I'm I'm Dr P Montgomery and in this video I'll be working through a problem that involves an RLC circuit uh here's basically a kind of parallel combination of of those elements and what we see here in this problem is that initially I have a switch that is open Okay so this inductor is initially disconnected I have a voltage source here that's of course providing some currents and charge to the rest of the circuit and at time T equals z I'm going to shut that I'm going to close that switch thereby connecting up this inductor I'll probably be charging that up to some degree and then basically evaluating uh kind of what's going to happen uh with the voltages and currents that we have in in the circuit as defined here uh as we go through the problem so if I going to make one addition here some current I so what we're going to do for the first part here is go through again all our initial conditions so again before I close that switch we assume we're going to assume that circuit is already at a steady state condition and so therefore evaluate what each of these uh voltage across the capacitor the uh inductor current the capacitor current and then again because this RLC circuit it's a second order circuit so we need to also know the derivative of the voltage at time 0 plus and this is what we'll go through here immediately okay so the first step here looking at the voltage across this capacitor time 0 plus well again we know that the voltage here cross the pass time 0 plus has to be the same as what it is at time 0 minus so those are equivalent quantities so again let's think about before I've done anything with that switch I have this capacitor in parallel with this 4 ohm resistor uh with another 4 ohm resistor here and is 24 volt Supply so I know that my capacitor at at steady state is acting as open circuit so the voltage across that um of course we just have to be equal to the voltage across this 4 ohm resistor uh which we could apply uh the method of the voltage divider in order to to determine that so the voltage divider would just tell us that uh this quantity four over the sum of these two 8 * 24 Vol would give us the voltage here we made a quantity of 12 volts okay that immediately tell us this first part that we have right here okay so now looking to the second part what is the IND uor current at time 0 plus well with the current through an inductor we know that at time 0 plus because we cannot have any instantaneous change in the inductor current this again has to be equal to whatever the current is at time0 minus well at time0 minus meaning before I've closed that switch I know that this inductor is more or less disconnected this is a disconnected part of the circuit so um we know that I cannot have any current flow through that inductor before I've close that switch so because we know that I have no inductor current here at time 0 minus and I know that that has to be equal to the inductor current at time 0 plus uh we know that this would just have to be uh zero n no current flow all right now thinking about the current flow through the my capacitor right here um let me just take a quick minute eras this part up here so I can have a little more room to work and then we'll go through uh that part well okay so for finding the capacitor capacitor current at time 0 plus now we can't make any definitive statement just by saying oh well it has to be equal to zero because I know a capacitor at steady state can't have any uh current through it and that's true but the current in the capacitor at time 0 minus okay which we know would definitely be zero this is not the same as the capacitor current at time 0 plus so even though we cannot have a instantaneous change in those voltage across the capacitor uh there can and there usually will be an instantaneous change in the current flow at time0 minus compared to 0 plus so in order to figure out what's going on at time 0 plus um we would have to evaluate the circuit with that switch in the Clos POS closed position and then maybe then we see what we have a variety of elements and I can maybe apply some KCl type of rule uh expression to figure out what the current would be through the capacitor right here um to make that a little bit easier though I'm going to do a source transformation of this voltage source and this resistor that would then allow me to combine these two resistors and probably make our lives a little bit easier so if I were to do that that would change my circuit just to look like this have a resistor parallel with my capacitor parallel with my inductor again this is after I've closed that switch so after T equals z so my resistance here this resistance is going to be the parallel combination of these two 4 ohm resistors here so I'm kind of skipping one step in that I'm going to do a source transformation and then I'm also going to combine these two resistors uh to the same resistance value and so thereby giving me a resistance of uh just 2 ohms and four and parallel with four would give me two you know that you just split that divide that in half in order to get the parall combination here two yeah just clarify this is four parallel with four ohms these two resistors are parallel this current Source then uh let's say this is I so is we can find by relating the voltage source here and the resistor that I'm transforming so I know that this would have to be given by 24 Vols divided by this 4 ohm resistor here so that's going to give me a current of 6 amps so that would be this current 6 amp all right so now uh we know this is current i c this is current I sub L now we can write a KCl expression that would help me to determine what this current IC is again at time t equal 0 plus meaning immediately after I poose that switch so if I write that KCl equ equation at this node here that would look like this so I have minus 6 because the current coming in I'm going to count as a negative current um the current then going down through my uh resistor here is going to be 12 Vol ided by the 2 ohms so where did this 12 volts come from well it came from the fact that I knew or that I know that the voltage across this capacitor at time 0 minus is equal to 0 plus and we already determined that that was 12 volts so if I have I know I know that I have 12 volts across this capacitor and because these are now in parallel I also know that this resistor has a voltage of 12 volts across it so this is just ohms law to tell me the current that I have uh down through this capacitor so that's that uh quantity there um then we have added to that whatever this i 0 plus is and then added onto that the final current here is uh I at 0 plus and again this is fly K tells me that all these have to sum to equal to zero okay so a quick thing here is that I already again determined that my inductor current at time 0 plus has to be equal to zero and more or less just get rid of that term there so then just rearranging here at Min - 6 plus this will just divide down to 6 amps right so then basically those are going to cancel out and so then I can definitively say that my my capacitor current at time 0 plus is going to be0 amp now in this case it just it did in fact work out that this happened to be zero amps and that is the same as the current before I Clos the switch but that would not a automatically be the case I mean if this was a different resistor value here then those the current from this uh Source would not uh counterbalance the current through that resistor and such uh so that would not universally be true it's just in this case with the numbers that worked out to be Essen right um so now to get to the next part of the problem looking at the derivative of the voltage at time 0 plus we take a quick minute to clean up there here and uh that we'll work on that part okay so now we need to if we want to know the derivative of the voltage of VC the voltage across the capacitor specifically at time 0 Plus in order to figure out this we could just apply the expression that we know know relating the um current through the capacitor I at 0 plus is going to be equal to whatever the value of C is time DVC DT again at time 0 plus so for this part of the problem it's basically just asking what is this component of this expression going to be as a function of what we already found to be the current here I see at 0 plus so because we already found that to be 0 amps um of course very easily we can just state that the derivative here of the voltage would also have to be equal to zero uh units here would be volts per seconds so now let's maybe think about what would be the final condition of the voltage so let's say VC as time goes to infinity or just uh more specifically saying we've reached five time constants by Pals the the final what would that look like well if we see that once this switch is closed here I do have this voltage source connected um as sort to the Circuit as it were um but we see that my inductor here is in parallel with my capacitor well what do we know about an inductor that is fully charged and is already in steady state well that would tell us that the um voltage across that inductor has to be equal to zero because it more or less ends up acting as a short circuit and so because this uh inductor is acting as a short meaning no voltage across it and this is in parallel with my capacitor that would thereby telling tell me that my final voltage across my capacitor would also have to be equal to uh zero volts again after we've already reached steady state in that part of the circuit um from here now to write the more General expression which we won't get into because this was actually an exam problem and we wouldn't be able to fully evaluate the uh final response um just on on an exam type of question but we could evaluate what uh our various frequencies are so-called damping frequency um and Omega KN radiant frequency as it were and so for these we can just apply the general formulas that we have for each of these uh 1 over 2 RC and 1 over the root of LC again these are specific for the fact that we have have a parallel combination of an RLC circuit and just plugging in what we have here again for R in this case that would be again the parallel combination of the two 4 ohm resistors as I showed in that previous part of the problem so that was 2 ohms so plugging it out for our resistance and the capacitance is going to be 1 over8 keep in mind units here for each of these frequencies is radians per second uh and then L and C of course those we can just pull directly from the circuit itself that would be 1 over two also radians per segs then if we wanted to know what is the damping condition specifically for this RLC circuit well there what we do is compare the square of the two frequencies that we've just determined and here we see that Alpha square is going to be less than Omega squ and thereby using that to tell us that this is a condition of an underdamped response um then from there using these frequencies we can plug it into our uh general form of the solution to find out what the actual voltage looks like as a function design so that's going to wrap up what I'm going to do for this problem hope you enjoy and uh look I'll look forward to seeing you on the next video