AC Circuits Basics, Impedance, Resonant Frequency, RL RC RLC LC Circuit Explained, Physics Problems

in this video we're going to talk about AC circuits and impedance so the first thing you need to know is that the RMS voltage is equal to the peak voltage divided by the square root of 2 and RMS current is equal to the peak current divided by root 2 now see if we have an AC source connected to a resistor in this circuit energy is transformed into heat the current and the voltage are in phase so voltage is equal to their current multiplied by resistance and the average power consumed in the circuit is simply voltage times current so let's work on the problem let's say if we have a 120 volt AC source let's say that the resistance is 100 ohms what is the current that flows in the circuit so using the equation V is equal to IR the voltage is 120 the resistance is 100 so 120 divided by 100 the current in the circuit is 1.2 amps now how much power is dissipated by the resistor so power is equal to voltage times current the voltage across the resistor is 120 volts and the current that it uses up is 1.0 so 120 times 1.2 is 144 watts now let's say if we have an AC signal that is an alternating current attached to an inductor which has this symbol the inductor has the inductance let's say 50 milli henries what equations apply in a circuit inductors present inductive reactance to an AC circuit which acts like resistance the inductive reactance represented by X L is equal to 2 pi F L XL is measured in ohms like resistance so just as V equals IR V is equal to I times XL if you were to plot a graph between the frequency and the inductive reactance these two are directly related as the frequency increases the inductive reactance increases as well so it s close up XL goes up and as the inductive reactance goes up the current in the circuit decreases current and adoptive reactions are inversely related so high frequency AC signals will yield left current in the circuit and low frequency signals to pass more current in the circuit now going back to this equation where inductive reactance is equal to 2 pi FL you need to know that Omega the angular frequency is equal to 2 pi F so Excel can be represented as Omega times L so just keep that in mind also for an inductor the current lags to voltage by 90 you need to know that as well so whenever you have an AC signal passing through an inductor the AC signal the current in it is changing its increase in and decrease in is that it's not constant and because the current changes you can have a change in magnetic flux inside the inductor as the current increases the energy stored in the inductor increases as well and as the current decreases the energy stored in the inductor decreases so energy is stored in the inductor temporarily the inductor stores energy in this magnetic field and then it releases it back as the current in the circuit increased in decrease so in the case of a resistor energy in the resistor is dissipated as heat so that energy is gone but in the case of an inductor the energy sauvé it is stored temporarily in the magnetic field in the inductor and then it is released back to the source therefore for all practical purposes the inductor doesn't consume energy it simply stores it and gives it back so there's no loss in energy now granted no inductor is perfect it's going to be some resistance in the coils of the inductor and then that resistance will just pay mgsv but assuming if the inductor has no resistance then it dissipates in no energy it simply stores it temporarily now let's talk about frequency let's say if the frequency of the AC signal is 60 Hertz what does that mean this means that the alternating current changes direction 60 times in one second frequency is the number of cycles for a second so the current changes direction 60 times per second let's say if the frequency was 5 kilohertz this is equal to 5,000 Hertz that means that the AC signal changes direction five thousand times and the single second so whenever you see the word frequency in terms of an AC signal that's what it means now let's move on to a capacitive circuit capacitive reactance is 1 over 2 pi FC and we know that Omega is 2 pi F so capacitive reactance is 1 over Omega times C so as the frequency increases the capacitive reactance decreases notice that F is on the bottom of the fraction so F and X see the inverse River later if we were to plot a graph between the capacitive reactance that our frequency here's how it will look like as the frequency goes up the capacitive reactance goes down so it's an inverse relationship and the voltage in the circuit is equal to the current multiplied by the capacitive reactance the capacitive reactance like the inductive reactance and resistance they all have the same unit and that is ohms so just keep that in mind by the way for a capacitive circuit and the current leads the voltage by 90 for an inductive circuit the current lags the voltage by 90 now just as an inductor can store energy temporarily in its magnetic field the capacitor can store energy temporarily in its electric field in an AC circuit the voltage is constantly increasing and decreasing it's basically a finally as the voltage goes up then it goes down and so forth whenever the voltage is increase in the capacitor is stored energy is being charged it's stored energy in its electric field and when the voltage decreases the capacitor releases that stored energy because it absorbs and releases energy just like an inductor the capacitor doesn't consume energy so only the resistor consumes energy it loses energy as heat it doesn't give it back but capacitors and inductors they store energy temporary so they dissipate no energy now let's work on some problems number one the peak voltage of an AC signal is 170 volts and the RMS current is five point four amps what is the RMS voltage what is the peak current to calculate the RMS voltage it's simply equal to the peak voltage divided by the square root to the peak voltage is greater than the RMS voltage and that's always gonna be the case so it's 170 volts divided by root two so this is equal to 120 point 2 volts now Part B what is the peak current the RMS current is equal to the peak current divided by the square root of 2 so if we rearrange the equation by multiplying both sides by root 2 the peak current is equal to the RMS current times root 2 so it's going to be the square root of 2 times 5 point 4 so this is going to be about 7.6 for M so the peak current is always greater than the RMS current and the RMS voltage or the RMS current is always less than the peak value number two at what frequency will a 250 milli Henry inductor have an inductive reactance of seven hundred ohms so feel free to try this from the inductive reactance is equal to 2 pi times frequency times the inductance the inductive reactance has DNA ohms so it's 700 ohms we're looking for F the frequency and L is the inductance of the inductor which is measured in henries or in this case in the Henry's so 250 milli henries is basically 0.25 Henry's you got a divided by a thousand the current milli henries to Henry's so the frequency is going to be 700 divided by 2 pi times 0.25 so it's going to be about four hundred forty five point six Hertz this frequency will produce an inductive reactance of seven hundred ohms number three a sixty Hertz AC signal at 120 volts is connected across a five hundred milli Henry inductor calculate the inductive reactance so let's begin by drawing a picture so here's the AC signal and here is the inductor so it's the voltage is 120 and the frequency is 60 Hertz and the inductance is 500 milli henries so to find the inductive reactance let's use this equation it's 2 pi times F times L so the frequency is 60 and down 500 milli henries that's the same as 0.5 Henry's if you divided by thousand so 60 times 0.5 is 30 times 2 pi the 60 pi which is one hundred eighty eight point five homes so that's the inductive reactance once we have that we can find the current in the circuit so V is equal to I times XM which means that the current is equal to the voltage divided by the inductive reactance the voltage is 120 and the inductive reactance is 188 point five so the current in the circuit is point six three six six amps now what about Part C how much power is dissipated in the inductor the inductor doesn't dissipate anything so it's zero watch it simply stores energy temporarily in its magnetic field and then it releases it back and it's constantly doing it it's constantly absorbing energy and releasing energy so it doesn't consume energy number four a 1.5 kilo Hertz AC signal at 30 volts is connected across a 100 micro farad capacitor calculate the capacitive reactance so here is the AC signal and it's directly across a capacitor so the signal is at 30 volts and 1.5 killers and the capacitance is 100 microfarads so let's begin by calculating the capacitive reactance which is XC and is equal to 1 divided by 2 pi FC so it's 1 over 2 pi times the frequency 1.5 kilohertz is 1,500 Hertz you got to multiply it by a thousand to convert it to hertz and 100 microfarads is basically 100 times 10 to the minus 6 ference so any time you hear the word kilo it's 10 to the 3 mega 10 to the 6 milli 10 to minus 3 micro add 10 to negative 6 so if we multiply 2 pi times 1,500 times 100 times 10 to the minus 6 you should get point 9 4 to 5 and 1 divided by that number is 1 point 0 6 1 so this is the capacitive reactance the circuit now what is the current in a circuit so V is equal to I times X gene which moves I is V / XE so it's 30 volts divided by 1 point is 0 6 1 own so this yields the current of 2823 amps so that's being answer to Part B now Part C how much power is dissipated by the capacitor like the inductor the capacitor doesn't dissipate anything so it's zero watch it stores energy and it's electric field and then temporarily releases it back so this is the answer Part C number 5 a 500 Hertz AC signal at 60 volts draws the current of 2.5 amps when connected across an inductor what is the inductive reactance so let's make a list of what we have we know the frequency is 500 Hertz and we have the voltage the voltage is 60 volts and we also know the current which is 2.5 and so what we're looking for is the inductive reactance and also the inductance well we know that V is equal to I times XL so XL is the voltage divided by the current 60 volts divided by 2.5 amps so 16 divided by 2 point 5 that gives us an inductance or an inductive reactance rather of 24 ohms so that's the answer to Part A Part B what is the inductance so once we have X zone we can use this equation to find out so 24 is equal to 2 pi times the frequency of 500 find them 2 pi times 500 is a thousand pi so L is going to be 24 divided by a thousand PI and so L is point zero zero seven six for Henry's to convert that to milli henries multiplied by thousand so it's about seven point six for milli henries so that's the inductance now there are some other circuits that you need to be familiar with let's talk about three different types one circuit contains a resistor and inductor and also a capacitor so this circuit is known as an RLC circuit because it has all three elements the other two looks like this so on the Left we have a resistor and a capacitor so that's the known as an RC circuit on the right we have a resistor and an inductor so it's known as an RL circuit for each of these circuits z which is the impedance is equal to the square root of R squared plus XL minus XC squared so for an RLC circuit that's the equation that you're going to have for an RL circuit there is no XC XC is zero so it's just going to be R squared + XL squared + 4 and our C circuit there is no XL Z is going to be R squared plus XC squared now let's go back to the RLC circuit the voltage across the resistance is known as VR and this is equal to the current multiplied by the resistance the voltage across the inductor this is known as V out and it's equal to the current times the inductive reactance and the voltage across the capacitor VC that's going to equal to the current in the circuit times the capacitive reactance now if you want to find the current that flows in the circuit the current is going to be the voltage divided by the impedance Z R is the resistance in ohms XL is the inductive reactance which is also in ohms XC the capacitive reactance is announced z connects our XL and XC together as we've seen in this equation so Z is a type of it has the same unit of resistance so just as V equals ir z is equal to i z so that's how you can find the total current in the circuit just take the voltage divided by the impedance now there are some other equations that you need to know so if you want to find that phase angle its XL minus XC over R well tangent of the phase angle and the phase angle is going to be the arc tan of the inductive reactance minus the capacitive reactive over the resistance you can also use this equation cosine of the phase angle is equal to the resistance divided by the impedance cosine of the phase angle is also known as the power factor now to calculate the power dissipated by the circuit is equal to I squared times R keep in mind only the resistor dissipate energy the inductor and capacitor they both temporarily store it and then release it so the power that is absorbed by the circuit is I squared R you can also use this equation it's equal to the the average power is the RMS current and the voltage RMS value times the power factor so you can also calculate the average power using that formula and that's basically it these are the main formulas that you need to know for the RL circuit the RC circuit and the RLC circuit now whenever you have a circuit with a capacitor and an inductor be it an LC circuit or an RLC circuit you can calculate the resonant frequency which is 1 over 2 pi square root LC at the resonant frequency the capacitive reactance and the inductive reactance are equal to each other so in that case the impedance which is a square root of R squared plus XL minus XC squared since XL minus XC ated is same these two will cancel XL minus XC is zero so Z is equal to R so at the resonant frequency make sure you know that capacitive reactance equal inductive reactance X C equals XL and C the impedance is just a resistor the resistance of the resistor number six a 800 Hertz AC signal at 100 volts is connected across an RL circuit consisting of a 300 ohm resistor and an 80 mil ahem the inductor calculate the inductive reactance so let's draw the circuit first so we have a resistor and an inductor so the frequency of the signal is 800 Hertz and the voltage is 100 volts now we have a 300 ohm resistor and an 80 million REE inductor so to find the inductive reactance we can use this equation X L is equal to 2 pi SL so that's 2 pi times the frequency of 800 times L 80 million resor 80 times 10 to the minus 3 if you see Millie you could just add 10 to negative 3 to it that's a quick way to convert so X out is equal to 402 bones or for 2.1 ohms now let's calculate the impedance the impedance is going to be the square root of R squared plus XL minus XC squared since there's no capacitor in the circuit XC is zero so are the resistor has a value of 300 ohms and XL is 402 point one so the impedance of the circuit is 500 and 1.7 ohms now let's move on to Part C what is the current in the circuit to find the current it's simply equal to the voltage divided by the impedance the voltage is 100 volts the impedance is 500 and one point seven ohms so this is equal to point one nine nine amp so that's the current in the circuit now what about Part D what is the phase angle to calculate the phase angle we can use this equation R 10 XL minus XC over R so in this case XL we can see that it's 400 and 2.1 XC is 0 and r that's 300 ohms so arctan 402 point 1 divided by 300 make sure your calculator is in degree mode by the way this will give you a phase angle of fifty three point three degrees now there's another way in which we can calculate the phase angle we can use this formula as well it's equal to the arc cosine of R divided by Z so the resistance is 300 and the impedance that was 500 1.7 I believe that was the number that I used before so if you type in arc cosine 300 divided by 5 or 1.7 this will give you the same answer sixty three point three degrees so that's the phase angle now what about Part II calculate the power dissipated by the circuit a simple way to do that is to use this equation it's I squared times R so we know that the current is point one nine nine amps then the resistance is three hundred ohms so point one nine nine squared times three hundred that's about eleven point nine watch so that's the power dissipated by the circuit now in addition to using this equation you can use this equation it's equal to the RMS voltage times the RMS current multiplied by the power factor cosine of the phase angle so the voltage is 100 the current in this circuit is 0.199 times cosine of fifty three point three and this will give you the same answer of 11.9 watch and so that's it for part ii number 7 a 2 point 5 kilohertz AC signal at 60 volts it's connected across a RC circuit consisting of a 50 ohm resistor and a 1.5 micro farad capacitor calculate the capacitive reactance so we have a resistor and a capacitor and we have a 2.5 kilohertz signal that's the frequency and the voltage is 60 volts now we also have a 50 ohm resistor and a 1.5 micro farad capacitor so let's start with Part A the capacitive reactance we know it's 1/2 PI FC so that's 1/2 pi times the frequency 2.5 kilohertz that's basically two point five times ten to the three and then the capacitance C that's one point five micro farad's or 1.5 times ten to the negative six of ferrets so type it in your calculated exactly the way you see it and you should get forty-two point four ohms so that's the capacitive reactance now let's calculate the impedance so the equation is R squared + XL minus XC squared in this problem XL is 0 there is no inductor so it's just going to be the square root of 50 squared + 0 - 42.4 squared but once you square as a negative sign will disappear so the impedance is about sixty five point six ohms now once we have the impedance we can calculate the current the current is simply going to be the voltage divided by the impedance so it's sixty volts divided by sixty five point six ohms which is about point nine one five amps so that's the current in the circuit now to calculate the phase angle it's going to be arctan x l- x c XL 0 XC is forty-two point four divided by r which is fifty so it's our ten negative forty two point four divided by fifty so that gives us some angle of negative forty point two degrees now let's also calculated the other way let's use this equation R cosine R / z R is 50 ohms Z is sixty five point six and this will give you a positive angle which is about forty point three so which one is correct is the answer negative forty or positive forty it turns out that the phase angle is negative for any RC circuit for a circuit that's more capacitive than inductive the phasing will be negative for an RL circuit the phase angle should be positive now let's focus on Part II let's determine the power dissipated by the surgeon and let's use two equations to get the answer so first the power is going to be I squared times R we have the current its point nine one five and we know the resistance is fifty so point nine one five squared times fifty is about forty one point nine watch now let's use the other answer I mean the other equation to get the same answer so let's multiply it by the RMS voltage times the RMS current times the power factor the voltage is sixty the current is point nine one five and cosine of negative forty point two cosine negative forty point 2 and cosine forty point to dizzy you this will give you the same answer of 41.9 watch so you can use both equations to calculate the power dissipated by the circuit number eight a 1200 AC signal at 50 volts is connected across a RLC circuit consisting of a hundred ohm resistor 8.5 micro farad capacitor and a 12 milli Henry inductor calculate the inductive reactance so if you wish to draw a circuit we have a resistor we have an inductor and a capacitor so that's the circuit that we have in this problem so the inductive reactance XL that's going to be a 2 pi FL so that's 2 pi times the frequency which is 1200 times the inductance 12 milli henries so that's 12 times 10 to the minus 3 so that's going to be about 90 point 5 ohms so I'm going to put the inductive reactance right next to the inductor now let's calculate the capacitive reactance XC this is equal to 1 divided by 2 pi FC so it's 1 divided by 2 pi times the frequency of 1,200 times the capacitance which is 0.5 times 10 to the minus 6 and so that's going to be about two hundred sixty five point three ohms so now we can find the impedance Z is going to be the square root of R squared plus XL minus XC squared so let's plug in it r is 100 XL is 90 point 5 XC is 265 point three so first let's subtract ninety point five minus two sixty five point three that's negative one seventy four point nine one C squared is going to get a positive number and then add 100 squares to it so you should have forty thousand five hundred fifty-five and then take the square root of that result so the impedance of the circuit which I'm going to write it here it's a two hundred and 124 ohms so now let's just make some space let's get rid of this stuff we don't need any more Part D what is the current the current is the voltage divided by the impedance the voltage of the circuit is fifty volts so it's going to be fifty volts divided by two hundred and one point four ohms so that's point two four eight and now that we have the current let's calculate the phase angle so the phase angle is our tan XL which will know it to be a ninety point five minus XC which is two sixty five point three divided by r and r is 100 ohms which is given to us in a problem so notice that XC is larger than XL so the circuit is capacitive which means that the phase angle is going to be negative so our ten ninety point five minus two sixty five point three that's a negative 174 point so it's art can negative 174 20 divided by 100 and you should get an angle of negative sixty point two degrees so that's the answer to Part II now the last part the power dissipated by the circuit is simply equal to I squared times R we know the current is point two four eight amps since we have it here and a resistance is 100 ohms so point two four eight squared times 100 that's going to be about six point one five watts now just to confirm our answer just to make sure that it's correct let's calculate the power using the other equation if the two answers match that means that we've done everything correctly if they don't match we messed up somewhere so the RMS voltage is 50 volts the RMS current 0.24 eight amps and then it's going to be cosine of the phase angle which is negative sixty point two so fifty times one two four eight times cosine negative sixty point two that's equal to the same thing six point one six now the slight difference is due to round in intermediate answers so if you plug in exactly answers these two should be exactly the same but they're very close enough so that's it for this problem now let's go ahead and calculate the voltages across every element that's one thing I almost forgot to do in this problem now 50 divided by 2 or 1.4 we rounded to point 2 4 8 but let's use point 243 to get a more accurate answer so first let's find the voltage across the resistor then we'll find the voltage across the inductor VL and also the voltage across the capacitor the voltage across the resistor is simply I times R the current is point 2 4 8 3 and the resistance is 100 ohms so that's going to be twenty four point eight three volts so that's the voltage across the resistor next let's find the voltage across the inductor it's going to be I times XL so the same kind of point two four eight three times the inductive reactance of 90 point five so this will give us a voltage of twenty two point four seven one volts so that's the voltage across the inductor now the voltage across the capacitor the magnitude is going to be the current times Exene so the current is point two four eight three and the capacitive reactance is two hundred sixty five point three here it is so that's going to be sixty five point eight seven volts in a DC circuit with resistors the voltage drop across each resistor typically would add up to the voltage of the source but in an RLC circuit it's different if we add up these three voltages they do not add up to 50 notice that the voltage across the capacitor is greater than the source voltage of 50 so these voltages are related by a different equation vs the voltage goes to 50 does not equal the RF plus VL plus VC that doesn't happen in the circuit so rather they are related to each other by this equation VF squared is equal to CR squared plus VL minus v/c squared the same way as Z squared is equal to R squared plus X L minus XC squared now let's go ahead and check to make sure that all of our answers are correct so VF is 50 squared VR is twenty four point eight three squared VL that's twenty two point four seven one minus VC which is sixty five point eight seven and then twist so first let's get rid of some stuff 50 squared is 2,500 24.8 three squares that's about six hundred and sixteen twenty two point four seven one minus sixty five point eighty seven that's negative forty three point three and ninety nine now let's go ahead and square that number so this is going to be eighteen a three point five and if we add it to six hundred sixteen point five you see that twenty five hundred equals twenty-five hundred so the voltages across the resistor the inductor and the capacitor is related to each other by this equation number 9 a LC circuit consists of a 50 milli Henry inductor and a 100 micro farad capacitor what is the resonant frequency let's draw a circuit so here's our AC source and there is an inductor L and the capacitor C so that's a typical inductor capacitor LC circuit now the equation to calculate the resonant frequency is 1 over 2 pi square root LC so all we need to do to find it the answer Part A is just plug in the data that we have so we know the inductance 50 milli henries is 50 times 10 to the minus 3 Henry's and then the capacitance is 100 times 10 to the negative 6 ferrets so I'm going to do is type this in but let's take it one step at a time let's multiply 50 times 10 so minus 3 by a hundred times 10 to the minus 6 that's five times ten to the negative six if you take the square root of that and then multiply by two pi you should get a point zero one four and zero five and if you raise it to the negative one power this will give you the resonant frequency which is about seventy one point two Hertz now keep in mind at the resonant frequency X L is equal to XC and the inductive reactance is equal to the capacitive reactance and in fact we use in this expression you can derive this equation let's go ahead and do that so X L is equal to XC X L is 2 pi F L XC is 1 over 2 pi FC now we're going to do is let's multiply both sides by F so on the right side these two a plant them on the left side F times F is f squared so we're going to have 2 pi L times s squared and that's equal to 1 over 2 pi C now let's multiply both sides by 1 over 2 pi L on the left side 2 pi L will cancel so we're going to have F squared is equal to 1 over 2 pi times 2 pi is 4 pi squared and then times L C so now let's take the square root of both sides the square root of s squared is f the square root of 4 is 2 the square root of pi squared is PI and the square root of LC is just square root LC and so that's how you can derive this equation now let's work on Part B what capacitance is required to produce a LC circuit with a resonant frequency of 1,500 Hertz if lb inductor is 15 million Rees so let's start with this equation and our goal is to solve for C so let's begin by squaring both sides if we take the square root of both sides we see that the resonant frequency the square the residence frequency is equal to 1 over 4 pi squared times LC which we have this equation a not too long ago so now we're going to do is multiply both sides by C and so on the right side these will cancel and so we have C times F squared is 1 over 4 pi squared times L now let's multiply both sides of the equation by 1 over F squared so these two will cancel and on the right which is going to multiply this by F squared so I'm not going to have F squared on the left side anymore so C is equal to 1 over 4 PI squared times F squared times L so that equation will help us to calculate the capacitance and in LC circuit at residence so the frequency is 1,500 Hertz n L is 15 times 10 to the minus 3 so 15 times 10 to minus 3 times 1,500 times 4 PI squared and that's equal to 888 point to 6 so 1 divided by that result will give us a capacitance of 1 point 1 to 6 times 10 to minus 3 farik's 10 to the minus 3 is the same as milli so this is about one point 1 3 milli forage or if you want to in micro farad's multiplied by thousand that's 1130 micro farad's so that's the capacitance required to produce an LC circuit with this resonant frequency number 10 a 12 volt AC signal is connected across a RLC circuit consisting of a 50 ohm resistor a 7.04 nano farad capacitor and a 900 milli Henry inductor determine the resonant frequency of the circuit so let's draw a picture so we have a resistor a capacitor and an inductor so the voltage is 12 volts and the resistance is 50 ohms let's calculate the resonant frequency to begin it's going to be 1 divided by 2 pi times the square root of LC in this problem L is 900 milli henries or 900 times 10 to the minus 3 and the capacitance is seven point zero 4 nano farad where seven point zero four times ten to the minus nine ferrets so you should get a resonant frequency of if we round it 2,000 Hertz or simply to kill Urtz so let's put that here now is focused on Part B what is the inductive and capacitive reactance at this frequency so let's start with the inductive reactance which is 2 pi s times L so that's 2 pi times a frequency of 2,000 Hertz times the inductance which is about 0.9 Henry's or 900 times 10 to minus 3 so XL is about if we round it 11 thousand three hundred and ten ohms so let's put that here we're going to say it's eleven point three kilo ohms now let's calculate XC XC is one over two pi FC sets in your calculator you can type like this one divided by parentheses two pi times the frequency of 2000 Hertz times the capacitance of seven point zero four times ten to the minus nine so you're going to get also eleven thousand three hundred and four I'm going to round that to eleven point three kilo ohms so as you can see these two are very close if the circuit is perfect at the resonant frequency the capacitive reactance should equal the inductive reactance and on this example we have some rounding errors but is close enough we can work with that now what about Part C what is the impedance and the current in the circuit as a resonant frequency the impedance is equal to R so it's going to be 50 ohms and is how you know we know that Z is equal to the square root of R squared plus X L minus XC squared so this is going to be the square root of 50 squared X L is 11300 if we round it XC is also 11300 because it's same XL minus XC is zero so the square root of 50 squared plus zero which is the square root of 50 squared that's simply equal to 50 so under perfect conditions at the resonant frequency the impedance is simply equal to R so now that we have the impedance we could find the current the current is going to be the voltage divided by the impedance so we have 12 volts and the impedance is 50 12 divided by 50 will give us a current of 0.2 for imps now let's calculate the phase angle so this is going to be arc tan or inverse Chan XL which is eleven thousand three hundred minus XC divided by R so basically this is inverse tan of zero arc tan of zero is zero so that's a resonant frequency the phase angle is zero now what is the voltage across each element so we have a current of point two four amps let's use that information to find the voltage across the resistor the capacitor and the inductor the voltage across the resistor is going to be I times R so the current of 0.24 amps times the resistance of 50 ohms and that's equal to 12 volts so that's the voltage across the resistor notice that it's equal to the voltage of the source because the other two voltages they're the same and they're going to cancel out now to find the voltage across the capacitor it's going to be I times XC so that's the crown of 0.24 amps times eleven thousand three hundred ohms and this will give you a voltage of twenty seven twelve now the L which is I times XL it's going to give you the same answer you have a current of point two four and capacitive I mean an abductive reactions of eleven thousand three hundred so you get twenty seven twelve volts notice that the voltage across the capacitor and the inductor is very very high as you can see it's miss point Li higher than the voltage of the source the voltage of the source is only twelve volts the voltage across the capacitor and the inductor is two thousand seven hundred volts it's much much higher in fact more than one hundred times higher so very high voltages can be produced in an RLC circuit so just keep that in mind

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Keywords: ac circuits, ac circuits basics, ac circuits problems, impedance explained, impedance circuits, physics, inductive reactance, capacitive reactance, power factor, resonant frequency, RL circuit, RC circuits, ac, RLC circuit, RLC circuits explained, LC Circuit Tutorial, LC Circuit Resonance, LC Circuit, RL Circuits Explained, RL Circuits, Impedance vs Resistance, Impedance, AC Circuits Explained, AC Circuits physics, Tutorial, problems, explained, examples

Id: C8o2UpqzuKI

Length: 60min 12sec (3612 seconds)

Published: Tue Feb 28 2017