### Video Transcript

The circuit shown in the diagram contains capacitors connected in series and in parallel. The 65-microfarad capacitor is moved to be in series with the 55-microfarad capacitor. By how much does the total capacitance of the circuit change?

Okay, so in this question, we’re shown a diagram that’s got a circuit with three capacitors in. The circuit has two parallel branches containing capacitors. Initially, on this branch through the middle, there’s a 75-microfarad capacitor in series with a 65-microfarad one. Then this branch down here contains just a single 55-microfarad capacitor. We’re asked how much the total capacitance of the circuit changes if this 65-microfarad capacitor is moved to be in series with this 55-microfarad one. So that’s removing it from the branch containing the 75-microfarad capacitor and adding it into the one containing the 55-microfarad one.

To answer this question, we’ll first find the total capacitance of the circuit in this initial configuration. Then we’ll find the total capacitance once the 65-microfarad capacitor has moved to be in series with this 55-microfarad one. And then we’ll find the difference between these two values.

In order to find the total capacitance of this circuit, we’ll need to recall how we combine capacitances in series and in parallel. If we have multiple capacitors connected together in series, then if we add the reciprocals of the individual capacitances, this gives us the reciprocal of the total capacitance. That is, if we connect a load of capacitors in series with capacitances of 𝐶 one, 𝐶 two, 𝐶 three, etcetera, then one over the total capacitance 𝐶 subscript 𝑇 is equal to one over 𝐶 one plus one over 𝐶 two plus one over 𝐶 three and so on.

For multiple capacitors connected in parallel, we simply add the individual capacitances to give us the total capacitance. In the circuit, we’ve been given, we’ve got two parallel branches, and one of those branches contains two capacitors connected in series. So let’s first use this equation for capacitors connected in series in order to find the total capacitance on this branch. Then once we know the capacitance on this branch, we also know the capacitance on this other parallel branch. So we’ll be able to use this equation for capacitors connected in parallel to find the total capacitance of the circuit.

Let’s label this branch with two capacitors on as branch A and this branch down here with one capacitor as branch B. We’ll label the capacitance of branch A, which is the capacitance of these two capacitors connected together in series, as 𝐶 subscript A. And we’ll label the capacitance of branch B as 𝐶 subscript B. And this is simply equal to the value of 55 microfarads.

If we then label the 75-microfarad capacitance as 𝐶 one and the 65-microfarad capacitance as 𝐶 two, then we can use our general expression for capacitors connected in series to say that one over 𝐶 subscript A is equal to one over 𝐶 one plus one over 𝐶 two. To add together two fractions, we want them over a common denominator. And we can do this by multiplying the first fraction one over 𝐶 one by 𝐶 two over 𝐶 two and multiplying the second fraction one over 𝐶 two by 𝐶 one over 𝐶 one.

We can then rewrite the right-hand side to say that one over 𝐶 subscript A is equal to 𝐶 one plus 𝐶 two divided by 𝐶 one times 𝐶 two. We can then take the reciprocal of both sides of the equation. So that’s doing one divided by the expression on each side. On the left-hand side, one divided by one over 𝐶 subscript A is simply equal to 𝐶 subscript A. On the right-hand side, one divided by 𝐶 one plus 𝐶 two over 𝐶 one times 𝐶 two comes out of 𝐶 one times 𝐶 two divided by 𝐶 one plus 𝐶 two.

So we’ve now got an expression for the capacitance 𝐶 subscript A of branch A in terms of the individual capacitances 𝐶 one and 𝐶 two. Let’s clear ourselves some space so that we can substitute our values for 𝐶 one and 𝐶 two into this equation.

If we substitute in that 𝐶 one is equal to 75 microfarads and 𝐶 two is equal to 65 microfarads, we end up with this expression for the capacitance 𝐶 subscript A. In the numerator, we’ve got 75 microfarads multiplied by 65 microfarads, which works out as 4875 with units of microfarads squared. Then in the denominator, we have 75 microfarads plus 65 microfarads, which is 140 microfarads. In terms of the units in this expression, we can cancel one factor of microfarads from the numerator with the microfarads from the denominator. And this leaves us with units for 𝐶 subscript A of simply microfarads.

Evaluating the expression gives a result for 𝐶 subscript A to two decimal places of 34.82 microfarads. This value of 𝐶 subscript A that we’ve just calculated is the capacitance of branch A in our circuit. And we also know the capacitance of branch B. That’s our value of 𝐶 subscript B. If we label the total capacitance of the circuit as 𝐶 subscript i, where the i stands for the fact that the circuit is currently in its initial configuration, then from our general equation for capacitors connected in parallel, we know that 𝐶 subscript i is equal to 𝐶 subscript A plus 𝐶 subscript B.

Substituting in our values for 𝐶 subscript A and 𝐶 subscript B, we get that 𝐶 subscript i is equal to 34.82 microfarads plus 55 microfarads, which works out as 89.82 microfarads. This is the value of the total capacitance of the circuit before the 65-microfarad capacitor is moved.

We now need to consider the case where the 65-microfarad capacitor is in series with the 55-microfarad one. So let’s redraw our circuit in this new configuration. In this new configuration, we’ll keep the parallel branches labeled as branch A and branch B just as before. However, now branch A has just a single, 75-microfarad capacitor, while branch B contains two capacitors connected in series with capacitances of 55 microfarads and 65 microfarads.

So this time the capacitance of branch A, which we’ll again label as 𝐶 subscript A, is simply equal to 75 microfarads. Meanwhile, the capacitance of branch B is given by the total capacitance of these two capacitors connected in series. We’ll label this as 𝐶 subscript B. Labelling the 55-microfarad capacitance as 𝐶 one and the 65-microfarad capacitance as 𝐶 two, then from our general expression for capacitors connected in series, we have that one over 𝐶 subscript B is equal to one over 𝐶 one plus one over 𝐶 two.

Using exactly the same process as we did earlier, we can rearrange this to get the 𝐶 subscript B is equal to 𝐶 one time 𝐶 two divided by 𝐶 one plus 𝐶 two. If we then substitute in that 𝐶 one is 55 microfarads and 𝐶 two is 65 microfarads, we end up with this expression for 𝐶 subscript B. In the numerator, that’s 55 microfarads multiplied by 65 microfarads. And this works out as 3575 microfarads squared. Then the denominator is 55 microfarads plus 65 microfarads, which is 120 microfarads.

We can cancel one factor of microfarads from the numerator and denominator. Then evaluating the expression gives a result for 𝐶 subscript B to two decimal places of 29.79 microfarads.

Now that we’ve worked out the capacitance 𝐶 subscript B of branch B and we know the capacitance 𝐶 subscript A of branch A, then we can use our expression for capacitors connected together in parallel to say that the total capacitance of the circuit is equal to 𝐶 subscript A plus 𝐶 subscript B.

We’ve labeled this total capacitance as 𝐶 subscript f, where the f stands for the fact that the circuit is now in its final configuration after moving the 65-microfarad capacitor. If we now substitute in our values for 𝐶 subscript A and 𝐶 subscript B, we get that this total capacitance 𝐶 subscript f is equal to 75 microfarads plus 29.79 microfarads, which comes out as 104.79 microfarads.

Now that we’ve got values for both the initial capacitance before moving the 65-microfarad capacitor and the final capacitance after moving it, we are ready to calculate the change in the total capacitance of the circuit. This change is equal to the final value of the total capacitance, that’s 𝐶 subscript f, minus the initial value of the total capacitance, 𝐶 subscript i.

Substituting in our values for 𝐶 subscript f and 𝐶 subscript i and evaluating the expression, we find that the change in the total capacitance is equal to 14.97 microfarads. Since all of the capacitance values that we were given were given to two significant figures, then we should also quote our answer to the same level of precision. Rounding our result to two significant figures, we get our final answer for the change in the total capacitance of the circuit as 15 microfarads.