 Let's solve a couple of questions on combination of capacitors. For the first one, we have four, five microfarads capacitors, which are combined as shown below. What is the equivalent capacitance of this combination? We need to report the answer in microfarads. As always, hit pause and try this question on your own first. Alright, hopefully you have given us a short. Now before we get into figuring out the equivalent capacitance, first let's try and label, let's label all the nodes and all the capacitors. So we can call this C1, this is C2, C3 and C4. And we can label all the nodes as well. We can call this A, this is, let's say this is B, this is C, B and you can call this node as well. We can say this is E. Okay, now our strategy would be to identify sets of capacitors which are in series or parallel because we know how to calculate the equivalent capacitance of those capacitors which are in series and parallel. So first let's try and understand the potential difference across C1. Let's see what is the potential difference across all of these capacitors so that we know if they are in parallel. Now across C1, the potential difference across C1 that would be, so let's write C1 here. Across C1, the potential difference is VAB, this is VAB. Across C2, across C2, this would be VBC, VBC. But now if we look closely, if we look at the terminals, so we have the terminals XY, potential at E will be the same as the potential at A because there is no circuit element in between so there is no potential drop between A and E. Similarly potential at E should be the same at potential at C. Again there is no circuit element in between so there should be no potential drop which means we can write this if VA, if VA is equal to VE and if VC is also equal to VE then we can write that VA is really equal to VC. And if this is true then VAB is really equal to VBC because we can write either as in terms of the other, we can write VBC as in place of C, we can write A and these two become equal. So turns out the potential difference across C1 and C2 is the same which means that they are connected in parallel. Now let's look at the potential difference across C2 and C3. Across C2 we already know we have written it, this is VBC and across C3 it's VCD. Now when we look at the diagram we see that B, the point B is connected to point D with this wire. Between points B and D there is no potential drop so VB should be equal to VD, there is no potential drop between them and if VB is equal to VD then in place of D we can write this, we can write this as VCB. So turns out the potential difference across C2, potential difference across C2 and C3 is also the same VBC or in terms of VCD. But that's the same since there is no potential drop across B and D, it is connected by this wire which has no circuit element so no potential drop. So C1, C2, C3 all of them are really connected in parallel because the potential drop across each one of them is the same. So for C1, C2 and C3 we can find the equivalent capacitance by simply adding, by adding them C1 plus C2 plus C3. Because if capacitance are connected in parallel this is how you find the equivalent capacitance. And C4 in place of now when you find the equivalent capacitance of C1, C2 and C3 we can replace all of that with just one capacitance and then you have C4 which is connected in series. So the circuit looks like this, you have XY, these are the terminals and then this is the equivalent capacitance of C1, C2 and C3 and then you have C4 which is connected to the equivalent capacitance. This right here, this right here is C4 and this right here is the equivalent capacitance of C1, C2 and C3. So equivalent capacitance of these three should be 15 because one is 5, this is 15 micro farads and now this equivalent is connected in series with C4. So now we need to figure out the equivalent capacitance of these two. Then the relation for that, the relation for that and I'm writing it on the top over here. It is 1 upon C final, I'm writing the final equivalent capacitance. This is equal to 1 upon C equivalent which is this, that is 15 plus 1 upon C4 and that is 5. Now we need to work this out. So I would suggest you pause the video and work out this calculation. Here C equivalent is really 15 and C4 is 5. And when you work this out, you should get C, the final equivalent capacitance has 3.3.75 micro farads. All right, let's look at one more question. Now here we have 6 3 micro farads capacitors which are combined as shown below and they are connected across PQ. What is the equivalent capacitance C of this entire combination? Okay, like we did in the previous video, so we will start off by labeling all the capacitors. So let's call this C1, C2, this is C3, C4 and let's call this C5 and C6. And now let's label all the nodes that we see. So we can call this as A, B, let's call this C and let's call this D. Okay, now again our approach would be to identify those sets of capacitors which are in series or parallel to each other because we know how to calculate the equivalent capacitance for those sets of capacitors. So let's try and identify them in this combination. Now this is connected across PQ. We can start off by trying to understand the potential difference across each capacitor. So for C1, across C1, the potential difference really is VAB, we can see that is VAB. And across C4, across C4, it is C4, this is VAD, VAD. Okay, now the good thing is that the capacitors are of the same magnitude. It's 3 micro farads all throughout. So the amount of potential drop in this branch over here should be the same as the amount of potential drop in this branch over here. And that should mean that the potential at B should really be equal to the potential at D because the amount of potential drop in this branch is the same as the amount of potential drop in this branch. So from VA, the potential starts dropping and it drops by the same value because the two capacitors are the same. So VB is really equal to VD. And what does that mean? That means that there is no current flowing, there is no charge flowing in this branch. The branch that has C5 and C6. So these two capacitors, they play no part. They play no role in the overall capacitors. They are not even the part of the circuit because the potential across B and D, it's the same. So no current, no charge flows between B and D. So now the problem just has these four capacitors. C1, C2, C3 and C4. And with them, let's try and draw the new circuit. So the new circuit, that would look like this. So we have C1, now we have C2. And these two are connected in parallel with C4 and C3. And then you have them connected to B and Q. So this is C1, this is C2, this is C4 and this is C3. Now these two are connected in series and these two are connected in series and they themselves are connected in parallel with this combination, C4 and C3. Now we can figure out the net capacitance. So for the top branch, the net capacitance would be, we need to find the equivalent capacitance for the top branch that would be one by three plus one by three. Similarly for the bottom branch, it would be one by three plus one by three. But this is one upon C equivalent. So when you find C equivalent, this is called this dash. This would be three by two. That is 1.5 and similarly C equivalent here, that would be 1.5. And you can take your time and work out this calculation. And now those two combinations are connected in parallel to each other. So now the circuit, now the circuit that we have is we have this, we have this capacitance of 1.5, 1.5 and they are connected to PQ. Here you have P, here you have Q. Okay, now this is 1.5 and this is 1.5. Now they are connected in parallel and with parallel the equivalent capacitance is figured out by adding the capacitors. So this is 1.5 plus 1.5 and that would be three microfarads. All right, you can try more questions from this exercise in the lesson. And if you're watching on YouTube, do check out the exercise link which is added in the description.