence band okay let me consider this as a valance band right Valance band next yes okay now this is a veence band and this is the conduction band right if you don't know compare B conduction band compare for okay for for conductors for conductors what is e value and for semiconductors semiconductors semiconductors G what is e value and for insulators for insulators e what is the value do you have any idea do you have any idea about these values okay now so first one conduction Gap energy gap that is measured in electron Vol okay for cond almost zero EEG value is a zero okay and next one for a semiconductor semiconductor it is a one electron Vol one electron Vol near one electron related it is less than elect Less Than 3 electron right insulators insulators semiconductors 0° C insulators nothing but semiconductors at a 0 de semiconduct what is the energy gap between conduction and Valance band more than three electon more than electon easy value what of material energy gap is less than 3 electron volt what is your answer semiconductors energy gap is zero that means it is a conductor and if energy gap is greater than three electron Vol that is a insulators very good okay now next one next one okay concept of holes holes in a semiconductor so first first of all semiconductors semiconduct semiconductor ex semond semicond semiconduct but EXT semond it is a impure semiconduct EXT semiconduct right doing so do B again semond based on doing p p semond okay so first of all semiconduct semiconductor material first one semiconductor semiconductor okay now the semiconductor divide into two parts one do intrinsic semiconductor intrinsic semiconductor okay semiconductor and you can call it as a pure semiconductor pure semiconductor and this one extran semiconductor extran semiconductor impure semiconduct impure semiconductor imp it is a doped this one is doped semiconductor doped semiconductor dop semiconduct adding doing semond do doing and adding impu the process of adding impurities impurities okay process of adding impu is the semiconductor so for example inic semiconductor ex semicond based on what type of doing it is now again ex P Tye semiconductor P type semiconductor semiconductor okay semond what is okay this is the doing material element pass okay pass okay pass B so what are the meaning of for this Bal so Boron indium aluminium phph aren right so phosphorus and many okay yes clear yes these are the doping elements type semiconduct doed with these elements n type semiconductor doed with these elements clear right off for doubt off got the doubt okay right addition of imp impure material yes clear right okay and in semiconduct concept Tri delete that is the transistors delete okay clear yes all B in game yeah good okay arenic as yes clear okay yes fine now so next one concept of holes and it is the vacant Place vacant vacant place for Po and vacant place of electron vacant place of electron okay holes concept of holes energy gap is equal to zero and nearly equal to one less than three electon for Semiconductor okay other is greater than 3 electron volt for a insulators okay insulators one important semiconductors at0 de sorry Z insor semiconductor for example silicon and germanium at a 0 Dee 0 kin temperature what is the energy gap 0 de sorry 0 mention it is not less than 3 electron volt it is greater than 3 electron volt okay except Z kin less than electron clear right yes next one next one mass action actually difference between and this is a extr semiconductor EXT semiconductor okay EXT semiconductor ex first of all in semiconductor conduction veence band conduction band okay AC group of element n 15 group actually semiconductor 14th group element right so less than 14 and N sorry okay that is aent El comb free electrons okay free electrons positive that is the donor ions so acceptor are the impurities okay next on whole negative acceptor ion holes are in a majority and electrons are in a minority in Cas of y n type semiconductor okay next one in intrinsically current is due to both electrons and holes pure semiconductor majority Carri electrons majority charge Carri electrons holes mple times next one number of electrons and number of hes semond semiconduct right yes y okay okay fine okay next one number of electrons is greater than number of holes Don number of donor number of electrons number of electr so number of electrons is greater than number of hes number of hes is greater than number of electrons okay next one current due to electrons as well as holes okay due to only electrons due to only holes because of a majority carriers in a P that is a holes if in case of a n type that is a due to only electrons got it okay now next one [Applause] entirely charge Neal this one is very important one okay entirely neutral entirely neutral enely neutral yes question same question right neutral now quantity of electrons and hes are equal already discussed in this case electrons are majority and minority that is a holes in PP holes are majority and minority electrons clear yes so just the theory part and this table is very important okay one second very important very important okay next huh now effect of imper semiconductors L that is a p as and SB donor impi any number of electrons is greater number of holes valent P Tye trivalent that is a BAL right Boron Indi and aluminium and number of holes is greater than number of electrons right number of holes is greater than number of electrons but in intrinsically number of electrons number of holes okay now next one important [Music] right by 12 morning okay by yes yes of course I know that you need notes so it will take you right okay huh because of conductivity okay yoube description it is a large download any okay now okay okay fine fine fine okay now next one statement statement one statement one yes po statement one by doping silicon semiconductor with the pentavalent material the electron density increases right doing with semond yes semond is it a p type or n type P type or n type yes okay don't worry very good n Tye so first one it is correct okay first one statement it is true next one the end type semiconductor has net negative charge in the light of the above statement net negative charge net negative charge no because it is a neutral electrically neutral therefore second statement is wrong therefore first statement right second statement is false yes correct cor false wrong wrong statement first and second true correct but wrong and statement one and statement two both are false wrong wrong okay so therefore which is the right answer that is option A that is option A right neutral exactly option A got it now next one biasing so first of all simple word biasing and unbiasing biasing biasing and connecting battery connecting battery connecting battery unbiasing simple simple English no confusion okay forwarded bias forwarded bias and conney for example for example diode okay this is a diode this side is a p and this side is a n type so this sidei connect that means this is a forward bias p p side it is connected to positive terminal of a battery forward same diode okay same diode P type and N type this one we call it as a reverse bias okay positive terminal is connected to Tye region connecting external voltage to the PN Junction simple right okay yes next one right so forward reverse reverse clear yes positive terminal that is the forward bias positive terminal conect that is the reverse bias clear now next one p junctional part reason reason electric field that is from n Ty to P type electric field develop because electrons jumping from P type region to n type region electrons jumping from P type to PP region to npe electrons are jumping so drift drift okay so electri field without connecting into battery okay now what happens when PN Junction diode is not connecting to external battery so there will be a electric field there will be electric field okay right so that is the only important concept here next one biasing forward bias positive terminal connected to positive terminal I mean p type semiconductor P type regon connected to positive n type P type is connected to negative type is connected toi according to that you can understand now next one potential barrier redu forward potential I mean right next one width of a dep layer decrees yes potential barrier definitely this layer also decreases but in case of a reverse increas because rever yes resistance is a low in a forward forward bias but resistance is high in case of a reverse bias okay yes next one PN Junction provide very small current PN Junction provides very small resistance resist not current so small resistance current is more PN Junction provides High Resistance therefore current is less therefore in Reverse bias current will be very small but forward bias current will be maximum because it provides low resistance this one provides large resistance clear yes between infation reason immobile ions are present in a forward bias good okay now next one so again forward current flow in a current forward reverse bias reverse current flow in a circuit next one order of forward current is a m that is in terms of a m butro okay second one micro ere nanoampere very small amount of current flow reverse bi okay very good okay next one mainly majority current flows mainly minority current flows okay majorly majority current flows okay mainly minority current flows majority and minority difference okay now next one forward characteristic graph which type of Okay negative negative that is reverse bias Okay negative sign negative that is reverse bias okay next one reverse bias that is negative potential is connected to a p type reason other positive terminal is connected to a n type reason fine okay now yes this is highly important okay this is highly important highly important highly important okay now taking of a biasing whether the given Di is a forward bias or reverse bias based on the external battery based on external battery okay so it is simple trick second okay so VP is greater VP minus VN if it is greater than zero and VP minus VN if it is less than zero then it is a okay then it is a forward bias and it is a reversed bias okay so first one calculate man first one calculate man okay the first one you see it is a p type and this is the n type this is a p type and this is a n type this is a p type and this is the n type this is a p type and this is the n type okay now P side VP okay VP side VP volage 5 Vol minus 10 Vol equals to -5 Vol Which is less than zero now tell me what type of bias it is what type of bias it is first one negative value yes Yu reverse bias or forward bias reverse bias very good reverse bias okay now second one no second one VP potential across P typee that is a zero minus potential across n type region that is a minus 2 so minus of Z minus of minus plus that is a plus two old now tell me so this is greater than zero this is greater than zero now tell me what type of a bias it is forward bias very good forward bias forward bias okay now third one third one potential letter P type Reon 6 minus n type Reon minus 3 that is a plus + three and it is greater than zero now tell me what about this and fourth one fourth one p side it is a min-2 minus n side that is a minus 5 this is min-2 minus ofus + 5 that is a + three it is greater than zero now tell me what are the last two condition third and four third and four third and four very good very good third and four it is third and four this is a forward bias and this is also forward bias clear one question fixed okay one question fixed clear right do you have any questions related to this forward checking of a biasing checking of a biasing do you have any questions yes okay clear fine now next one next one representation of p and Junction in a circuit okay representation of a PN Junction I mean p and Junction diode in a circuit P Junction this side p and this side n okay so circuit you this is a high potential this is a low potential so potential is connected to P type this is a forward bias okay forward bias and same di same this is low potential low potential is connected to P type that means it is a reverse bias reverse biased is it clear okay representation of a circuit representation of a p Junction diode in a circuit next fine these values but still next p and Junction diode in electrical circuit fine ideal diode ideal diode ideal diode so suppose diode is a forward bias okay diode is a forward bias germanium Okay g that means it is ideal diode IDE this is just wire just wi for the current okay now reverse reverse okay reverse this PN Junction this PN Junction acts as a acts as a you know just open circuit open circuit okay open circuit is it clear open circuit correct simple suppose real di silicon and germanium silicon and German so for example real real real cases okay real cases for example diode is a forward diode play okay for example this is a diode okay this is a p type and this is a n type as a battery opposing battery okay opposing battery and opposite the flow of current okay opposite to the flow of current so this is battery battery so opposing battery opposing battery opposing battery clear suppose reverse bias reverse bi same diode PN Junction diode reverse bias again it acts as a open open circuit open circuit open Circ because where is small amount current okaying yes okay ignore them right now clear topic clear yes topic clear clear why open circuit that is the characteristics of that is the characteristics of p and Junction in a reverse bias yes clear very good very good ide a okay CC it acts as a wire cleever okay conne yes maximum 10 slides okay 10 slides maximum fine very good now let's move on to the next concept yes already explain IDE behavior of diod in a circuit right next yes figure shows a diode of okay integer type say 90 seconds yes number fine now figure shows a diode connecting to an external resistance this is external resistance and EMF assuming that barrier potential developed in a diode is a 0.5 Vol 0.5 obtain the value of a current in a okay real case scenario okay real case this is high potential and this is low potential and this is a resistance right this is a resistance r that is a 10 ohm clear 10 ohm and here this is which diode huh yes for bias follow bias yes yes 18 Mar 18 questions okay now okay 0.5 0.5 potential barrier potential barrier okay potential barrier do am I right so this is 0.5 okay 0.5 opposing battery right opposing battery 0.5 0.5 so what is the net EMF okay what is the net EMF okay what is the EMF what is the value here 4.5 and here opposing 0.5 that is a four okay 4 and here they're asking about a current right current that is equal to okay current that is equal to potential divided by resistance potential divided by resistance okay potential potential divided by resistance what is the potential 4 divided by resistance s to 10 0.4 okay so 0.4 yes 0.04 resistance yes yes soly divide by multiply and divide by th000 yes multiply and divide by th000 multiply multiply and divide multiply and divide 400 4 mamp 400 your answer must be 400 okay clear yes no idea is 0.5 okay yes right next one ah which is the right answer that is 400 400 400 so multiple choice question start The Bard yes so correct a PN Junction diode shown in the figure can act as a rectifier rectifier it converts AC to DC and alternating current Source V is connected in the circuit the current I in theist can be shown by so positive connected to positive p and p and side okay first try to understand so POS keep onu because poity keep on reverse we are we are getting current Okay so this one becomes a negative negative becse right reverse it acts as a open Circ it acts as a open CC so therefore current will be zero so current will be Z yes therefore it is eliminated and Z it is also eliminated therefore option A and process repeat you'll be getting complete cycle half W rectifier very good half W rectifier okay now next one okay consider this as a homework so like a same what we have done a previous case okay same way you have to do it okay now applications of a junction the first one halfwave rectifier next one concentrate party okay important okay this one is important this one is important this one is important okay so half only half ccra Rectify half cycle Rectify working working EXC working working principle explain because already have done with exam no yes you're done with exam why option D is not correct huh ofal to positive reason that becomes reverse I mean yes yeah beta okay fine beta right full wave rectifier okay full wave rectifier already full is this still AC but not completely AC or not completely DC partial AC partial DC DC elimin call it as a DC okay yes uh VI what is that ATP what is that ATP ah yes okay next H yes a full w retif with a diode full wave rectifier with a diode okay full wave rectifier with a diode full wave rectifier with a diode D1 and D2 is used to rectify 50 hertz alternating voltage the died One conducts okay died One conducts times in 1 second so what they're asking frequency what they're asking frequency which is the right answer okay okay okay fine fine yes Mah very good very good very good very good mahes wow nice nice wow yes yes correct correct my is correct yes option D now yes they're asking about a frequency hence option D is the right answer option D is the right answer good keep it up now that is 16% answer times in 1 second what is the frequency of alternating volage alternating current okay what is the frequency 50 Herz in India okay us the L to 60 Herz this is in India this is in USA 50 clear yes now next question half rectifier same explanation so next one working principes yes what is the answer for this what is the answer for this question what is the answer for this question a 220 okay a 220 Vol AC Supply is connected between point a and point B okay point a and point point B right as shown in the figure what will be the potential difference across capacitor what is the potential difference across capacitor yes what is the potential difference simple like what is the voltage here providing VMS AC current VMS that is 220 and what they're asking across SOC what will be the peak that is a peak value that is the potential difference okay SOA what is the relation between V not and VMS this is the equation okay V Sirac but yes got it okay now < tk2 vrms now V not that is equal to < tk2 what is the value of VMS that is a 220 220 or 200 into < tk2 old right which is the right answer which is the right answer yes option D option D is the right answer option D is the right answer 31% okay lead lead leader board answer Jaguar we right now next one sorry yeah so po start for 30 seconds a positive hole in a semiconductor is positive hole starting what is the importance of concept of holes it is a vacant Place electron okay so anti particle positive hole is the okay a positive hole in a semiconductor is an anti particle of electron a vacancy created when electron leaves a calent band and absence of a free electron an artificial created particle which one which one is the right answer that that is option b direct okay that is option b right option b is the right answer yes yes very good option b is the right answer it is the vacancy created by a electron in a coal band okay that should be Valance band okay now conductivity starting what is the importance of semond it is the one of the property that is a conductivity conductivity of a semiconductor increases with increase in temper because y yes reason is first one number of density of a charge carrier increases relaxation time increases both number density and charge carriers and relaxation time increases the last one number density of a current Carri carrier increases relaxation time decreases collision between particles okay time taken by electron to collide consecutive two times okay but effect of sorry effect of a decreas in relaxation time is much less than increase in number of a density this is the proper answer okay this option D this proper answer this is option D is the correct answer okay yes got it now so here option D is the right answer option D is the right answer not 14 and 14% just 14% okay now try to answer this one resistivity resistivity conductivity resistivity and conductivity what is the answer for this resistivity of a semiconductor at a room temperature room temperature Plus or 25 plus or minus 2 25 plus orus 2 temperature what is the resistivity of a semiconductor 10us 3 into 10us 3 to 10^ 6 ohm CM 10^ 6 to 10 Power 8 ohm CM 10 power 10 to 10 10^ 12 ohm CM 10 power - 2 10 power - 5 ohm CM which one H Omar yes yes yes it is possible okay okay yes death wi it is possible dedication depend okay so this is option A is the right answer okay it is direct question from textbook okay so that is option A only 5% answer my okay now forbidden energy gap of germanium Crystal ah question at0 kin at0 Kelvin yes at know the Forbidden energy gap for a germanium Crystal at 0 kin 0 insulator semiconductor behaves as a insulator at a 0 Kelvin temperature insulator do what is the at0 degree I mean 0 Kelvin temperature G this Crystal acts as a okay germanium Crystal acts as a insulator insulator so insulator what is the energy gap for insulator what is the energy gap of insulator it is greater than three electron Vol so greater than 3 electron Vol it is less than 3 electron volt next one it is less than 3 electron volt and it is greater than 3 electron volt and 0.071 it is less than 3 electron volt so option D is the right sorry option C is the right answer go okay right option C is the right answer that is 31% answer next question minority carriers in a p type y minority carriers in a p type minority carriers in a p type y free electrons holes neither hold nor free electrons both holes and free electrons y yes free electrons majority Carri majority Carri that is holes minority charge carriers that is the option a free electrons option A it is a free electrons okay very good now next one in a PN Junction diode not connected to a battery p and Junction diode is not connected to battery right in a p Junction diode not connected to any circuit battery not connected to any circuit battery right yes potential difference okay generally poti difference amplitude Peak Val got it now the potential is same everywhere the P type side has higher potential than n type side option C there is an electric field at the junction directed from n Type n type side to P type side option D there is an electric field at a junction directed from P type P type to n type which side which side so this one n type to P type Electric fi okay oh start option C option C is the right answer okay next yes huh yes in a reverse biased diode when answer m in a reverse bias diode when the applied voltage changes by okay changes by 1 Vol the current is f to change by 0.5 microamp the reverse bias resistance of a diode is reverse bias resistance reverse bias resistance how to find it out so not resistance reverse bias okay reverse bias resistance reverse bias resistance that is Delta V divided Delta I what is Delta V 1 what is Delta I 0.5 microamp 10 powerus 6 that is a 10 power + 6 divid 0.5 0.5 1 by 2 right 1 by 2 0.1 10 5 1 by reverse bias resistance is 2 into 10 power + 6 okay plus 6 that is option b that is option b 60% very good clear yes next question to a germanium Crystal equal number of aluminium and indium atoms are added impurities are added alum these are the dopant dop yes which one is right it Rems an inic semiconductor it becomes n type semiconductor it becomes a p type semiconductor it becomes an insulator which one end P type confusion confusion huh really n type P type option b 27 indium and aluminium Y type P type Y type P type yes this is p type it becomes a p type that is option C start okay next one huh not so 154 members okay 154 members are the best students of this live session beginning morning 11 morning 11 almost 9:15 morning 11 okay 9:15 almost almost 9 hour M hour 11 hours right so lead leader board welcome yes congratulations okay next one pades okay pades and next one dri congratulations srishti it's me viru Karthik next Mula Samar Shalini radika Kupa jagadish G anaa marshmallow answer okay fine abishek kwami bti so these are the 20 top students okay parti congratulations okay first place in okay next time okay yes yes so yes how was the session yes how was the session is it useful yes Nots to okay by morning 12:00 yes thank you mes thank you so much yes ah part second place congratulation yes useful very good super sir top five Super very very best really good super okay this was good session huh next chapter my favorite chapter that is Optics okay Optics s creation very useful okay opposite of worst that is the best only right yeah ah second place next time only first place yes okay thank you thank you so much thank you so much everyone okay okay okay yes thank you thank you so next class YouTube Community right yes yes okay fine no problem okay okay so next time better per okay so revise Optics electromagne waves okay next topic I mean next session next not like right yeah yes thank you thank you thank you thank you thank you so much sir yes uh it's me one time by UK language yes uh actually sir our College considered as one of the best college in out of the town there are 3 99.8 percentile students but C coaching we oh is it okay okay fine fine akbm I'm really very happy for you because okay so by the end of the session you are on top right so next time okay if possible okay that will be very helpful next s until take care bye-bye

Channel: PW Kannada

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Keywords: emi in one shot, 2ndpuc physics 2024, Physics KCET, KCET 2024, alternating current class 12, kcet 2024 preparation, kcet preparation 2024, PW Kannada, pw kannada kcet, kcet physics important chapters, alternating current important questions, dual nature of matter and radiation class 12, dual nature of matter and radiation class 12 important questions, atoms and nuclei class 12, atoms important questions class 12, semiconductor class 12, semiconductor electronics class 12

Id: oEI_HAerGcY

Length: 620min 2sec (37202 seconds)

Published: Wed Mar 27 2024