Video Transcript
The circuit in the diagram contains two capacitors connected in series. The total capacitance of the circuit is 12 microfarads. What is the capacitance 𝐶?
We can see that we’ve got a diagram here which shows a circuit where there’s a cell connected in series with two capacitors. This capacitor on the left is labelled with a capacitance of 45 microfarads. And this one here is labelled with a capacitance of 𝐶. The value of 𝐶 is what we’re asked to find. The question tells us that the total capacitance of the circuit, which we’ve labelled as 𝐶 subscript T, is equal to 12 microfarads. In order to answer this question, we’ll need to recall the formula which tells us how to combine capacitors in series.
If we have several capacitors connected in series, then if we add the reciprocal of their individual capacitances, we get one over the total capacitance of the circuit. That is, if we have capacitors with capacitances of 𝐶 one, 𝐶 two, 𝐶 three, etcetera, then when we connect them in series, 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 all the capacitors connected in series.
In the circuit from this question, we’ve just got two capacitors. So we just want the first two terms on the right-hand side of this equation. This equation connects the total capacitance of the circuit to the values of the two individual capacitances. In our case, we know the value of the total capacitance 𝐶 subscript T. We’re trying to solve for this quantity 𝐶, which is going to be one of these two capacitances on the right-hand side. For the sake of clarity, let’s say that 𝐶 is our quantity 𝐶 two. And then, that makes our other capacitance of 45 microfarads our value for 𝐶 one. This means that we want to take this equation and rearrange it to make 𝐶 two the subject.
The first step is to subtract one over 𝐶 one from both sides of the equation. Then on the right-hand side, we’ve got one over 𝐶 one and negative one over 𝐶 one. So these two terms cancel out. This gives us that one over 𝐶 subscript T minus one over 𝐶 one is equal to one over 𝐶 two. On the left-hand side, we can rewrite the first fraction by multiplying the numerator and denominator by 𝐶 one. And we can rewrite the second fraction by multiplying the numerator and denominator by 𝐶 subscript T. Then, since both fractions on the left now have the same denominator, we can rewrite the left-hand side as a single fraction. When we do this, the left-hand side then becomes 𝐶 one minus 𝐶 subscript T divided by 𝐶 one times 𝐶 subscript T.
We can then multiply both sides of the equation by 𝐶 one, 𝐶 subscript T, and 𝐶 two. On the left, the 𝐶 ones in the numerator and denominator cancel out, as do the 𝐶 subscript Ts. Meanwhile, on the right, the 𝐶 two in the numerator cancels with the 𝐶 two in the denominator. Once we’ve got rid of the terms that cancel out, what we’re left with is that 𝐶 two multiplied by 𝐶 one minus 𝐶 subscript T is equal to 𝐶 one times 𝐶 subscript T. Finally, we can divide both sides of the equation by 𝐶 one minus 𝐶 subscript T. On the left, we’ve then got 𝐶 one minus 𝐶 subscript T in both the numerator and the denominator. So this cancels out.
We end up with an equation where 𝐶 two is the subject. And we have 𝐶 two is equal to 𝐶 one times 𝐶 subscript T divided by 𝐶 one minus 𝐶 subscript T. In this question, we know the values of the quantities 𝐶 one and 𝐶 subscript T. And the quantity 𝐶 two is the value 𝐶 that we’re trying to find. Substituting those values into this equation, we have that 𝐶 is equal to 45 microfarads multiplied by 12 microfarads divided by 45 microfarads minus 12 microfarads. In the numerator, 45 microfarads multiplied by 12 microfarads comes out as 540 microfarads squared. In the denominator, we’ve got 45 microfarads minus 12 microfarads, which is 33 microfarads.
In this expression then, in terms of the units, we can now cancel one factor of microfarads from the numerator and denominator. That leaves us with units of microfarads for the capacitance 𝐶. Evaluating the expression gives a result of 16.36 recurring microfarads. Since the values of capacitances we were given in the question were both given to two significant figures, then we should quote our answer to the same level of precision. Rounding to two significant figures gives us our answer for the capacitance 𝐶 is 16 microfarads.
