 So, let us now in the remaining time what I want to do is, want to go through some of these other topics in particular RC circuits, because I have seen that there is a lot of confusion about RC circuits in students. Some students when they come for our M. Tech interviews for example, they are not quite clear about how to handle an RC circuit or RL circuit so important. So, I thought I will just share some of these slides with you. So, in the meanwhile I will maybe I should just say something about this RC circuit. Simple things like what why is the capacitor voltage, why does it not change instantaneously, ok. Things like this I have asked in at several places which I visited when I gave talks on circuit simulation and so on. And I have not really got satisfactory answers most of the time. So, that is why I thought it is important to just clarify these some of these issues. So, that when you teach you should explicitly point out why that happened. I have asked whether I mean what exactly will happen if the capacitor voltage changed suddenly and I did not very often I did not get a satisfactory answer. So, let us look at some of these things. Also the other thing that I observed is when I when we give up a time domain problem to students on RC circuits or RL circuits, they immediately jump to Laplace transform, ok. And that is actually the last thing one should do because it is not at all convenient if you are talking about time domain. Laplace transform is ok if you are looking at some periodic kind of input, but it is not ok if you are looking at a general time domain problem, ok. So, all these things must be made very clear to the students and that is why I thought I will just take up this particular presentation, ok. So, this is this has to do with RC and RL circuits and I have said RC and RL circuits with DC sources, but you will see that this is not only with DC sources, but it is also with sources which can change from one value to another. So, in strictly although this is DC sources it is it also includes abrupt changes in the DC sources. So, of course the first few slides are just introduction to capacitors and so on. What is the capacitor? There is an there are conductor and conductors and there is an insulator in between and although we show this parallel plate structure in class, the actual capacitor construction is anything but this. Normally what they do is they have these films and they roll this, they have two films of metal and they put an insulator in between, they roll this and then they package it. So, that is what these capacitors look like. So, there is a wide variety of capacitors available, this I have taken from Wikipedia. The important thing is that the capacitance is assumed to be constant that is given by epsilon A by thickness, where A is the area of this plate equivalent. In these cases there is no plate, but there is a sort of equivalent area of that plate. T is the thickness. In practice capacitors are available in wide range of shapes and values and so on, I will skip this because this is sort of well known. But one thing that is often not very clear to students is that there are capacitors which are not constant and so what we can say in general about a capacitance is that the charge will be like C times V and so therefore d cube by dt is we say it is C dv dt. But this is only if the capacitance is taken as a constant. If it is not taken as a constant, then the current is not given by C dv dt, but it is given by the more general term d cube dt. And why we are partnering this out? That is because in almost all semiconductor devices for example, the p n junction capacitance is not a constant capacitance, but it itself depends on what is applied voltage and therefore in that case C dv dt is not appropriate because C is not constant. And even in MOS devices C is not constant and then we talk about the charge rather than the capacitance. And the derivative of the charge then gives us the current rather than C dv dt, alright. But in this basic course we talk about capacitances which are constant and all of these, these are constant capacitance not voltage dependent. So in that case the d cube dt or the current is simply C dv dt. So that is the sort of starting point of this whole lecture and the other extremely important point which students often do not get it is if V is constant the current is 0, right. And therefore the capacitor V is like an open circuit in DC conditions. Many students are just mocked up this point like capacitor is an open circuit in DC. But if you ask them why is that? I mean they just cannot give a satisfactory explanation and the explanation is so very simple, right. That the current is C dv dt, the derivative becomes 0, V is constant and therefore the current is 0. The current is 0 the capacitor is an open circuit. It is as simple as that, but students are not even able to tell us this simple reason very often and that is disturbing. And all this has to do with the way we use our textbooks and the way we design our exams and just expect the students to just reproduce certain things which they just mock up. So the whole thing has to change actually and that is what this course is all about. All right, go ahead, plot V, P and W versus time for the given source current. So this is just to sort of reinforce the students thinking about the relationship that I is equal to C dv dt. So if I is C dv dt then V is nothing but 1 over C integral I of t dt and if I is given then they simply need to find this integral and then they will get the voltage. So this is just that exercise. You can go through these plots later, the font is too small for you to read right now, but you can go through these plots and verify that what we have said is right. So an important plot is this power. If the voltage is, the power is simply given by V times I and if you do that for this particular case the power is 0 here then it rises then it becomes actually negative right and then it becomes finally when the current becomes 0 the power becomes 0 again at this point. Now so this, it is very important to point out that the power absorbed by capacitance can be either positive as in this region or it can be negative. What it means is when the capacitor is absorbing power that means it is getting charged right the current is going into the capacitor if your polarity is like that. What it means for power to be negative is that the capacitor actually is actually getting discharged. So all of these things are very important. So in other words in this region the capacitor is losing energy. In this region the capacitor is gaining energy with respect to time. And this in fact shows the energy as well. So in this when the power is positive the energy is actually getting is energy of the capacitor is actually increasing and the power is negative the energy of the capacitor is actually decreasing. So all of this such a simple figure will actually bring out so many important points and if the student has just seen this before his or her understanding can actually improve quite dramatically. Homework there is some homework that student can do. Inductors again this is all preliminary now inductors basically there is a core around which there is a coil and the magnetic field lines inside the inductor look like that there is a uniform density of the magnetic field and then that is a symbol. Again in this course we are not so concerned with how a capacitor operates or how an inductance operates what is flux what is charge and so on. But they are more interested in what does it do as a circuit element and that is all you need really to understand RC circuits, RL circuits and of course in addition if students know a lot about flux and charge and physics it is even better. But that is not really a prerequisite. An inductor is basically a conducting coil bound around a core and in fact it is good to derive this when you are talking about inductors it is good to give this very simple explanation that the voltage across an inductor is nothing but the number of terms times d phi dt where phi is the flux that is linked with the inductor. Flux is nothing but b dot a where a is the area of this coil and then and so on and so forth you can relate that with permeability and ampere turns and finally what you get is b equal to L d i dt where L is mu times n square times area and mu itself is mu r times mu 0. Now these are interesting formulas and you can actually bring in some interesting information which will sort of motivate students a little bit to gain further inside. So what do we do if you want to make the inductance large what we can do is we can make n large that means increase the number of turns or we can increase the area and both of these things will become the make the inductors bulkier bulkier or heavier. But in addition you can do you can increase the relative permeability of this material and that is why we do not use air but we use iron or some other materials like that. Otherwise the inductance size will have to be increased hugely to get a given inductance value. So now for a 99.8 percent pure iron the permeability relative permeability is 5000 and that is very large much larger compared to air and that is why we have iron cores and not air cores. So therefore an inductance you inductor you make with iron will be 5000 times larger than an inductance inductor that you make with air with other dimensions may be same. There are some other special alloys for which the mobile the relative permeability can be as high as 10 to 6 much larger than even iron. So all these things are sort of interesting facts and figures which students will find interesting. So now from the circuit point of view all we need to do is all we need to know is P is L D I D T that is it that is simple as that and if I is constant V is 0 and that is following simply from this equation. If I is constant the derivative becomes 0 therefore the voltage becomes 0. So an inductor behaves like a short circuit in DC conditions. Now this much people know but if you ask them why they cannot relate this fact with this formula and that is the P T that should change. So and this is something that of course we would expect from a highly conducting coil with DC voltage across it just a 0 resistance. This really comes from B equal to L D I D T alright now finally if before we leave this slide it is important to point out that this B equal to mu H that we have put over here is really an approximation in a normal in a in many of these materials B may be a non-linear function of H and that depends on the core material. But for all our purposes we will treat this to be valid and that gives us constant L which does not depend on the applied voltage or the current. So now we come to RC circuits with DC sources and I want to really start from scratch here and I would also encourage or I would urge all the teachers to really start from the basics rather than just stating formulas in class and so what is the most basic thing that they should know about RC circuits with DC sources. So here is an RC circuit and I have taken RC circuit with a single C and that single C has been taken out and the remaining circuit has got resistors maybe you cannot read so I will read it out resistors voltage sources current sources and it could have controlled sources like CCVS or CCCS or VCCS or VCCS you know all you know what that means so that is a starting point. So the capacity there is an RC circuit the RC could be many but the C is the single one and that has been taken out here. So the important thing now is to know what this C what the capacitor will see between A and B and we know we all know the answer to that question and that is this whole box can be now replaced by an equivalent thevenin voltage and an equivalent thevenin resistance. So this whole thing will reduce simply to thevenin voltage source and thevenin resistance that is it. So this no matter how complicated this circuit is no matter how many connections it has got as long as it has got only R's and voltage sources and current sources and dependent sources it can this whole black box can be represented by this simple representation and that is the key to understanding RC circuits without using any Laplace transform kind of a mumbo jumbo. So let us go ahead all sources are DC that is what we are assuming and therefore Vth is a constant. So this is just a constant number why because all the independent sources here are assumed to be constant. Let us see what happens next because of this fact then what happens is we write KVL which says Vth is equal to Rth times I the current times this voltage the capacitor voltage and then we replace the current we know that the current through a capacitor is Cdv dt and therefore we replace this current here by Cdv dt and we get a differential equation Vth equal to Rth times Cdv dt plus V. Now this equation has got two solutions one is a homogeneous solution one is a particular solution the homogeneous solution is the solution of this equation dv dt we can actually divide this whole thing by 1 over RC because that is an important quantity called the time constant. So after we do that we get dv dt plus 1 over tau V equal to 0 and this 0 of course this 0 is there because we want to separate the homogeneous and a particular solution. So homogeneous solution if I solve this ordinary differential equation I get the homogeneous solution to be some constant times exponential minus t by tau very simple this is something that the students would have already done by the time they come to this class they would have done mathematics and they would have done differential equations simple equations like this and they would know that this is the solution. So there is no black magic next what is the particular solution we still need to figure out what is the particular solution to this problem a particular solution is a specific function that satisfies the differential equation and there could be many particular solutions and how do we get one of these particular solutions very simple what we do know is that if I just keep this circuit in this configuration what will eventually happen is that all derivatives will vanish as t goes to infinity that will make the current equal to 0 and so in other words we will get a steady state the current will go to 0 all derivatives will go to 0 whether it is a voltage or a current all derivatives will go to 0 and we get this voltage V since the current here is 0 this voltage will be the same as V 7n and that is our particular solution in this case. So Vth equal to Vth is a particular solution which happens to be simply a constant. So this is very simple but it must be emphasized that students must be made aware of the simplicity of this whole thing otherwise they just think that this whole thing is all falling from the heavens and therefore it has to be mocked up and reproduced in the exams. So putting all of these things together we have the total V as the homogeneous solution plus a particular solution and that is the homogeneous solution is k exponential minus t by tau plus particular solution which is a constant in this case that constant happens to be the 7n voltage here. Next in general so now we can actually generalize this and this happens to be true V of t is equal to a exponential minus t by 2 tau plus b in this case this a is k here and b is Vth here. But in the general case a and b can be can be treated as just constants which are to be obtained from known conditions on v and we will see some examples of doing this. So if all sources are dc like we have been assuming we have V of t equal to this expression which we have just seen that tau here is r times c where this r is actually the Thevenin resistance as seen from the capacitor. Now what is the current through the capacitor it is c dv dt and that turns out to be some other constant times exponential minus t by tau. As t tends to infinity i tends to 0 and the capacitor behaves like an open circuit since all derivatives go to 0. Now since the current in then since the circuit in this black box in this box is linear we can say that any variable any current or any voltage okay in this particular circuit can be expressed as a constant k1 exponential minus t by tau plus k2 and that is only because this circuit is linear where k1 and k2 can be obtained by from suitable conditions on x of t okay. Same thing we can do with inductance an RL circuit again the inductance has to be a single one but the resistors can be any number the voltage sources current sources dependent can be any number and all of those things are in this box. So again we represent this box with this Thevenin representation and write a kcl write a kbl loop equation which gives vth equal to r Thevenin times i plus the voltage across the inductor that is ldi by dt and this again is an ordinary differential equation again with homogenous equation homogenous solution particular solution and so on. The only thing that now differs is the time constant the time constant is now given by l over rth where rth is the Thevenin resistance. Particular solution again the same philosophy to obtain one particular solution we can just look at this circuit as t tends to infinity and as t tends to infinity all of these derivatives will vanish ldi by dt will become 0 therefore v will become 0 and therefore the current will become vth divided by rth. So that is a particular solution that we can use. So again in the general case we can write the current or in fact any quantity in this circuit any current or any voltage in this circuit as a constant times exponential minus t by tau plus another constant. So that is what this slide says and then let us now take an example very simple example. It is a very simple R C circuit with a single source a single resistance and a single capacitor and now we want to explain why this vc cannot change suddenly. The sky going to fall down the answer is no but so what will happen if vc actually changes suddenly. So here is a very simple example showing a step change in vs and we want to convince ourselves and the students that even if there is a step change here the capacitor voltage cannot change suddenly and let us see why. vs changes from 0 at t equal to 0 minus 2 pi volts at t equal to 0 plus as a result of this vc will rise. Question is how fast can this vc change that is the question. For example let us just take some concrete numbers to make the discussion a little easier. For example what would happen if vc changes by 1 volt in 1 microsecond. Suppose this vc changes rather fast 1 volt in 1 microsecond at a constant rate as an example of 1 volt by 1 microsecond that is this turns out to be 10 raise to 6 volts per second. What will happen if that happens? What will happen then is the current in the capacitance current through the capacitance is C dv dt that is 1 micro farad the capacitance value times the rate of change of the current rate of the voltage which we have assumed to be 1 volt in 1 microsecond or 10 raise to 6 volts per second. So if I do this simple multiplication I get 1 micro farad divided by 10 raise to 6 volts per second or 1 ampere. So that is the current that I will need for a change of that kind 10 raise to 6 volts per second. Is this visible? Let us see now with i equal to 1 ampere and there is a resistance of 1k here the voltage drop across R would be 1000 volts right 1 ampere 1k to 1000 volts. And this is not allowed by KVL because this is only 5 volts. So if you have 1000 volts here KVL in the loop will simply not be satisfied and that is really the reason why we see cannot change certainly. And I have seen that like 90 out of 100 students do not answer this question properly and that is why it is very important to explain why this is not possible. Textbooks very many text books that are used in engineering colleges they just say we see cannot change certainly and then they just stop at that and no explanation is given. So that is absolutely the worst thing that one can do to teach students about RC circuits. So therefore we conclude that Vc at 0 plus must be the same as Vc 0 minus and otherwise if it is not the same then this current will be very large KVL will be violated which is not possible. So capacitor does not allow abrupt changes in Vc if there is a finite resistance in the circuit that is very important if there is a finite resistance in the circuit. If you have an ideal voltage source and an ideal capacitor then it is possible for the capacitor voltage to change because it will simply follow this voltage but there is no such thing as an ideal source because and even if you do not then connect this resistance externally a voltage source will always have a small series resistance whether it is 1 ohm, 10 ohms, 50 ohms or whatever and therefore this statement will always hold in practice. Similarly for exactly for the same logic an inductor does not allow an abrupt change in the current because for an inductor you will have V equal to L di dt and if your current changed abruptly di dt will become very large and at some point the KVL will not allow that okay. What do we do in case the source is not a constant but piecewise constant okay. Then what happens? Now if I ask this question to a student who is appearing for our MTech interviews they immediately jump to Laplace transform and that is absolutely the worst thing that can happen okay whereas there is no need to really jump to frequency domain when the problem is in time domain and when you can actually do it do the problem in time domain it is even more convenient to do it in time domain. So Laplace transforms are just simply uncalled for and should not be used okay. So how do we handle this problem? Although this source is varying with time it is actually piecewise constant okay. So there is a VES here is 0 here piecewise constant so in this region it is a DC circuit in this region it is another DC circuit in this region it is a third DC circuit and so on right. So we need to first identify intervals in which the source voltages or currents are constant. For example here we have just identified these three intervals. Next for any quantity of interest X of t we can write general expressions such as X of t equal to a1 exponential minus t by tau plus b1 in this region. We can write a similar expression in this region right only thing is the constants will not change so instead of a1 and b1 we will have a2 and b2 and so on okay and then we need to use boundary conditions to connect all this. Workout suitable conditions on X of t at specific time points using what facts? Using these facts if the source voltage or current has not changed for a long time long compared to tau then all derivatives are 0 okay so that is one thing that we can definitely use. So that means the capacitive current is 0 if things are not changed for a long time and the inductor voltage is 0 if things are not changed for a long time alright. Then when a source voltage or current changes say at t equal to t0 the next useful thing that we can use in deriving this constant is vc of t or isle of t cannot change abruptly and that we have already explained in the previous slide very important. So vc at any t0 plus is the same as vc at t0 minus iL at any t0 plus is the same as iL at t0 minus okay these things are these things serve to these things will help us to go from t1 minus to t1 plus or t2 minus to t2 plus and so on right whenever there are abrupt transitions in the source voltage or source current. Then a compute a1 b1 etc using the conditions on x of t okay so let us take this example in some detail and see how we can apply all these things. So you cannot probably see this thing very clearly because of resolution problem but go through these slides find okay all everything is self-explanatory you finally end up with this explanation at this expression then there is a discharging transient and so all of this follows very logically from what we have discussed so far and that is the way it should really be done in class rather than just giving this expression out of thin air or this expression for discharging from nowhere it is always good to derive this systematically knowing using things that the student has already known at this point. So this is the same thing but now it talks about the current and since the font is very small we will not go through this okay a very important point here charging and discharging transient so here is a charging transient and what is the duration of this transient that is an important question and that must be answered so this is a charging transient this is the current an interesting problem is interesting point to note is that here the current is actually discontinuous the voltage across the capacitor is continuous because as we have seen the voltage does not change suddenly giving rise to violation of KVL but the current can so the current does indeed change abruptly here from 0 to 5 milliamp and then it decreases down to 0 similarly here the voltage is continuous but the current is not so the all this plots are actually very useful helpful in terms of fixing some ideas in the students minds okay now what is the significance of the time constant always good to give a little table like this this shows x this shows e raise to minus x this shows 1 minus e raise to minus x okay and you notice that if x is 5 then e raise to minus x is very small number is like 0.007 or something and 1 minus e raise to minus x is very close to 1.993 okay so that is why we say that and normally our x would be t by tau so when t is 5 times tau the charging or discharging process is almost complete and that is why we say that 5 tau is the is the period of is the time period in which the transient has settled right transient has died down so this is the figure that shows these numbers in plot form okay so to continue further so the significance of time constant is shown in this figure in this case or the time value of resistance is 1 K the capacitor resistance is 1 micro so therefore the time constant is 5 milliseconds and you can see that in about 5 the time constant is 1 millisecond and in about 5 millisecond that is 5 times tau this transient has become as almost managed if I change the resistance to 100 ohms instead of 1 K the transient of course will settle down to a steady state value in a much shorter time again 5 times tau but now the time has become 10 times less same thing for discharging okay now here is an example which illustrates what to do if there are if there is a source which is piecewise constant okay so here we have an inductor and we are looking at this current inductor current and the voltage is given by this figure here so 0 here then 10 and then 0 here note that these transitions are shown as ideal sharp and therefore we have three regions we have region 1 here in which the source is 0 region 2 here in which the source is constant 10 volts and region 3 here where the source is again 0 okay so let us see how we can tackle this kind of problem so in each of these intervals our differential the nature of the differential equation actually is not going to change and therefore the same expression A times e raise to minus t by tau plus b is going to be valid in all of these regions only the constants will change so how do we start attacking this problem you first of all we need to look at the circuit as the inductor will see it okay so let us take the inductor out redraw the circuit and then replace it by its Thevenin equivalent okay so that is what is shown over here so the Thevenin equivalent turns out to be 8 ohms here you can go through all these numbers that is not very important the time constant turns out to be 0.1 second okay all right so now what this so up to this point things are steady and we say that this inductor current is the inductor voltage is 0 and therefore we can actually calculate the current knowing r 1 r 2 everything actually will be 0 up to t 0 beyond t 0 we enter a region where the voltage has now changed to 10 volts and now this is the key right how do you how do we evaluate these constants now A and B for that what we need to do is we already have one condition at this point and that condition is that the inductor current which was whatever value at 0 minus must now be the same at 0 plus that is one condition the second condition comes from this fact okay so now we are in this state now the circuit actually does not know whether this voltage is going down to 0 or is it going to stay at this value infinitely right so suppose we assume that this voltage is going to actually stay there indefinitely now we can figure out what will happen if t went to infinity and the voltage remained at 10 volts so that gives us some of the constants okay and that is all that is shown in this slide alright now in reality of course what happens is this voltage will go down to 0 and so we need to worry about what happens here at this point so go through so the third phase in the third phase of the transient we get this particular result and finally what we do is we put all together all of these this and this and this and then get the final transient okay which looks like this okay so that is one example of how to handle RC circuits or RL circuits with piecewise constant sources okay now another commonly encountered problem is if there is a switch in the circuit which is opening at some specific time and then what happens to the various quantities in the circuit so how do we tackle this kind of problem okay so here first of all let us let us figure out regions in time or the intervals in time for which the circuit can be represented by a circuit in which we have a capacitor resistors and DC sources in this particular case there are two distinct time intervals T less than 0 in which the switch is closed giving us this circuit and T greater than 0 in which the switch is open and giving us this circuit okay alright so that is so there is a there is one problem for T less than 0 another problem for T greater than 0 both of these problems we have a single capacitor and we have a circuit in which there are resistors and voltage sources current sources control sources in this particular case there is only a voltage source here again we have a circuit in which there is a capacitor and there is a circuit in which there are resistors sources and control sources and in this particular case the all the voltage sources and current sources are 0 okay but these are just special cases of the very first slide that we saw in which we showed a black box with DC sources resistors and therefore our expression and the A exponential minus T by top of B will apply in both of these cases so where do we begin we begin at t equal to 0 we will assume that we have the circuit has been in this state for a long time what does that mean that means this capacitor current is 0 because the circuit is in steady state once that is known we know that this is an open circuit we can actually find this current I that turns out to be 1 milliampere and so on okay so once we have the situation at t equal to 0 minus we use the capacity the continuity of the capacitor voltage to connect with this figure so the capacitor actually provides us the connection between 0 minus and 0 plus right so we see we can calculate vc at 0 minus that turns out to be 5 volts so therefore we can see that vc at 0 plus which is in this situation which is this situation here is also 5 volts and that is really the connection between t less than 0 and t greater than 0 circuits okay so the rest of the slide is just more details and you should actually go through this and convince yourself that you will get this kind of expression or this kind of plot for the current note that the current is discontinuous this particular current so it is there is a discontinuity here whereas the capacitor voltage will be continuous that is the capacitor voltage so the capacitor voltage is actually continuous but the other quantities in the circuit need not be right and that is what we have just shown over so make sure you actually can justify all these plots based on the expressions and there is a simulation file called e101 or c2 that you can use as well there is some homework all right so this is something that is that you can give students as homework and then if you do not want them to submit the homework they can you can say that okay you can try out this simulation and see if your results are actually correct or not yeah so I think that is all that we have for this RC and RL circuits in piecewise constant or constant sources okay so I wanted to cover this because I see many misconceptions in this topic and this is the topic that we are not going to cover in any other lectures next week so I thought it is important to just go through this let me see if there is something else I would like to cover in the time that is remaining okay there are topics which we are going to be covered in the next week so I do not want to take that up but let us look at some RLC circuits and I am doing this because I want to also show you an applet which will be useful in understanding RLC circuits and resonance and quality and bandwidth and that kind of stuff okay so I thought I will just talk a little bit about RLC circuits and then we will look at one Java applet that can help you to illustrate some of these things in class okay all of these things of course you know the current phaser is given by Vm by the total impedance in the circuit and so on okay and all of this is all standard expressions the magnitude of the current and the angle of the current okay here is an example of what that looks like so here we are plotting Im the magnitude versus frequency and whenever you are talking about frequency it is always good to plot it on log scale otherwise you would not get good resolution like we were saying yesterday and that is the phase okay and then there are these definitions of the bandwidth and so on there was one question about why this bandwidth is defined as I max by square root 2 and the reason for that of course is like we were talking about when we were discussing body plot we said that power normally we talk about dB as a power ratio right but if it is voltage or current ratio then since power goes as voltage squared or current squared we talk about ratio of square and that is where that root 2 comes so if the power becomes half that means the current has become 1 over square root 2 times okay so the important result that I wanted to point out is this bandwidth is r over l irrespective of c so it does not depend on p in this case quality is omega 0 divided by bandwidth so it is given by omega 0 l by r okay now this is a very important point and that is something the students often do not realize show that at resonance the magnitude of vl and magnitude of vc are both equal to quality times vm where vm is the maximum vm is the peak of this input voltage okay so what it means is that actually all the inductor and capacitor voltages at resonance can be actually quite large very large right because the quality can be like 10 or 20 or even some circuits even be larger than that and that is a very important point so that omega 0 square root omega 1 omega 2 all these things are interesting exercises to give students okay what happens if r is increased if r is increased then the current will the peak current will decrease also it will change the shape of the resonance the curve okay one important point for omega less than omega 0 the net impedance is capacitive and the current leads the applied voltage that is a very important conclusion for omega equal to omega 0 the net impedance is impendence is purely resistive and the current is in phase with the applied voltage and for omega greater than omega 0 the net impedance is inductive and the current lags the applied voltage so all of these things are actually very important point out to students so that they can understand this topic better okay here is a time domain example so here what we have three frequencies this is the center one is at resonance center one is at resonance this is before resonance and this is after resonance and then you can show the current leading the voltage or lagging the voltage in these plots okay here is a phasor diagram so this is at resonance so vs and we are actually the same and notice that vl and vc here are actually large but they are opposite each other and therefore they cancel each other exactly and these two figures this is before resonance and this is after resonance okay so this is a very important to discuss this phasor diagrams and this phasor diagram is what I wanted to talk about when I show the applet okay so let me just do that part and something that you can use in class share this okay alright so let me look at this RLC circuit alright so what this applet does is it allows you to choose R and L and C and you can compute the response it will compute the resonance frequency it will compute the quality and it will compute the maximum value of the current okay and all of that will be plotted here okay and when you start the animation this scale is not too the resolution is probably a little too large the size is a little too large but anyway let us see when you start the animation okay you get these plots here and let me just pause and show what how we can use this in class okay alright so here we have three four different quantities one is the inductor voltage that is the inductor voltage capacitor voltage the green one and the resistance the voltage across the resistor now the voltage across the resistor resistor and the current through the circuit basically are in phase so therefore the inductance voltage is leading the resistance voltage and the capacitor voltage lagging the resistor voltage right this blue line is VL plus VC and that is what is shown over here and it turns out that this particular phasor VR actually traces a circle because this angle between angle between VL plus VC and this voltage is always 90 degrees and therefore it traces a circle okay so you see that as we approach resonance the these magnitudes actually start growing and at resonance they can be they can be quite large it depends on the quality of the circuit and VR the resistor voltage at resonance it is actually equal to this source voltage and away from resonance it is smaller but it is always it always falls on this circle and that is because that angle is always 90 degrees so that is the animation that I wanted to show you and you can use this in class if I change this you can play with these values and even the formulas actually displayed on this same applet so the students can actually connect all these calculations with these formulas here as the animation is going on alright so I think we have reached the end of this session and it is also my last lecture and let us go through some question and answer session if you have any questions go ahead and ask KJ Sumeya, Sumeya college your question. Hello sir in case of RS flip-flop when you are feeding R equal to 1 and S equal to 1 it is written that in table it is not allowed not allowed in what sense if I tried physically giving R and S it is harmful to the circuit or. Okay the answer has to do with like we when we were talking about the RS latch right so the first latch that we took the NAND gate in that R equal to 0 S equal to 0 was not allowed in the second latch that we took the NOR latch R equal to 1 S equal to 1 was not allowed and later on we saw one reason why that is so so briefly the answer is suppose you allow that condition for example take a NOR latch let me just write this on paper okay so suppose I make this equal to 1 and this also equal to 1 so what will happen because it is NOR gate this will become 0 and this will become 0 okay alright and we say that this is not allowed so let us see what why it is not allowed so let us let us take let us draw this plot okay let me just sketch it here so let us say R has been 1 S has been 1 up to here and then Q has been 0 and Q bar is also 0 right now suppose this becomes 0 the question is what happens to Q or Q bar right now if R and S are 0 our table says that the state the Q is the previous state okay now in this case what is the previous state there is no such thing as previous state because our Q is 0 Q bar is also 0 and what the circuit will settle down to is not very clear we might have Q equal to 1 or we might have Q equal to 0 all that will depend on what are the delays of these two gates okay so that is and so there is a there is an uncertainty when we are going from 1 1 to 0 0 and that is the uncertainty that we do not like so that is why R equal to 1 X A is equal to 1 is not allowed okay any other questions sir in case of capacitor you are having the insulator between the two plates of the capacitor so of course the conventional current will not flow through the capacitor it is given that the displacement current will flow through the capacitor so please sir do not give answer orally some schematic and tell me how exactly the charges will move from one plate to another plate so that you can say that some displacement current will flow over. So the question is about displacement current in a capacitor so let us take a capacitor so let us say take let us say the voltage at a particular time is positive V right and so there is some charge here and there is a negative charge here okay now let us say with time there is a positive current flowing inside from this terminal to that terminal so what is actually happening in this capacitor right because there is these charges really cannot go through this material because there is dielectric and that is insulating so what is actually happening so what actually happens is as time goes by this charge will increase this charge will also increase okay and how what makes it possible to have this change all that is happening because of the flow of current in the external circuit not internal circuit right and I cannot do that right now but if you write Maxwell's equation you can actually show that this current will be the same as the term DDDT in this in the parallel plate okay and so inside the plate the the current is not a physical current but it is just the displacement current which is exactly the same as the external current so that is what happens in a capacitor. Sir, my question is related to the initial condition when you are hello and we can find out the initial condition whenever we connect the DC source across any RLC circuit and that time we can find out any current or any voltage across the capacitor we have to find out so that time we can there is any need to find out the initial condition at every time. I did not get the question fully but from what I understand the question is if you have an RLC circuit with DC sources is it necessary to find the initial condition every time okay that is what I understand as the question so the answer is let us just take a simple example let us say you have something like this L, C, R1, R2 and so on and let us say this voltage is like this okay so there are regions in which the voltage is constant another region here which where the voltage is constant and a third region here where the voltage is constant. Now when you come to 0 minus let us call this 0 when you come to 0 minus you know that this current has been 0 will be would have been 0 because this situation is there for a long time and things are very steady state derivatives are 0 therefore this voltage is 0 across this inductance and we also know that this current is 0 through the capacitor right. This will allow you to calculate I L at 0 plus and V C at 0 plus which are actually the same as I L at 0 minus and V C at 0 minus alright. So now we have another source in the circuit right so using this value you find your I L or V C of T okay and what do you do in this case you assume that this voltage this voltage source actually is going to continue all the way up to infinity and use those conditions to find all the constants that you need. Now when you come to this point right you need to go from let us call this T0 you need to go from T0 minus to T0 plus. At this point again you need to calculate I L at T0 minus using the expression you found V C at T0 minus using again the expression that you found for V 0 V C of T and then this is the same as I L at T0 plus and this is the same as V L at T0 plus okay. So the indirect recurrent the capacitor voltage or will provide you the continuity between transitions and that is how it has to be used so in that sense the initial conditions must be calculated every time. Sir one more question is related to the where at the time of K map when you can design the K map that time there is any if you can change the order of the column means just like when you are we can design the K map the order is 0 0 0 1 1 1 1 0 so can you change the order of that column instead of 1 1 we can change the 1 0 for 4 variable. Okay the question is about K maps is the speaker on the question is about K maps and we actually had answered this question in the lecture but let me just take it again let us say you have 4 variables A B C and D normally what we do is we write the we label this column so that there is only one change of variable allowed from column to column or row to row right that is how we actually do this things okay. So let us let me take an example sorry this is 1 0 so let me take an example of a term like let us say what term called x equal to let us say B okay so what will this look like in this map x is equal to B so it will look like let me take x equal to B C bar for example okay so that will look like B is 1 right so B is 1 C is 0 so that will look like 1 here 1 here 1 here 1 here okay and then therefore if I am given this pattern I know I can combine these two and I can derive this term from those two those four ones alright so now suppose I did not order it did not order this columns like that but like you are saying order the columns like this okay let us see what happens now to a term like this okay B C bar so B is 1 C is 0 so 1 here 1 here B is 1 so 1 here and 1 here right so now you see that although this they are supposed to be adjacent like in this one in this map they do not appear as adjacent I do not I cannot combine these I do not know whether these to be these should be combined or not whereas if I order it like this I know that they are adjacent they form a rectangle of size 2 raise to n where n is 2 in this case and therefore I can combine them so that is the reason we can we have to label the columns so that there is only one change between any two columns we will go to another center now where you are well over what is your question hello good afternoon sir so my question is given a resistor capacitor under DC for constant DC source how do you measure or how do you find the time constant of the RC circuit practically my next question is the second question is how to model a Norton's current source or how to prove a Norton's theorem practically what is the correct procedure to prove the Norton's theorem practically sir thank you sir over to you sir okay the first question is if there is an RC circuit how do you find the time constant in practice okay so let us take some example we will only take circuit in which there is a there is a single capacitor okay so let us say there is a circuit like this and there is a voltage source okay there are some capacitors resistors R1 R2 R3 okay now let us say we apply a VS of this kind and measure any of these quantities for example this V C or V or I measure this V R2 or whatever right so any measure any current or any voltage we will do something like this either like that or like that or whatever where from this I can extract the time constant okay from the shape of this I can see how fast this is rising I can actually just compare this with the exponential minus t by tau and find out from there what should tau be so that this expression matches this one but now in practice if you apply this you cannot see this on an oscilloscope okay so what you do is you apply a signal like this make sure that the frequency of the period of the signal is much larger than the time constant and now you can see this waveform or something like this on the oscilloscope so now you will be able to see this trace on the oscilloscope and now you can concentrate on this part and get your time constant okay so that is what you will do in practice the second question I do not quite understand fully it has to do with Norton theorem how do you prove Norton's theorem right so for that matter how do you prove Thevenin's okay so how do you prove Norton's theorem or for that matter how do you prove Thevenin's theorem now proof of these would be given in any network theory book and I know at least one of these books which is called Balabanian I am not sure about the spelling but it is something like that right so it but if you take one of these classic network theory books you will find a proof in that it is not something that I can do right now and between Thevenin and Norton the transformation actually is very easy to see this is your Thevenin equivalent circuit this is your Norton equivalent circuit and let us say it is the same circuit that we are looking at now suppose I will measure this open circuit voltage in this case I am going to get an open circuit voltage equal to Vth because the current is 0 in this circuit in this case I am going to get an open circuit voltage which is In multiplied by Rn okay now let us say I short these what is the short circuit current that is going to flow it is going to be Isc is equal to Vth by Rth suppose I short these two here what is the current that is going to flow this is 0 resistance so all of this current will actually go through this short circuit there is no voltage drop here so Isc is equal to In okay so from these two I will get Rn equal to Rth and In equal to Vth by Rth okay so that is how that is the conversion between Thevenin and Norton okay we will go to another center mufakum college my question is what are the differences between a latch and a flip flop sir over to you what is the difference between what is the difference between a latch and a flip flop okay so they are actually very similar latch is if there is a clock in a latch it is given it is level sensitive that means clock equal to one will make things happen whereas in a flip flop clock is equal to one is not enough as we have seen but there has to be an edge in the clock the edge the active edge could be either a positive edge or a negative edge so that is the basic difference between latch and a flip flop yeah my next question is among the NAND and NOR which gate is preferable in the formation of a flip flop over to you sir okay the question is among NAND and NOR which gate should be prefer for a flip flop this is very technology dependent question in some technologies may be it is easier to make a NOR gate in some other technologies easier to make a NAND gate so it really depends on which technology you are using also it will depend on whether you are using static gates or dynamic gates and so on so it is not it is not an easy question to answer you can just you can just post it on moodle maybe professor Sharma can explain if there is if there is a single answer my next question is sir what is the importance of preset and clearing a flip flop okay the question is what is the importance of preset and clear in a flip flop so as you said these inputs actually can be used to design counters and in fact we have seen a counter in which we let the counter count up to a certain state and we made sure that the next state will reset all the flip flops so that is just one application but there are other things like for example when you start up a digital circuit you do not know what state each flip flop is going to enter into right and that will decide its subsequent behavior now if you would like each flip flop to wake up in a known state and that is where preset and clear inputs can be used because it they will they will force the flip flop to be either 0 or 1 depending on what preset and clear are and that is irrespective of the clock input we will go to Pruba college Hello Open is formed by connecting two diodes into telephone holi why open is formed by connecting two transistors in parallel why one cannot make open by connecting to transistors in series okay the question is why opamp is formed by connecting two transistors in parallel and not in series. The answer has to do with the basic differential pair and that is like this. I am just drawing the BJT part, even you can have similar things in MOS as well. So the way this circuit works is of this voltage, if V1 is greater than V2 then this current I1 is greater than I2, V1 is less than V2 then I1 is less than V2, I2 such that I1 plus I2 is I0, that is the basic principle of a differential amplifier. And then once you have I1 and I2 change then you can take this to an appropriate stage, the next stage and enhance the difference between these two. Now you cannot do the same thing if you connect these two transistors in parallel, so that is the answer. Capacitance, an inductor and capacitor are considered as linear component or non-linear, at what condition they will be considered as linear? The question is whether capacitance and inductance is linear under what conditions? So as we have seen for a capacitor the current here is actually given by I equal to dq dt. Now whether the capacitor is linear or not linear will depend on whether this q is a linear function of the voltage or not. So if q is simply Cv divided by, so d dt of Cv where C is a constant then this is a linear element because then the C comes out. If q is sort of non-linear, it is a different function for example, if q is square root of v then your derivative of the current will itself become 1 over 2 square root v. Now this is not a dv dt, so this would be considered a non-linear capacitor. Same thing holds with a method inductor, if the flux is a linear function of the current then it is a linear conductor otherwise it is non-linear conductor. Yeah, we can go to another center. K. J. Sumaya College, you have a question? Sir, in case of capacitor the energy stored is one of Cv square. Definitely the energy will get stored in the dielectric medium between two plates. So exactly in which form? Definitely we will answer that the ill-studied field get created over there in that energy gets stored, but I do not want that answer. I read in one book the dipole formation takes place inside the dielectric medium. So can you explain in that context? So the question is how is energy stored in a capacitor in what form? So when we say C energy is equal to half Cv square, what exactly is this? And so it is really sort of physics question more than engineering, but let us see if we can. So in a capacitor, a simple parallel plate capacitor, we have these kinds of charges. Now let us just take hypothetical condition where we have a plus Q charge here and we are bringing in a minus Q charge from infinity to some place close to this. So now definitely there is we need to do some work to make this happen and that work is equivalent to the energy that is stored in this capacitor. So that is what it is. That is probably the best answer I can give you right now. Sir, what the dipole formation takes place over there? Dipole formation, whether the dipole formation take place over there? Well a dielectric will have dipoles, but I am not sure if you change the charge on the parallel plate capacitors if the dipole moment will change and so on. So I must pass this question. I will refer to you, refer you to this book by Resnick and Halliday and they talk about electrostatics in detail. So it is called Physics Part 2 by Resnick and Halliday. So just look up this book you might have might find the answer to that question over there. Jay Chum Rajendra College, what is your question? My question to you is, can you explain the working of a dual emitter transistor in a circuit and its significance over there The question is about dual emitter transistor and its significance. Let me, this really has to do with bipolar technology more than anything else. So the way transistors are made is there is a collector, let us say an NPN transistor with the collector will be N type. Then there is a base, there is a base contact here P plus and then there is a emitter. So that is the normal transistor configuration. Now if you want to have two emitters then it is very easy to do because you can just introduce one more of these N plus regions. So this is your one emitter, this is the other emitter. Since the base and the collector region are common that will, this single device with a little additional area will actually serve as two transistors and this is better than making two separate transistors, one like this and one more like that in terms of area. So that is why this configuration is sort of favored. The other thing is then you are absolutely sure that these two emitters will see the same conditions like temperature and so on. So that is the other reason. Sorry we could not take all the questions but please do post them on modal and we will get back to you. So we will break for lunch now. Thanks.