24-Step Response of Parallel RLC Circuits

okay we're going to do step response for parallel RLC circuits and we're going to set it up this way we have Sam current source here and we'll put a switch right after it and this switch is closed but is open at t equals zero and then in parallel with that we have capacitor inductor and resistor like that and they all are going to have the same voltage across them resistor inductor capacitor and let's go ahead and identify these currents that's I sub C I sub L I sub R okay now the switch is originally closed there and the current is flowing but because the switch is closed right next to the to this current source here all of the current is going to be passing through that let's call this current capital I so fixed current does it change over time so there's essentially a short circuit across the current source so there's going to be initially zero current through each of these elements here but at T equals zero the switch is opened no longer short circuit and now currents can begin to flow okay so let's do KCl for the top node I sub C plus I sub L plus I sub R has to equal I okay so that's the top essential node up there that's just KCl for the top essential node okay now let's go ahead and simplify or or substitute in for each one of these for example I sub C that is C DV DT okay now for I sub L I'm just going to leave that as I sub L for the moment and then for I sub r that's V over R and all that is equal to AI okay now I want to pause just for a minute and look at something with just this equation right here I want you to notice that I could write I sub L is equal to big I whatever that is that's the the current source here - V over R minus C DV DT now that's just a relationship that should hold for any time whatsoever so keep that in mind right there because we may return to this later that's that's just a relationship you you need to keep in the back of your head at this stage though here's what I want to do I want to write this a little bit differently than I have before I wanna write it in terms of I specifically I sub L now you know that V equals L di DT that holds for the inductor everything has the same V so this I is specifically I sub L now if I take a derivative of that that's DV DT that's going to be L times the second derivative of I spell DT squared okay so let's substitute these two things in for up here and when we do that we're going to end up with a C L d squared I DT squared plus I sub L plus now this is V over R so we're going to have 1 over R times L times di sub L DT and that's equal to VI now I'm going to rewrite the order here d squared I sub L DT squared just rearrange the terms plus 1 over R C di sub L DT plus 1 over L C I sub L and that's equal to I over LC like that ok now here this is a non-homogeneous second order differential equation a little bit more difficult to solve but I want you to notice what we did here is we wrote this in terms of I sub L so keep in mind it's sort of like AI sub L keeps popping up here and it is it's an important term so the secret to solving for this type of circuit here is to keep in mind the importance of I sub L now I wrote this in terms of a current I could have gone back here and done the same sort of thing that we did before where we write it in terms of the voltage so I'm going to stop here for the moment and I want to go back and the reason I want to go back is because I want to show you something here let's start once again with I sub C plus I sub L plus I sub R equal to big I okay now make the same substitutions that we did before this is C DV DT plus I sub L plus V over R and that has to equal big I okay now I sub L can be written in terms of voltage so DV DT plus 1 over L times the integral of V from 0 to T DT plus I at 0 plus V over R equals I ok now you have seen this type of equation before at least the left-hand portion of this equation you've seen before previously when you saw it though it was equal to 0 and what did we do you recall what we did was we took this because it's got a derivative and integral both in its time as we took a another derivative of this so we took a second derivative with respect to time and when we do that we end up with C d squared V DT squared plus this is just V now plus 1 over R DV DT and now when we take the derivative of the right hand side that turns out to be 0 ok so I'm going to rearrange the terms here d squared V DT squared plus 1 over R C DV DT plus 1 over LC v equals 0 now if you recognize that equation you should just be tickled pink at this point and the reason is because you know the solutions to that one right there we could go through the process of solving it but we've already done it this ordinary differential equation right here has three possible solutions and you know the forms the general forms of those solutions are going to be V equals a 1 e to the s 1 T plus a 2 e to the s 2t or V equals B 1 e to the negative alpha T cosine Omega sub D T plus B 2 e to the negative alpha T sine Omega sub D T or v equals d 1 t e to the negative alpha t plus d 2 e to the negative alpha t one of those three solutions is the answer to this second-order differential equations here okay but you know we still have to go through the process of evaluating the coefficients to figure out what's going on here and to get a complete solution nevertheless we've already been through this you know you in other words this is the voltage across all three of these components and the current source as well so again across all of those components there they have the same voltage so you solve for the voltage across all of those exactly the same way that we've done before so you know how to do that then then once you've got the voltage you can get I sub L because I sub L is just going to be big-eyed minus V over RC DV DT so what is that ice Appel equals big I - the over R minus C DV DT okay so yep I did write it correctly so the key here is to figuring out voltage so that you can then get I sub L okay and once you've got V you can also get this right here what is that well that's the current through the capacitor what is this term right here well that's the current through the resistor and what is this term right here well that's just the source and by the way after a long period of time that's going to be the current passing through the inductor remember an inductor after a long period of time behaves like a piece of wire so when the switch is open all the current begins flowing into the right-hand portion of the circuit here let's see you could have some current going that way some current going this way some current going that way but after a long period of time after there's no more change in current so you don't have a DI DT through the inductor then all of the currents going to pass through there and this is going to be a piece of wire in and it's another short circuit across the inductor so there is no more current going here there is no more current going here all of the currents going here and at that point the voltage should be zero so after a long time the voltage is going to be zero and all of the current big eye is going to go through the inductor okay well what about right at T equals zero well right at T equals zero imagine that switch was just open and we're talking about the fraction of a second right after T equals zero the inductor is going to create a back EMF and not allow any of that current to go through here remember you cannot have an instantaneous change in current for the inductor so it's going to be zero for that fraction of a second now some of the current could go down here but if it does it's going to meet some resistance however remember right at T equals zero plus this capacitor is uncharged so the current can go through the capacitor it'll be a displacement current but it'll appear as if it's going completely through the capacitor and it's going to meet zero resistance at least for that first fraction of a second and then as it builds up a charge here then it'll be more and more difficult to get that current to flow through there so you may start seeing more current over here as well as over here but after a very long time and you know after five tau of time whatever tau happens to be for this case all of the currents going to go here and any charge potential difference that existed on this capacitor is going to bleed off and also go through here until ultimately all of bigeye is going through the inductor and the voltage is zero so right when that thing is open what we expect the voltage to also be zero so at T equals zero the voltage is going to be zero as well okay so you see what there's there's a lot of things going on here nevertheless that I sub L is sort of an important piece to the puzzle because you can see how it's going to end up being sort of well it's just an important piece of the puzzle here okay now you know you need to find V and after V you can get that but there's more to this that helps you and I'll show you this and then we'll start working a problem or two let's say that we have one of these cases here let's start with this case right here that's the over damped case okay so we're gonna start with the over damped and I want to show you something i sub l equals big I minus V over R minus C DV DT okay well V for our over damped case is going to look like this a 1 e to the S 1 T plus a 2 e to the S 2 T and then DV DT is going to be take the derivative of that a 1 s 1 e to the S 1 T plus a 2 s 2 e to the s 2 T like that ok let's take these two values and plug them in over here for ice belt ok well first of all that the guy is going to be the same - now this is going to be V over R so that's going to be a 1 over r e to the S 1 t minus a 2 over r e to the S 2 T and then minus C so C a 1 s 1 e to the S 1 t minus C a2 s 1 I'm sorry s to e to the S 2 T like that ok now I want you to notice what we've got here these things here are essentially const so we've got these constant coefficients in and you can include the the negative sign in there if you want to but here's my point I could rewrite this just taking all of those constants and calling them some other constant I plus some a one prime e to the S one T plus a 2 prime e to the S 2 T so I've got this s 2 here and this s 2 here this e to the S 2 T both of those cases add those two together it's just going to be a bunch of constants added together and I'm just calling that a to the same with the e to the S 1 T I'm doing the same thing I'm just calling it a 1 so I've created what looks like the natural response now if I were to go through this same sort of exercise for the under damped and say I sub L is equal to I minus V over R minus c dv/dt and use V equals V 1 e to the negative alpha T cosine Omega sub DT plus B 2 e to the minus alpha sine Omega sub DT and then take the derivative of that and plug that in as well as this so that we could go through the same general process here's what I would discover or what you would discover that I is equal to I'm sorry I I sub L is equal to big I plus some coefficient which we'll call B 1 prime e to the negative alpha T cosine Omega sub D T plus some B 2 prime e to the negative alpha T times the sine of Omega sub D T and if I were to do the same sort of thing with the critically damped case and go I sub l equals being I minus V over R minus C DV DT and use V equals D 1t e to the negative alpha T plus D to e to the negative alpha T and then take the derivative of that as well and then plug those two back in over here guess what I'm going to end up with I'm going to end up with an equation that looks something like this big I plus some D 1 prime T e to the negative alpha T plus some D 2 prime e to the negative alpha T so you do you see what's going on here for each one of these cases I have a general solution that looks like this ice Appel is going to be whatever big eye is and that's the final current that's going to be passing through it excuse me that's the final current that's going to be passing through the inductor so I'm going to put a subscript F on there plus each one of these cases is the natural response so the natural response that is a much more general solution for what the current through the inductor is going to look like it's going to be the final current plus whatever the natural response has to be and you judge what the natural response is exactly the same way we did for you know the previous problems that we did nevertheless this is the step response now I will say this and you could go through this if you wanted to the same sort of thing holds for the voltage the voltage is going to be the final plus the natural response like that so in our case for our specific example that we have here for the specific example V final is going to be 0 so that's the reason that we didn't write a V final here because it's 0 so that's the natural response 4 over natural response 4 under the natural response we're critical but regardless of which of those three cases you have V final is going to be 0 for this specific case now you there may be some circuit that you encounter in the future perhaps that that's not the case that V final is not going to always be that way now that holds for I sub L but if you were to go through the same exercise to figure out what I sub C is or to figure out what I sub R is remember I sub R is just going to be V over R and V is given by this expression here so an R is a constant so this boils down to the same sort of thing that I sub R is going to be I final plus whatever its natural responses so that's the step response for the current you could do the same sort of thing for the capacitor it's equal to C DV DT now the derivative here you know C is a constant but the derivative of V doesn't change the form of of these it does change the the coefficients but it doesn't truly change the form it in other words when you take the derivative of this I'm sorry we take the derivative of this you're going to end up with an e to the s 1 T multiplied by some constant plus some e to the s 2 T multiplied by some other constant and the same down here if you take the derivative of these terms you're going to end up of the same sort of thing the coefficients are going to be different though and if you do the same thing here you're going to end up with the same sort of thing and that's because of the nature of these the way different derivatives work for these things for these functions okay so here we can make a general across-the-board statement here for all the voltages and all of the currents for this step response then I regardless of which I you're talking about is going to be equal to I final plus the natural response okay now V also is going to be equal to V final plus the natural response so regardless of what I you're talking about or what V you're talking about how it well in our case for the parallel there's only one V but regardless of what you're talking about these aren't going to hold for every step response okay now what's going to be different in the cases for our circuit is I final I final for I sub L is just I I final for I sub C is 0 I final for I sub R is also 0 so that's because all of the currents going to be going down through the inductor after a long time so you've got to have enough circuits knowledge to figure out what I final is and you got to have enough circuits now let's figure out what V final is as well okay but after that you solve for the natural response and you just tack that on and there's your answer for I ok now this will hopefully be a lot more clearer whenever we get into some problems here so let's see if I can come up with a problem here real fast

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Id: MuP9jbi2zUQ

Length: 24min 5sec (1445 seconds)

Published: Fri Mar 27 2020