Video Transcript
The circuit in the diagram contains capacitors connected in series and in parallel. The total capacitance of the circuit is 36 microfarads. What is the capacitance 𝐶?
We’re given a diagram that shows a circuit where we’ve got a cell connected to this arrangement of capacitors. We can see that this circuit has two parallel branches with capacitors on. There’s this branch here that contains two capacitors: one with a capacitance of 𝐶 and the other with a capacitance of 95 microfarads. The other branch is this one here that just has one capacitor on it, with a capacitance of 25 microfarads.
In order to answer this question, we’re going to need to recall how we combine capacitances in series and in parallel. When we combine capacitors in series, we add the reciprocals of the individual capacitances, and this gives us one over the total capacitance. That is, if we connect capacitors in series with individual capacitances of 𝐶 one, 𝐶 two, 𝐶 three, et cetera, then one over the total capacitance of these capacitors, 𝐶 subscript T, is equal to one over 𝐶 one plus one over 𝐶 two plus one over 𝐶 three, and so on. When we combine capacitors in parallel, we just add the individual capacitances directly. That is, if we have parallel branches in a circuit with individual capacitances of 𝐶 one, 𝐶 two, 𝐶 three, et cetera, then the total capacitance 𝐶 subscript T is equal to 𝐶 one plus 𝐶 two plus 𝐶 three, and so on.
Now, in the circuit from the question, there are only two parallel branches. And the largest number of capacitors in series on a single branch is also two. That means that in both of these two expressions, we’re just going to want the first two terms on the right-hand side. That gives us these two equations for combining two capacitors in series and combining two capacitors in parallel. Let’s label the two branches in our circuit as branch A and branch B, respectively. Since branch B contains just a single 25-microfarad capacitor, then the capacitance of branch B, which we’ve labeled as 𝐶 subscript B, is simply equal to this value of 25 microfarads.
Let’s also label the capacitance of branch A as 𝐶 subscript A. And this is equal to the total capacitance of these two capacitors connected in series. We’re told that the total capacitance of the circuit is 36 microfarads. We’ve labeled this as 𝐶 subscript Tot to avoid confusion with the 𝐶 subscript T we’ve put in these generic equations. 𝐶 subscript Tot is equal to the total capacitance of these two parallel branches in the circuit, which have capacitances of 𝐶 subscript A and 𝐶 subscript B, respectively. Using our general equation for combining capacitances in parallel, we can then write that 𝐶 subscript Tot is equal to 𝐶 subscript A plus 𝐶 subscript B. We know the values for both 𝐶 subscript Tot and 𝐶 subscript B. So we can use those values in this equation to work out the value of 𝐶 subscript A.
To do that, we need to make 𝐶 subscript A the subject of the equation. We can do this by subtracting 𝐶 subscript B from both sides so that on the right the positive and negative 𝐶 subscript B terms cancel each other out. So we’ve now got an equation that tells us how to find the value of 𝐶 subscript A given the values of 𝐶 subscript Tot and 𝐶 subscript B. Substituting in our values for 𝐶 subscript Tot and 𝐶 subscript B, we get that 𝐶 subscript A is equal to 35 microfarads minus 25 microfarads, which works out as 11 microfarads. We’ve added this value for 𝐶 subscript A onto our circuit diagram so that we can now clear ourselves some space on the board.
Now that we’ve found this capacitance 𝐶 subscript A, then to work out the value of the capacitance 𝐶, we need to consider these two capacitors connected in series on branch A. With reference to this general equation for combining capacitors in series, we’ve labeled our unknown capacitance 𝐶 as 𝐶 one and our capacitance of 95 microfarads as 𝐶 two. We know that the total capacitance of these two capacitors combined is the total capacitance of branch A, which is our value of 𝐶 subscript A. That is, in this equation for capacitors connected in series, the total capacitance 𝐶 subscript T is our value for 𝐶 subscript A. So then our version of this equation for this specific circuit is that one over 𝐶 subscript A is equal to one over 𝐶 one plus one over 𝐶 two.
We’ve already worked out the value of 𝐶 subscript A, and 𝐶 two is a known value of capacitance that we were given in the question. To find the value of 𝐶 one, which is our unknown capacitance 𝐶 that the question is asking us to find, we need to rearrange this equation in order to make 𝐶 one the subject. The first step is to subtract one over 𝐶 two from both sides of the equation. Then, on the right-hand side, the positive one over 𝐶 over two and negative one over 𝐶 over two terms cancel each other out. This gives us an equation that says one over 𝐶 subscript A minus one over 𝐶 two is equal to one over 𝐶 one.
On the left-hand side, we can rewrite the first fraction by multiplying numerator and denominator by 𝐶 two. And we can rewrite the second fraction by multiplying numerator and denominator by 𝐶 subscript A. Then, since both these two fractions have the same denominator, then we can rewrite the left-hand side as 𝐶 two minus 𝐶 subscript A divided by 𝐶 two times 𝐶 subscript A.
We can then multiply both sides of the equation by 𝐶 one, 𝐶 two, and 𝐶 subscript A. On the left, the 𝐶 two in the numerator cancels with the one in the denominator and the same thing for the 𝐶 subscript A’s. Meanwhile, on the right, we’ve got a 𝐶 one in the numerator which cancels with the 𝐶 one in the denominator. Getting rid of these canceled terms, we have that 𝐶 one multiplied by 𝐶 two minus 𝐶 subscript A is equal to 𝐶 two times 𝐶 subscript A. The last step is to divide both sides by 𝐶 two minus 𝐶 subscript A so that on the left the 𝐶 two minus 𝐶 subscript A in the numerator and denominator cancel out.
We end up with an equation that says 𝐶 one is equal to 𝐶 two multiplied by 𝐶 subscript A divided by 𝐶 two minus 𝐶 subscript A. In our case, this quantity 𝐶 one is the value of the capacitance 𝐶 that we want to find. Then, substituting in our values for 𝐶 subscript A and 𝐶 two, we find that 𝐶 is equal to 95 microfarads multiplied by 11 microfarads divided by 95 microfarads minus 11 microfarads. In the numerator, 95 microfarads times 11 microfarads works out as 1045 with units of microfarads squared. In the denominator, we’ve got 95 microfarads minus 11 microfarads, and this works out as 84 microfarads.
In terms of the units in this expression, we’ve got two factors of microfarads in the numerator and one factor in the denominator. That means that we can cancel one factor from numerator and denominator, leaving us with overall units of microfarads. Evaluating the expression gives a result for the capacitance 𝐶 of 12.44 et cetera microfarads. If we round this to the nearest microfarad, we get an answer for the capacitance 𝐶 equal to 12 microfarads. As a final side note, since this value of 12 has two digits neither of which are leading or trailing zeros, then we could also say that we’ve given our result to two significant figures.
