Hello. In the second video of the series, we learn about equivalent capacitance. So, we would have a network of capacitors connected in series or connected in parallel, and we'd like to replace them with one equivalent capacitor. So, as a reminder, a capacitor is a dynamic element that saves energy in the form of electric charge, and we studied the voltage-current relationship as follows: The current through the capacitor is proportional to the derivative of the voltage across the capacitor. And if we integrate this equation, the voltage over the capacitor is equal to the integral of the current through the capacitor times 1/C, where C is the capacitance of our capacitor.

Now, we start with a series circuit. So, let's assume here that I have a voltage source V_T supplying a current i(t). In the series network, I have C1 in series with C2, and I'd like to get an equivalent capacitance such that an equivalent circuit to the one on the left-hand side can be shown as follows: I have my supply V_T supplying a current I, and I'd like to replace these two capacitors with one equivalent capacitor. Let's call it here C_T. So far, so good.

Now, I'd like to study the relationship between V_T and i(t) in terms of C. Since the capacitor is in parallel to the supply source, the voltage across the capacitor V is equal to V_T, and the current through the capacitor is the same as the supply current. So, as a result, I can give an expression for the capacitor voltage V_C(t), which is the same as the supply voltage V_T. This is equal to 1/C integral of i(t) dt, using the physics of the capacitor.

Okay, so now I have a relationship between the supply voltage, supply current, and the equivalent capacitor that I wish. Let's do analysis on the circuit on the left-hand side to see how I can combine C1 and C2 such that I use this equation here.

So, to start, we have the circuit on the left. I have my power supply V_T. How about we write a KVL here? So, writing a KVL would give us V_T equal to V1 and V2, given that V1 is the voltage across capacitor 1 and V2 is the voltage across capacitor 2. I'm going to replace V_T using the equations that we derived in the previous slide—this one—so it has the total capacitance. It has I, so I have it here.

Now, I can write the voltage across capacitor 1 and the voltage across capacitor 2 in terms of the current in each. But since both are in series, they share the same current, and since this is the series network, the current going through the two capacitors is the same as I, the current supplied from our source. So, I have it here. So, V1 would be the integral of i(t) dt over C1, and V2 is the integral of i(t) dt over C2.

Now, for this equality to hold, we must have 1/C_T = 1/C1 + 1/C2. So, from there, I can give you an expression for the equivalent capacitance: C_T = C1*C2/(C1 + C2).

C_T = C1*C2/(C1 + C2). If you notice, this equation looks like as if we are combining parallel resistors. So, capacitors in series would follow an equation that looks like a parallel resistor. So, in general, if I have more than two capacitors in series, one over the equivalent capacitance is equal to a summation of the inverse capacitance of all of my capacitors.

Now, let's move to a parallel combination. So, assume here that I have two parallel capacitors, and I'd like to replace them with one equivalent capacitor. So, for the circuit, let's assume that I have a current source I_S supplying this network, and since all three elements here are in parallel, they have the same voltage V across all of them. I'd like to replace the circuit with one equivalent circuit that has my source I_S, and I replace the parallel combination with one equivalent capacitor C_T. And again, since they are parallel, they have the same voltage drop, and here's the current going through C_T as I_S because the two elements here make a series network.

Okay, so I can write an expression for I_S in terms of the capacitor because it is now the current going through the capacitor C_T. So, I_S can be written as C_T dV/dt.

Now, let me take this equation, keep it in mind, and go back to the original circuit on the left. Okay, so here we have our current source I_S. Okay, so here I have a node. The current input is I_S. I have two current outputs, I1 and I2. So, I can write this KCL equation: I_S is entering the node, I1 and I2 are leaving.

I have an expression for I_S from the previous slide in terms of the equivalent capacitance from here, and then I can represent I1 and I2 using the capacitance and the voltage across capacitor one and voltage across capacitor two. But when you look at the circuit, all three elements are in parallel, so they have the same voltage V. So, the current I1 can be represented as C1 dV1/dt, but V1 is the same as V, and V2 is the same as V. So, I2 is equal to C2 dV/dt.

Now, for this equation to hold, we should have C_T = C1 + C2 because this right-hand side here can be written as (C1 + C2) times dV/dt, and the left-hand side is C_T dV/dt. So, for equality, these two terms have to equal each other. So, the equivalent capacitance would equal C1 + C2.

So, if you have two parallel capacitors, I can replace them with one equivalent capacitor whose value is the summation of the two capacitances. In general, the equivalent capacitance of N parallel capacitors would be simply the summation of all these capacitances.

Let's apply an example for that. So, here I have a network of capacitors, and I'd like to get the equivalent capacitance C_T. First of all, you can see that these two here are in series, so I can replace them by one equivalent capacitor. Let me call it C_S. So, my circuit will become C1, and then I replace C2 and C3 by one equivalent capacitance. Let's call it C_S, where C_S is the series combination of C2 and C3.

When we do a series combination, we treat them as if they are parallel resistors, so we do multiplication over summation, and this gives us 150 pF. Now, to still get the equivalent capacitance, I have C1 in parallel to C_S. We reduce parallel capacitors by adding them together, so C_T would be C1 parallel to C_S. When we do the parallel, we add them together here, so the equivalent capacitance is equal to 3.65 nanofarads.

So, to recap, in this video, we learned about the series combination of capacitors and the parallel combination. The equivalent capacitance for a series combination follows this equation:

1/C_T = 1/C1 + 1/C2 + ...

Specifically, if I have exactly two series capacitors, the equivalent capacitance C_T would be the multiplication of both divided over the summation. For parallel capacitors, I can replace them with one equivalent capacitor whose value is the summation of those parallel capacitors. And this concludes this video. Thank you so much for listening.