okay now we're recording basic our LRC circuit source free circuit again notice there is no source here these two are connected in parallel so if I do KCl at the top it says the current entering that node should equal all the currents leaving that node well there is nothing entering that node we have I sub R leaving izybelle leaving so zero equals well the current through that resistor is the voltage divided by that resistor forms long current through the inductor current ah let's see what's the current there well if you go back to where's my cheat sheet from before v equals L di DT or I equals what one over L the interval of V DT I don't know where I put that sheet save the lifetime so I can refer to it I must have thrown away that's summary maybe it's in the book somewhere where you can check your note you'll see we have what one over L the integral of VDT from negative infinity to T in any time you have an integral what we're going to do is actually take the derivative of that the derivative of any constant is 0 this will be 1 over R DV DT plus 1 over L and we take the derivative of the integral they cancel each other out and if we multiply everything by our to get rid of the one over our in the front you end up with what DV DT plus R over L times V is equal to zero and the question how do we felt that this is a first order differential equation it's called differential equation because it has a derivative and it's first-order because what's only first degree first derivative so that's the first order differential equation and the question we need to solve it well you learn that when you take differential equation next semester most of you probably signed up for differential equation next semester right so that's where you can be learning in that how do you solve a differential equation that has first derivative in it notice there's no value here and the rule is really simple if you have I'll just make it for any differential equation if you have a differential equation that looks like this I call it alpha usually that's a function of time equals to zero the solution for that is always going to be some initial condition e to the power of alpha T I had a handout for that and I was looking forward to and I can find it I think negative alpha T we can research that one we can try to derive it but I will leave it it's negative alpha T and if this is not a constant here if this is the equation that looks like this is going to have a solution of k over alpha plus x sub 0 e to the minus alpha T so these have h of that the rules I'm going to use in solving differential equation first-order differential equation if I have a differential equation where this is 0 there's my answer K 0 is the initial condition some constant will get to that and if this is actually not a 0 its K over alpha plus some initial condition e to the negative alpha T if I if I compare this one to these two it matches the first equation so my solution for that the voltage across the inductor is going to be some initial condition we're called V 0 e to the power of negative R over L times T so if I know a bit about the circuit what was going on before when you had the source because if you never had a source of the circuit there is no initial voltage there so that's zero but apparently we had something attached to the circuit initially and that will be that value right there that will go right here we know what R and L we can plug it in and go from there let's see if I can take an example that has some numbers in it initially so let's take this circuit we have an inductor we put some numbers on it and then depta has 50 milli Henry that's the value of it going through a resistor of 200 ohms and there was something attached to that circuit before so this was attached to something else here there's a whole circuit was driving that but what happens at T equals to zero we opened that switch so once you open that switch this is no longer a factor this way you have left so before this was closed attached there and at T equals D or somebody came in disconnected that power source whatever that circuit here so that means something was driving that so let me give you some initial condition assume when this was closed before the switch opened the current through the inductor is 0 minus before we open that switch assumed that was at 2 amps normally we have to calculate that but in this case assume that was the case the question is find the current as a function of time find izybelle as a function of time the expression for it for T greater than zero and I also want to find the current through the inductor at T equals 200 micro second how much current was going through it two things I'm asking for so for T greater than 0 this will my circuit looks like that's it and if those current going through that there'll be the same printers called ISO Bell it's going to mark this voltage I mean this inductor plus demise record vo is going to mark this plus two - were called VR if I do KVL some of the voltages in a closed loop is zero what we have VL plus VR is equal to what 0 the voltage across inductor V equals L di DT plus VR which is what R times I I can plug in the numbers now in it and that will give me what what cell 50 milli Henry what's 50 is at 0.05 dril DT because that's the current through the inductor really as a function of time if you want to plus R which is what 200 times R sub L as a function of time is equal to zero let me divide both sides by 0.05 what is 200 divided by 0.05 200 divided by 0.05 is 4,000 right now that looks like the first equation I said if you have an equation that looks like this a derivative here plus a constant times that function equals to zero that's what I have then my solution for that I sub L as a function of time is going to be the current at 0 times e to the power of negative 4,000 T based on that equation now in this problem they stated that the current at 0 was what to M go that's 0 minus not 0 plus what you remember anything about inductor I sub L 0 minus equals what I sub L 0 plus the current does not change in 0 time so this is equal to 2 e to the negative 4 thousand T this is for T greater than 0 that is the current through the inductor at any given time and by the way if you just graph that to see what it looks like then we're going to ask what's the current to 200 microseconds if you have that's an exponential function it will look like this at T equals to 0 what's the value of the current to and what happens after that instead to decrease it doesn't take that long to go to zero notice we ask you what's the current at 200 microsecond well let's see e to the negative four thousand times 200 micro what's four thousand times 200 micro that's point eight it's a negative 0.8 so let's do that two times e to the negative point eight and this is what roughly 0.9 m in two hundred microseconds so we're talking about a short time actually it's not going to take that long for that after the actual time it takes to go to zero it's five over that number five tau we call that the time to go to zero here we call that roughly zero here this is approximately five tau and what's tile that's five times one over that number one over 4,000 tau is one over that number that's how long will it take for that current to go to zero so what's five over 4,000 1.25 million a millisecond so that's how long is going to take that inductor to discharge completely roughly 1.25 millisecond before is no longer a factor so we're not talking all is going to stay there for five seconds ten seconds a minute we talking about milliseconds microseconds that kept that inductor actually will work as a source staff through the value of two M's sad to decrease by the time we hit 1.25 milli second that inductor is almost empty oh the charge is gone let's try couple more examples will show you what the circuit like in this case we give you the initial condition as I said usually you're not given that initial condition you have to find it so here is the circuit I have a 40 ohm resistor here plus 2 minus here of 24 volts there is a switch that's attached right now that at T equals to zero that's which opens swings open there is 10 ohms and there is an inductor 5 Henry and let's say I want to find this country called a sub L what is that current and let's say I want to find this voltage we called V for T greater than zero that's what we have the one assumption I'm gonna make that this circuit been sitting like this for a while connected I walked in this morning I build that circuit is sitting upstairs been up there for at least five six minutes so you find well if it's been sitting there for a few minutes and since this is a DC source how does the inductor behave to DC source on that sheet I summarize everything I wish I have kept a copy of it maybe it's here then dr. behaves as what DC source I mean short circuit to DC source you see if I have that page and how I saved it I don't know whatever it everyone used the room so I'm sure it's gone so to DC the inductor behaves as a short that's Henry that's not resistor five Henry that look like a Henry now it looks like a so then inductors not behave as a short circuits for T less than 0 I have to look at that circuit this well my circuit will look like and this is 10 and this is 40 so I can tell you the voltage here for T less than 0 is going to be wet isn't 25 volts how the three of them connected in parallel well what's the current going down here what's R sub L well that's the same come through that resistor izybelle 4 zero - it's been sitting there for a while is going to be this voltage divided by that resistor 24 divided by 10 which is what - 0.4 m/s this this and that all connected in parallel they have the same voltage I'm only interested the question was the current through that so the 10 through that for T less than 0 2.4 what's the voltage here for T less than 0 that's for 24 volts done with it so now I have to look at the circuit for T greater than 0 so what's going to happen when T goes above 0 the switch is going to open once that switch is open this source is gone and that's what I have that's which opens this so it's not a factor now I'm not looking at T equals zero plus I'm looking in general for T greater than 0 at T equals zero plus that comes through the inductor doesn't change so at that instant that come through this still 2.4 then we'll start to shrink after that reduce then we go to 0 so I'm looking at it for any time for T greater than 0 so I can't use the 2.4 so only good for microsecond don't blink it's 2.5 and start to shrink so here's my current this is our sub L is going to mark this + 2 - it's going to mark this + 2 - it's going to mark this + 2 - so if I do KVL there sum of the voltages in a closed loop is zero let's begin with this one what's the voltage here Ohm's law says 40 times R sub L plus what's the voltage drop here 10 times R sub L + V equals all di DT in case you forgot that's the voltage across the inductor so the voltage there will be what 5 times di sub L DT all of that is equal to zero if I simplify I have five DRL D DT plus what 50 R sub l equals to zero divided by five plus ten I sub L again that fits that equation we mentioned early if you have a differential equation that looks like this and that's what we have your answer is going to be whatever the initial condition of that so a sub L as a function of time is going to be a sub L at 0 e to the power of negative 10 T what was my initial condition for the cont 2.4 two-point-four due to the power of negative 10 T for T greater than zero that is the current as a function of time notice that T equals to zero if you plug it in when T is zero what's eetu the zero one what are you good to point for we said the current that t minus and t plus are the same so zero minus zero plus your current is the same two point four so if you were monitoring what the current looks like again the comb through that inductor here was a two point four for T less than 0 as the two point four right here if you can see the four that's two point four then what happens after that it goes from two point four like this how long is going to take a tick for this to be zero well let's see it's five tau so that's five cows one over ten you have they'll take half a second before that current that inductor has no more charges left in it so half a second and after that the voltage or the current through that will be almost a zero again the point I'm trying to make it's not a long time it's a very short time you want to find the voltage now well if you remember the voltage was labeled in the problem where plus on the top - on the bottom right the way the current going is going to labeled what Plus - so what is V going to be the voltage across the resistor is going to be negative 40 times I sub L so that's negative 40 times 2.4 e to the power of negative 10 T I'll put that in a second because the voltage was labeled they wanted v+ on the top - on the bottom right and notice the way the current is going in this circuit the current traveling which way this was no market plus - - and they want the plus to be here the minus down so you can see these guys are backward that's where the minus comes in so it's negative 96 e to the power of negative 10 T sure if you want to find that vola be 10 times I sub L if they wanted this way of plus/minus so notice first we have to look at the circuit for T less than zero and calculate if you have an inductor you want to find the current through the inductor zero - if you have a capacitor you want to find the voltage across the capacitor as zero - not the current because for capacitor the voltage of the capacitor is zero - the same as zero plus so let me take one with the capacitor here there is my circuit I have a 2 ohm resistor on the left side we have a voltage source very similar to the previous one but nine volts here and is attached is connected at t equals to zero swings open I got a four here and I got a capacitor with the value of 10 micro farad and the question what is the voltage across the capacitor that's what I'm looking for find V sub T the voltage across the capacitor at any given time now again we have to look at the circuit for t less than zero and I need to find the voltage across the capacitor zero - because that's the same as the voltage across the capacitor zero plus so when you have a capacitor you don't want to find the current you want the voltage when you have an inductor you want to find the current through the inductor because that doesn't change so everyone assuming has been sitting there for a while how does the capacitor behave - a DC source yes that being open and what's that voltage equal to looking at that circuit what is it no zero that was a short-circuit nine volts while nine if this is ground what's this voltage here nine volts how much current go in that direction non zero so the voltage drop your arms law says V equals I times R four times zero which is what zero voltage drop here so that end is going to be the same as this nine volts so that's all I need what is the voltage across the capacitor is zero minus that's nine volts that's the whole purpose of looking at the circuit for T less than zero now let's look at it for T greater than zero for t greater than zero this which is going to open and my circuit will look like this now you can combine these two if you want these two resistors and they'll be what 610 microphone now again if I do KCl at the top this is I of are coming down this is I subsea coming down I define them I do KCl at the top it says some of the currents entering that mode equals some of the current leaving that node what is entering the node bless you zero what's leaving that node IR plus IC I is the voltage here which happens with the voltage across the capacitor divided by six and if you remember for capacitors i equals c DV c dt so there will be 10 micro DV c dt let's divide by 10 micro so DV C DT that's 1 over 6 divided by 10 micro 1/6 divided by 10 micro or is that II would you go here we go I got run 6 6 6 7 roughly VC equals to 0 16,000 667 or 1 over 6 micro if you want to if you don't like that number so if you don't like that number plus if you don't like that stupid number you can say run over 6 times 10 60 micro that's what that number is 1 over 60 micro this I look like a 60-year so if you go 1 divided by 60 micro 1/60 EE to the negative 6 notice the same number I know if you can see it hmm go find the spot 1 over 60 micro is the same number so if you don't like to use that number you can use this number so this is your alpha so what's the voltage across the capacitors that fits that phrase in there the voltage across the capacitor at any given time is the voltage across the capacitor is zero e to the power of negative this one times T so if you don't want to use one 16000 667 you can use that one and what's my voltage across the capacitor is in that mine e to the negative 1 over 60 micro times T that is the voltage across the capacitor at any given time what was my question for that one find the voltage yep that's it if they ask you for the current what would you do okay so what's the current through the capacitor well the current is what C which we know what C at in micro time is the derivative of that so if you want the turn through the capacitor what is C 10 micro let's see how good are you in calculus what's the derivative of this there's the 9 that's the derivative of each of the minus 1 over 60 micro T negative 1 over 60 micro each other with negative 1 over 60 micro T right and if you do the math the micro and the zero will cancel here w6 that's with negative 9 over 6 because then 10 cancels the 0 cancels the zero right the micro will cancel the micro the 10 cancels the 10 a zero here so nine times a negative 1 negative nine over six which was well 3 over 2 e to the power of negative 1 over 60 micro times T for T greater than zero that's the current through the capacitor at any given time so once you know what the voltage is you can find everything else you can find the current do they want to know what the voltage right here is four times I do they want the voltage here two times I you can multiply them through them get everything that you need let's try one more example maybe make a little bit bigger here and take advantage of everything that we learn here's my circuit 120 ohms 61 we have a switch here plus 2 minus 18 volts the switch was connected then swings open at T equals to 0 9 y arms here 1 milli Henry inductor here another 5000 ohm resistor here another inductor here of 2 milli Henry and finally one more inductor right there of what 3 milli Henry it's really not a big deal because you can combine a lot of these and the question for that is can you tell me what the current through this izybelle as a function of time that's what we're looking for find a Sibel correct we can combine just like resistors the product over the sum here but if I look at this for T less than 0 and again I'm assuming it's been sitting there for a while my source is right here that's 18 this is the nine this is web now short-circuit and doctors always a short-circuit there is 15 short-circuit short-circuit this is I so bail and there's what well there's hundred and two resistors combined together that's here at 180 and where they're connected with the source and the 90 yep nervous it's really right there attachment is right here at the same point right this all one node so that 180 is actually here if you want to you can put it up there big deal but that's the same thing as connecting right there so if you clean that circuit a little bit this is what you're going to have 18 these two resistors combined into one the product over the some 90 times 180 and I'm going to leave with that because I'm looking for that current over the some of them what's nineteen one a is that 270 that's a 60 and that's the 50 and I'm looking for that current R sub L so our sub L at 0 - which is the same as also bad as 0 plus notice this this and this are connected how parallel they have the same voltage the voltage is 18 the resistor is 50 so the Clarins 18 over 50 which is wet 0.36 M that is the current through izybelle 4 zero - you can move that resistor and put it right there you'll see they're in parallel through them are in parallel so I'll bet what I need out of it for T less than 0 now I gotta look at it for T greater than 0 this which opens which means that source is gone so if that source is gone there is my circuit for T greater than 0 I have the nanyue right here there is that 180 right here there's that inductor there is not done one millihenry there is the 50 and I can take these two inductors combine them together to make it with the product over the sum 2 times 3 6 6 divided by 5 plus I 1 point 2 product over the sum this is also Bell I can take these two resistors make them 1 which would be what 60 you can combine even more 60 and 50 which is with 110 the inductor will be with 2.2 million read by combine them does that change the current through this one the same current so I can combine them without effect in the now that's my circuit I can do a KVL here some of the voltage in a closed loop is zero so the currents going this way this way so V equals L di DT plus R times I equals to zero L is web 2.2 million di L DT plus R which is 110 I sub l equals to 0 what is 110 divided by 2 point 2 milli Henry 50,000 so now I know the current through the inductor izybelle as a function of time for t greater than zero is going to be I sub 0 e to the power of negative 50,000 T we know what I sub zero was wasn't 0.36 again if I was to graph that through that inductor for T less than zero was 0.36 so 40 lessons you're after uses set for a few seconds and let us sit there you build it last month the first two seconds I'm not looking at it after that it's set like this 0.36 you walk into the lab you disconnect to is going to discharge same things the charges like this yep so now this is 5 tau and that's 5 and tears with 1 over 50,000 what's that point 1 millisecond so that's how long that capacitor or inductor is going to behave as a source for just only 0.1 millisecond then after that the whole sort of gotten right that's 105 over 50 that's 1/10 and that's a milli 0.1 millisecond I think I'm right 5 divided by 50,000 yep point 1 milli second so that's how long again that that's as a source there again we can make the problems bigger add more stuff to them now instead of using switches a lot of times we walk away from this we start using unit step function so instead of a switch a lot number with the switch it looks like this if you have a source that looks like this let's say this is 5 volts and closed at T equals to 0 for T less than zero what's the value of that nothing it's not it's open right so this one kicks in when T equals to it zero you know what unit step function looks like the graph of it the graph of u sub T is called unit step function has a value of 0 for T less than 0 then it goes to 1 at T equals to 0 that's a unit step function called use of T now if you saw that math or nut so instead of using a switch is there we know it my source here is 5 u sub T that's the same as that one because for T less than 0 this is open it's not a no big deal for T less than 0 what's the value of that source 0 0 times 5 nothing what happens at T equals 0 this one closes and what do you have now 5 volts what happens when T equals to 0 u sub T has a value of 1 1 times 5 which is what 5 volts so instead of using switches because people don't want to draw the switches you'll see us actually use use of T to replace the switch and I'm going to do that next time next video on Thursday I'll stop here that's just what's coming up next instead of doing switches and all that we're going to look at use of tea
