 In the previous lecture, we looked at reciprocity of resistive networks. It is a very important property of resistive networks and it is also quite useful because when you interchange the stimulus and response that is if you have a two port network and you apply a stimulus on the left side, measure the response on the right side and then you apply a stimulus to the right and measure the response on the left, the ratios will be the same. That is effectively what reciprocity is. Any questions regarding what we did until the previous class? So, now we look at a couple of more circuit theorems and then move on to two port representations, ok. One of the interesting things is and a practical thing to that we need to calculate many times is that we have some component let us say resistor. It could be a voltage source or a current source also, ok. And let us say this is a and this is some network, ok which has let us say it is a linear network with some independent sources, ok. So, many times what we need to do is to calculate the effect of component changes, ok. So, let us say this R changes to R plus delta R, ok. So, the question is how can we calculate all the new solution to the circuit that is new branch voltage and branch currents. Now, one possibility of course is that we recalculate everything that is with the new value of the component we redo the circuit analysis. So, the whole point is to try to do it in a way that is easier than that, ok. Now, again as usual we exploit linearity to help us calculate this without going through the whole circuit analysis again, ok. So, that is the idea that is with the new value of the component it is just a new circuit and we could repeat the circuit analysis, but we will try not to do that and try to try to calculate the effect of changes without having to go through the circuit analysis all over again, ok. Now, instead of resistance it could be a voltage source or a current source also in that case it is easier and with resistance it is slightly more complicated, ok. So, let us say we have a linear network with independent sources and let me take a particular voltage source V s and then let us say I change only this voltage source value to V s plus delta V s, ok. So, in general all the branch voltages and currents inside this will change. So, is there an easy way of computing only the changes without redoing the circuit analysis? Please try to think about this. Is there a way of calculating only the change due to the change in the voltage source? How would we go about doing this? Please try to answer this. So, that is right as Sumit answered you use superposition and the way to do it is because this is a linear network. Now, we have already calculated the solution to this and this V s plus delta V s is nothing but V s in series with delta V s, ok. So, now in this particular case when we originally calculated everything is active except this one, ok. You can think of it as the case with delta V s equal to 0. So, what I have to do is to take the nulled circuit, this is nulled and V s is also nulled and I have only delta V s acting on the network, ok. So, now if you superpose these two if you take the sum of these two solutions you will end up with the solution to that one, ok. But so now the point is that analyzing this circuit with only delta V s is simpler than reanalyzing the whole circuit, ok. And in fact, if you look at the branch voltages and currents in this it will be the effect of delta V s that is only the changes to the branch voltages and currents due to delta V s, ok. Now, let us say instead of a voltage source, I have a resistor R, ok. I have a linear network plus independent sources here and then I change this one resistance to R plus delta V s, ok.