 The series RLC circuit we took to be this and I wrote the differential equation with the capacitor Vc as the desired variable and this Vs steps from 0 to VP but in general it could be anything okay. We are now looking at a constant input and a parallel RLC we could have this. I believe this is the example I took last week as well. Let me just check and again I will write the differential equation in terms of Vc as the variable okay. So we will have in the first case this is the differential equation we get okay. Mainly we are now interested in the left hand side of this. The right hand side when equated to 0 you get the homogeneous differential equation and the natural response can be obtained from the homogeneous differential equation and in this case the coefficients will come out to be slightly different in the parallel RLC case okay. So this is what we end up getting. Again the right hand side is not so important. The left hand side it has the same structure as this one. These two look similar to each other okay. Now a general differential equation of the second order will be of the form. I am assuming the homogeneous case okay that is the right hand side is 0. Now we know that a differential equation of this type has solutions of the form exponential Pt. Now we know this because the time derivative of exponential Pt is also exponential Pt times some number okay. So if you go on differentiating you will go on getting exponential Pt and it is quite easy to see that but could potentially satisfy this equation for the right value of P. So how do you find the right value of P? We substitute exponential Pt in the differential equation. All the exponential Pt's will get cancelled out and we will be left with a polynomial in P which is known as the characteristic equation of this differential equation okay. So what we get if we do that here if I substitute exponential Pt in this we will get a2 P square plus a1 times P plus 1 equals 0 okay. So all we have to do is the second derivative is replaced by P square first derivative by P and this Vc itself by 1 okay. So clearly this will have two solutions there will be two possible values of P that will satisfy this and they are given by P1 P2 are given by the solutions to this quadratic equation which is minus a1 by 2 plus minus this okay a1 square square root of a1 square minus 4 times a2 divided by 2. This comes from the very familiar solution to the quadratic equation. So now again there are three possibilities to this the term under the square root could be positive 0 or negative depending on that we will have two real roots complex conjugate roots or two identical real roots okay. So a1 square greater than 4 a2 or in other words square root a2 by 2 a1 less than 1 or let me put it in another way let me copy this over. So what I have done is to write the differential equation with some general coefficients a1 and a2 so that it is not tied to some specific circuit okay. Previously we got these coefficients in terms of LC and R we will do that when we go to the specific circuit but this will tell you that it is not specific to RLC circuit it will be the case for any circuit that has a second order differential equation describing it okay. We will connect it to the actual coefficients in different circuits okay. So now in this case we have three kinds of solutions one is when a1 square is more than 4 a2 this can also be written as square root of a2 by a1 less than half or equivalently a1 by 2 square root of a2 greater than 1 okay. So these are all equivalent statements I hope that is clear this inequality says exactly the same thing as this which says exactly the same thing as that okay. So now in this case we will have two distinct real roots okay that is this term under the square root is positive that means that the square root is a real number that means that both these p1 and p2 will be real numbers which will be different from each other because one comes from a plus sign other comes from a minus sign. Now what I would like the participants to do is to tell me whether both p1 and p2 will be negative or one will be positive and one will be negative or will both be positive okay. Please look at these numbers and what I will tell you is a1 a2 are both positive okay this we know for a series RLC case a1 was RC and a2 was LC because these are real component values which are positive these numbers will be positive okay. So please tell me if both of these will be positive both will be negative or one positive and one negative and for this particular case okay what do you think okay there is a correction to what I wrote here this is the roots are I should have a2 in the denominator okay thanks for providing the correction now of course this does not change this inequality in anyway so please let me know if the roots will be both positive both negative or one positive and one negative so it should be clear that both the roots will be negative because a1 and a2 are both positive so this number is positive sorry this number is negative and this the square root of this if you take plus the square root of this you can see that the second one will be positive but smaller than this negative number so the overall result is negative and when you have both minus signs naturally the whole thing will be negative also okay. So the answer is that p1 and p2 will both be negative in this if a1 and a2 are positive okay so we will have two distinct real roots and both negative what does it mean for these roots to be negative it means that exponential pt if p is negative this means that this response will die out with time okay exponential pt it decays with time okay so that is the implication of p1 and p2 being negative both are negative which means that eventually the natural response will go to 0 okay this was the case with the RC circuit and this is also the case with this one the RLC circuit okay I hope it is clear why both the roots are negative okay when both terms are negative of course it is obvious when the first term is negative and we take the plus sign here this entire term is smaller than this so the overall result is still negative okay these are all things that we evaluated last time we did it specifically with values of LC and R now I will do it generally with a2 and a1 and show how the two cases of series and parallel RLC are different from each other okay and this case when a1 a1 square equals 4 a2 what this does is to make the term under the square root 0 so both p1 and p2 will be equal to minus a1 divided by 2 times a2 okay and of course that will be negative because we have minus a1 by 2 a2 okay and finally when a1 square is less than 4 a2 in other words square root a2 by a1 is greater than 1 greater than half or a1 by 2 square root a2 is less than 1 we have the term under the square root is negative so the square root itself will be an imaginary number and we will have the two roots to be minus a1 by 2 a2 plus j square root 4 a2 minus a1 square I will reverse the sign inside the square root by 2 a2 and the other one will be minus a1 by 2 a2 minus j square root 4 a2 minus a1 square by 2 a2 okay so more than the exact values of this what you have to notice is we have plus j times some number here minus j times the same number here the real parts of the two are the same the imaginary parts of opposite signs so what we have are a complex conjugate pair of roots so we have three different cases and there is a reason I wrote the same condition in three different ways this is in terms of coefficients here I have defined some parameter less than half and here I have defined some other parameter greater than 1 okay the reason to do that is that these parameters this one and this one are some standard parameters that are very widely used okay. So if you write the differential equation like this let me copy over that if this is the differential equation a2 being the coefficient of the second derivative a1 the first derivative and 1 is the coefficient of Vc this is important then the characteristic equation of this of this form okay you have to normalize this to 1 that is important okay that is the way we have done it then this parameter square root a2 by a1 is known as the quality factor q okay it is usually denoted by q the quality factor and square root a1 by 2 square root a2 is known as xi the damping factor okay so these are constants that are used quite commonly to describe a second order system okay so square root of a2 by a1 is known as the quality factor and a1 by 2 square root of a2 is known as the damping factor okay so clearly you can write the three conditions these conditions in terms of either the quality factor or the damping factor okay so this same thing is written as q less than half or xi greater than 1 okay and here the conditions will be opposite okay so now we have the series and parallel RLC circuits okay let me copy these things over so please tell me what is the quality factor of the series RLC circuit in terms of the component values in terms of R, L and C okay please give me the expression for the quality factor based on the definition I just gave you the definition is here this is the definition corresponding to this differential equation so what I would like is the quality factor for the series RLC circuit okay what is the expression for the quality factor so this is a2 this is a1 and the coefficient of vc is 1 like I said if it is not 1 you have to normalize that by dividing it by dividing the whole thing by that term okay so the quality factor of the series RLC circuit which is square root a2 by a1 is given by 1 by R square root L by C and the damping factor xi is given by R by 2 square root C by L okay I think a couple of you were able to answer this quite easily it is just a substitution of terms okay so now what I would like is the quality factor of the parallel RLC what is the quality factor of the parallel RLC circuit yeah I got a couple of answers to this as well and it turns out the formula is the same as before it is square root of a2 divided by a1 and that would be R square root C by L okay the damping factor xi would be 2 by R square root L by C and also by the way it is pretty obvious that from these relationships from these two that the damping factor xi is 1 by 2 times the quality factor and vice versa okay so the exact damping or quality factor will depend on these coefficients in both these cases there are L R and C but you see that the expression for quality factor in fact is the reciprocal of one another in the two circuits okay so that is why I defined the quality factor and the damping factor based on the coefficients a2 a1 and 1 this term has to be normalized to 1 then the second order term is a2 first order term is a1 then the quality factor is square root a2 divided by a1 okay now depending on the circuit that you have it may not even have R L and C okay it can have two capacitors or two inductors and so on you can find the quality factor or the damping factor okay so especially what I want to emphasize is that the quality factor expression is different for a series RLC circuit and a parallel RLC circuit okay so do not try to learn any one of these formulas by heart depending on whether it is parallel or series the damping factor will be different okay now one sanity check you can use is the following a series RLC circuit looks like this and I have nulled the source okay and quality factor is given by 1 over R square root L by C okay now if R tends to 0 the quality factor tends to infinity okay and if all tends to 0 in the circuit what we will have is only L and C in parallel with each other okay now if I take a parallel RLC circuit okay let me close this and open it again there is some problem okay if I consider a parallel RLC circuit which is nulled quality factor of this is R square root C by L and if R tends to infinity Q tends to infinity okay and in the circuit we will have only L and C in parallel with each other so what I want to emphasize here is that when quality factor tends to infinity you will have only L and C in a loop okay so the quality factor can tend to infinity if R is 0 or R is infinity if R is 0 in a series RLC circuit then you will have a single loop of L and C and the quality factor of that is infinity if you have R equals infinity in a parallel RLC circuit you will have a single loop of L and C and the quality factor of that is infinity okay so do another sanity check that you can use at the circuit level when the quality factor is infinity that is let us say you get a certain expression for quality factor and you adjust the resistance value so that quality factor goes off to infinity then in the circuit what should be happening is that it should be left with a single L and single C in a loop okay it should be a lossless loop and Q equal to infinity means R equals 0 in the series case and R equals infinity in the parallel case okay so mainly I am emphasizing this over and over because the formulas for quality factor for series and parallel RLC circuits are opposite of each other but there is nothing contradictory or confusing there should be nothing confusing about this it is just that when the quality factor tends to infinity you should be left with pure L and C in a loop that happens when the parallel RLCs resistance R is infinity or the series RLCs resistance R is 0 okay I hope that part is clear now let us take the case of quality factor less than half which means the damping factor more than 1 basically means that A1 square is more than 4A2 okay so in this case P1 and P2 will both be real and negative now the natural response will be of the form A1 exponential minus P1 t plus A2 exponential minus P2 t okay and the constants A1 and A2 you can adjust using initial conditions now qualitatively if you look at these curves exponential minus P1 t might be like this okay and exponential minus P2 t might be like that when I plot them versus time so the combination A1 exponential minus P1 t plus A2 exponential minus P2 t okay it could be of depending on the initial condition I will assume it starts from there it could be of some form like that okay it will be just a combination of exponentials now let me take the other case when the quality factor exactly equals half or damping factor equals 1 and A1 square equals 4A2 in this case it turns out that P1 and P2 are real and identical the natural response it turns out because we have only a single value of P1 that is P1 and P2 are real and identical the natural response it turns out consists of this form A1 plus A2 t times exponential minus P1 t okay so this means P2 equals P1 so if you say exponential P1 t and exponential P2 t they will be the same as each other so the actual natural response will have A1 plus A2 times t times exponential P1 exponential minus sorry not minus P1 exponential P1 t okay I think previously also I perhaps wrote this wrongly it is exponential P1 t and exponential P2 t okay not minus because that minus sign is in the value of P1 itself okay and in this case when the damping factor is 1 or the quality factor is half the natural response will be A1 plus A2 t times exponential P1 t okay finally when the damping factor is less than 1 or A1 square is less than 4A2 P1 and P2 will be a complex conjugate pair okay so what does that mean the natural response will be of the form first of all P1 and P2 I will write as some real part PR plus or minus J times an imaginary part PI okay so the natural response will be of the form A1 exponential P1 t plus A2 exponential P2 t which can be written as A1 exponential PR t exponential J PI t and A2 exponential PR t exponential minus J PI t so this part is real and these parts are complex okay so these are complex parts and I will take out the real part this is what the natural response looks like okay now the natural response itself is a real number that is we are talking about a voltage here for instance in case of an RLC circuit the natural response of the voltage we see so what I would like from you is the conditions on these coefficients A1 and A2 so that the natural response is real okay what are the conditions on A1 and A2 so that the natural response is real what are the conditions on A1 and A2 so that the natural response is real it turns out that because these two are complex conjugates of each other if A1 and A2 are also complex conjugates of each other some will be a real number okay so this condition is A2 being complex conjugate of A1 okay so in that case the entire expression is real okay now the whole thing looks very complicated but mainly what I want to point out is that A2 is the when A2 is the complex conjugate of A1 and you know that the complex numbers can be described by their magnitude and some phase angle okay then this whole expression becomes exponential of PR times T times the magnitude of A1 times exponential J phi exponential J PIT plus exponential minus J phi because here we would have what A2 which is the complex conjugate of A1 times exponential J PIT okay so which results in A1 exponential PR times T exponential J PIT plus phi plus exponential minus J PIT plus phi okay you know that also exponential Jx plus exponential minus Jx is 2 times cos x so this sum is 2 times cosine of PI of T plus phi which basically gives us 2A1 exponential PRT exponential PIT plus phi okay so what do we have finally the entire natural response becomes a product of an exponential and a sinusoid okay so that is the qualitative difference I was trying to bring about okay there is a question about how to get the previously recorded lectures if you go to the NPTEL website you will see that all recorded lectures are available okay so you can go there and then get all of the lectures. So now let me just summarize the 3 cases the first one is when the quality factor is less than half or damping factor is greater than 1 in terms of the differential equation coefficients A1 square greater than 4A2 and the response is of the form A1 exponential P1T plus A2 exponential P2T A1 and A2 you choose from initial conditions and this particular condition where the damping factor is high that is damping factor is more than 1 is known as the over damped case and if I plot the natural response qualitatively okay it will look something like that let me plot all of them in the same plot later and the second case when the quality factor equals half and the damping factor equals 1 or in terms of the coefficients A1 square equals 4A2 this is known as the critically damped case and in this case the natural response again has 2 constants it is A1 plus A2T exponential P1T because P2 is identical to P1 okay and finally when the quality factor is more than half or the damping factor is less than 1 in terms of the coefficients A1 square is more than 4A2 this is known as the under damped case and in this case the P2 and P1 are complex conjugates of each other and the natural response is turns out of the form some constant exponential of PRT and exponential PIT plus phi okay. So again there are 2 constants here really there is the constant A1 and there is this constant phi and these have to be adjusted from initial conditions okay and what are PR and PI basically P2 and P1 are this P1 is PR plus j times PI it will be a complex number that is where the PR and PI come from okay these 2 constants and these 2 constants and this or that all of these are determined from initial conditions okay. So I hope this part is clear I will qualitatively show how these responses look like and why they are called over damped critically damped and under damped okay I will not go into much more detail about this but you can try it yourself I will try to determine A1 and A2 from initial conditions and get the total response okay by the way let me also describe what happens for the 3 cases in the 2 kinds of circuits over damped which means Q less than half or damping factor is I more than 1 okay and in case of series or we will see this means that basically you have a large resistance okay so if you work it out based on the formula for quality factor or damping factor this is what you will see parallel RLC value of R is small that is when you have over damped and critically damped means Q equals half or zeta equals 1 and under damped is when Q is more than half or zeta is less than 1 and the value of R is small in a series RLC case or large in the parallel RLC case okay. So let me take an example of some series RLC circuit okay so there will be a certain initial condition on the capacitor voltage and there will be a certain initial condition on the inductor current okay which basically can be thought of as initial condition on the derivative of the capacitor voltage okay and from these you can calculate all the responses I will assume the case with Vs equal to 0. So then let me imagine that for some value of RL and C the circuit is over damped that is the quality factor is less than half or the damping factor is more than half okay so I will plot Vc versus time starting from some initial condition Vc of 0 okay so what you would see is that for a certain initial condition on Vc of 0 and I love 0 the response may look something like this okay this is a typical response of an over damped case so this would be over damped and as you go on reducing the value of R okay let us say for some value of R it is over damped as you go on reducing the value of R at some point it becomes critically damped okay so for that the response tends to look like this okay so this would be critically damped and finally when it becomes over damped you see that again I think I made a mistake here this should be cosine okay this should be cosine Pi of t plus 5 and because of the presence of this cosine we will have a cosine sinusoidal response which has cycles right which goes alternately between positive and negative values and because of this exponential PRT the amplitude of the oscillation goes to 0 okay so what we would see would be something like this okay you will see some oscillations and then it can go off to 0 okay these are not to scale okay and also the response varies a little bit with the initial conditions but qualitatively this is what you would see and this happens in a series RLC circuit as R decreases it goes from being over damped to critically damped to under damped okay is this fine is there any questions I will take them now otherwise we can move on to the next topic then if you want to get the step response of a second order system you first calculate the steady state response okay this is based on open circuiting all capacitors and short circuiting all inductors and then you calculate the damping factor or the quality factor and from this you write the general form of the natural response the total response is natural response plus steady state response of the force response okay and the constants in the natural response either A1 and A2 or A1 and 5 these have to be adjusted from initial conditions this is fine so this is how the total step response of a second order system would be found what we will do next is to find out the response of these circuits whether it is RC or RLC to other kinds of time varying waveforms okay now we have analyzed them to some extent till now but we have considered only constant waveforms or steps which are of course varying but piecewise constants now signals in general can vary in more complicated fashion so what we will look at is the response of these signals to sinusoids which is a particular kind of time varying signal okay now we won't go into the details but it turns out that any signal of any shape can be constructed by combination of sinusoids okay that is by adding up sinusoids of different frequency and phase also now we are dealing with linear circuits so if the input is a sum of different sinusoids then we know that superposition applies we have to find the response to each of the sinusoids individually and then add up the responses to get the final response to the combination of the sinusoids okay so we can analyze the circuit for sinusoidal inputs with knowing fully well that for any other kind of input if we know its decomposition into sinusoids we will be able to tell the response also okay so that is why you find that most of the time either in analysis or in measurement the circuits are characterized using sinusoids okay so what we will do is we will look at the response to sinusoidal inputs now this can be done by solving the differential equation but we will take an easier route and show that what the relationships are show what the relationships are for each element when a sinusoid is applied okay and then from there what we will be able to do is directly from the circuit transform it into transform it in some way so that we can tell what the sinusoidal response is okay now there is one caveat here if we solve the differential equation we will see that it will have a certain natural response and a certain force response now it turns out that if you have a sinusoidal input the force response is also a sinusoid of the same frequency it is amplitude and phase can be different okay now with the method that I am talking about the natural response will not be computed at all okay the method I am going to outline where we transform each element R, L and C into something that is appropriate for sinusoidal steady state analysis it turns out that essentially we will be omitting the natural response completely so this will calculate only the steady state or the force to response to sinusoids but there is still okay it is still useful enough in practice because there are so many practical situations where you apply a sinusoid to a circuit and you wait for a while for the natural responses to die out and then you look at the output okay so it is still quite useful in practical context and it is also much much easier than solving the differential equation so that is why we do this thing okay and this entire business is known as either sinusoidal steady state analysis or phasor analysis the reasons for all of these things will become clear later okay we will talk about phasors later first I will show the sinusoidal steady state analysis okay so what is this all about so first of all let me apply a voltage which is just cos omega t across a resistor R okay what is the value of the current I please let me know I apply a sinusoidal voltage cos omega t across a resistor R what is the value of the current I so clearly a current I is V by R which is cos omega t divided by R okay now let us say I apply sin omega t across the same resistance R this let me call this I1 I1 will be V1 by R so V1 is cos omega t V2 is sin omega t so I2 will again be V2 by R okay we know the real relationship of a resistor this is sin omega t by R okay then now because this is a linear element okay superposition applies that is I applied V1 I have got I1 applied V2 I have got I2 so now if I apply let us say alpha 1 V1 plus alpha 2 V2 by linearity I expect the current to be alpha 1 I1 plus alpha 2 I2 okay now I will choose particular values of alphas which as you will see very easily will become will make it very convenient to analyze circuits I will choose alpha 1 to be 1 and alpha 2 to be j that is square root of minus 1 okay if I do that the applied voltage V which I call let us say V3 will be cos omega t plus j sin omega t right because it is cos omega t times 1 plus sin omega t times j which of course you know from basics of complex numbers is exponential j omega t okay so if you apply exponential j omega t the response will of course be I3 which is I1 plus j times I2 which is exponential j omega t divided by R okay now there is nothing surprising here I could have started off with exponential j omega t and said that the current is exponential j omega t divided by R because the current is simply voltage divided by the resistance but the interesting thing comes when we apply the same thing to capacitors and inductors right because initially we could always analyze resistance circuits somewhat easily right because we and I were proportional to each other and we had so many techniques to analyze even very complicated resistive networks like using nodal analysis and so on okay whereas once we had capacitors or inductors we ended up with differential equations which is definitely more difficult than solving algebraic equations so now this is trivial but it just illustrates the point I did it in a convoluted way instead of simply saying I will apply exponential j omega t because then you will not understand the motivation why I did that I applied cos omega t sin omega t and by superposition I constructed this exponential j omega t okay now let us do the same for capacitors and inductors and see what we get first let me take a capacitor and apply v1 equals cos omega t to it and what is the current i1 i1 is c times the time derivative of v1 which is basically minus omega c sin omega t and similarly if v2 is sin omega t that is I apply sin omega t across a capacitor a current i2 will flow okay which is c dv2 by dt which is basically plus omega c cos omega t now if I apply a third voltage v3 which is exponential j omega t and remember I do not have to solve for this separately I could I could have solved for this separately but I will first do it individually that is I imagine that this cos omega t and sin omega t are superposed I multiply cos omega t by 1 and sin omega t by j to get exponential j omega t the whole reason I am doing it this way is because I already told you that we will characterize the circuit with sinusoids we will use sinusoids to analyze and even in the lab to measure some things about our circuits so cos omega t and sin omega t are sinusoids and from that I will construct this some complex number exponential j omega t now that is an abstract thing which is really a mathematical quantity I cannot get exponential j omega t in the lab okay but it is still very useful to analyze using that we will see why now I will have this v3 which is exponential j omega t okay which is basically v1 plus j times v2 okay now the current I3 is I1 plus j times I2 okay which turns out to be omega c times minus sin omega t plus j cos omega t which you can see can also be written as j omega c times cos omega t plus j sin omega t which is basically j omega c exponential j omega t okay I could also have got this directly I3 which is c time derivative of exponential j omega t which is j omega c exponential j omega t okay now why did I do all this the point is the following first of all exponential j omega t can be thought of as superposition of cos omega t and sin omega t with some multiplying factors I multiplied cos omega t by 1 and sin omega t by j okay then the response to exponential j omega t is also a superposition of response to cos omega t and sin omega t okay now what I am trying to emphasize is that it is the response to cos omega t that I am interested in I apply cos omega t to some circuit and I want to find what the response is now in case of a capacitor I can still find it quite easily because it is very easy to differentiate v1 which is cos omega t but later we will see the circuits can get quite complicated okay now what happened I found this superposed response I3 by superposing this which is j omega c exponential j omega t okay or I could even differentiate it directly and I would get the same answer naturally okay they have to be consistent now the point is that if you look at this thing right the applied voltage is exponential j omega t and the current if you see it is proportional to exponential j omega t okay we have some constant it happens to be an imaginary number in this case but the point is it is some constant times exponential j omega t now this is not the case when I apply cos omega t when I apply cos omega t the response is sin omega t it is not something times cos omega t okay it looks more complicated I have to get it by differentiation but if I apply exponential j omega t I have shown using both ways that the response is some number times exponential j omega t okay it is proportional to the applied voltage and this happens to be just as in the resistors case okay in this case we have I3 to be 1 by R which is the conductance of the resistor times exponential j omega t okay the current is some number times the voltage the capacitor also has a current voltage relationship in the same form if the applied voltage is exponential j omega t okay so that is what makes it convenient but then what I am interested in is what happens when I apply cos omega t okay so I that is the thing that I will apply in the lab so that is what I want to know so now that is where this view point of superposition helps okay if you look at this excitation I have formed V1 plus j times V2 and let us say I am interested in the response to cos omega t okay this is my interest so now I have constructed V3 as superposition of V1 and V2 so naturally I3 the response to V3 is a superposition of this is the response to V1 plus j times the response to V2 okay let me in fact rewrite this I will write this as minus omega c sin omega t plus j omega c cos omega t okay remember I could get this answer much more easily by simply applying exponential j omega t and doing C dy dt of exponential j omega t but the point is I want to get back this quantity which is the response to cos omega t you see that the way this has been constructed V1 plus j times V2 okay the response also will be the response to V1 plus j times the response to V2. Now V1 and V2 are real quantities okay they are cos omega t or sin omega t and response to V1 is a real quantity it is some real voltage in the circuit response to V2 is also real quantity it is a some real voltage in the circuit okay so now this j this multiplier square root of minus 1 helps to keep the two responses separate okay you see what I am saying V1 is cos omega t V2 is sin omega t our superposition was V1 plus j times V2 so the response also will be response to V1 plus j times the response to V2 so the response to V1 will be the real part of the total response the response to V2 will be the imaginary part of the total response okay so I can calculate the response to exponential j omega t and take the real part to get response to cos omega t and if I wanted the response to sin omega t I could take the imaginary part we do not normally do that but we just take the real part I will explain why later I mean it is just an easy thing to do okay so we now have a much easier way of analyzing I will elaborate on this in the next lecture the reason it is easier is with exponential j omega t the relationship for the capacitor became somewhat like the resistor that is the current is just proportional to the voltage current is the voltage time some number okay instead of being this differential derivative and so on okay but from that we have to calculate the response to cos omega t and that is also easy okay the real part of that gives the response to cos omega t okay so from this we will be able to analyze any circuit consisting of R L and C when the applied voltages are sinusoid okay if there are any questions I will be happy to answer them now otherwise we will meet on Thursday I think also today there was a lot of lag between what I am saying and what was being received on the net hopefully you will be able to watch the recorded lecture and fill up any gaps that may be there okay thank you I will see you on Thursday