### Video Transcript

The circuit in the diagram contains capacitors connected in series and in parallel. What is the total capacitance of the circuit? Give your answer to the nearest microfarad.

The diagram that we’ve been given shows a circuit where there’s a cell connected to a combination of capacitors. We can see that this circuit has two parallel branches with capacitors on. There’s this branch through the middle which we’ll label as branch A. And we can see that it’s got to capacitors on it. There’s one with a 75-microfarad capacitance and another with a 55-microfarad capacitance. Then, there’s this branch down here which we’ll label as branch B. And this contains just a single 35-microfarad capacitor.

We’re being asked to find the total capacitance of the circuit. And in order to do this, we’ll need to recall how we combine capacitors in series and in parallel. When multiple capacitors are connected together in series, then the reciprocal of the total capacitance is equal to the sum of the reciprocals of the individual capacitances. That is, if we connect a bunch 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.

Meanwhile, if we have multiple capacitors connected in parallel, then we simply add the individual capacitances in order to get the total capacitance. In our diagram, we can see that the branch labeled as A consists of two capacitors connected in series. Let’s label the total capacitance of this branch, so that’s the total capacitance of these two capacitors together in series, as 𝐶 subscript A. We’ll also label this 75-microfarad capacitance as 𝐶 one and this 55-microfarad capacitance as 𝐶 two. Then from our general expression for capacitors connected in series, we know that one over 𝐶 subscript A is equal to one over 𝐶 one plus one over 𝐶 two.

Since we know the values of both 𝐶 one and 𝐶 two, then we can use this equation in order to find the value of 𝐶 subscript A. To do this, we’ll want to make 𝐶 subscript A the subject of the equation. We’ll start by multiplying both sides of the equation by 𝐶 subscript A, 𝐶 one, and 𝐶 two. By expanding the brackets on the right-hand side, we can rewrite the equation like this. In this fraction on the left, the 𝐶 subscript A in the numerator cancels with the 𝐶 subscript A in the denominator. In this first fraction on the right, the 𝐶 one in the numerator and denominator cancel. And in this second fraction on the right, the 𝐶 two in the numerator and denominator cancel.

Once we get rid of the canceled terms, we end up with this expression here. We can then factor out the 𝐶 subscript A, which appears in both terms on the right. The final step we need to make in order to get 𝐶 subscript A as the subject of this equation is to divide both sides by 𝐶 one plus 𝐶 two so that on the right the 𝐶 one plus 𝐶 two in the numerator cancels with the one in the denominator. If we then write the equation the other way around, we have that 𝐶 Subscript A is equal to 𝐶 one times 𝐶 two divided by 𝐶 one plus 𝐶 two.

So this is an equation which tells us how to calculate the total capacitance 𝐶 subscript A of two capacitors, 𝐶 one and 𝐶 two, connected together in series.

Let’s now clear ourselves some space so that we can substitute in our values of 𝐶 one and 𝐶 two from this circuit. Substituting in that 𝐶 one is equal to 75 microfarads and 𝐶 two is equal to 55 microfarads, we get this expression for 𝐶 subscript A. In the numerator, we have 75 microfarads multiplied by 55 microfarads. And this works out as 4125 with units of microfarads squared. Then in the denominator, we’ve got 75 microfarads plus 55 microfarads, which comes out as 130 microfarads.

In terms of the units, we can cancel one of the two factors of microfarads from the numerator with the microfarads from the denominator. This then leaves us with units for 𝐶 subscript A of microfarads. Then, evaluating the expression gives a result to two decimal places of 31.73 microfarads. So we’ve now found the value of 𝐶 subscript A, which is the capacitance of the branch that we labeled as A in our circuit diagram. Since branch B contains just one capacitor, then we know that its capacitance is just equal to this value of 35 microfarads. Let’s label this as 𝐶 subscript B.

We’re now in a position where we’ve got two parallel branches. That’s branch A and branch B. And we know the capacitance of each branch. That’s our values for 𝐶 subscript A and 𝐶 subscript B. If we now clear ourselves some space on the board, then we’ll be able to use our general expression for capacitors connected in parallel in order to find the total capacitance of these two branches.

This general expression tells us that for capacitors connected in parallel, we simply add the individual capacitances. In our case, that’s the values for 𝐶 subscript A and 𝐶 subscript B. So then the total capacitance 𝐶 subscript T is equal to 𝐶 subscript A plus 𝐶 subscript B. We can then substitute in our values for 𝐶 subscript A and 𝐶 subscript B and evaluate the sum to get a value for the total capacitance 𝐶 subscript T.

We’re asked in the question to give our answer to the nearest microfarad. Rounding to the nearest microfarad, we get our answer for the total capacitance of the circuit as 67 microfarads.