Question Video: Finding the Ratio of Capacitances in Series and in Parallel Physics

Video Transcript

A 135-microfarad capacitor and a 264-microfarad capacitor can be combined either in series or in parallel. Find the ratio of the total capacitance in parallel to the total capacitance in series. Give your answer to two decimal places.

Let’s begin by drawing circuit diagrams showing the two possible arrangements of these two capacitors. They can either be combined in series, which means one after the other on the same path like this, or they can be combined in parallel, which means that each capacitor is on a separate or parallel branch of the circuit. We can then add a cell and turn each of these sketches into a complete circuit. We’ll label the two capacitances as 𝐶 one and 𝐶 two. Let’s say that 𝐶 one is the 135-microfarad capacitance and 𝐶 two is the 264-microfarad capacitance.

We are being asked to work out the ratio of the total capacitance when the capacitors are connected in parallel to the total capacitance when they’re connected in series. If we label the total capacitance in series as 𝐶 subscript S and the total capacitance in parallel as 𝐶 subscript P, then this ratio that we’re asked to calculate is equal to 𝐶 subscript P divided by 𝐶 subscript S. In order to find the values of 𝐶 subscript P and 𝐶 subscript S, we’ll need to recall how we combine capacitances in series and in parallel.

Let’s begin with the case where the capacitors are connected in parallel. We can recall that if we connect multiple capacitors together in parallel, then the total capacitance is equal to the sum of the individual capacitances. That is, if we connect a load of capacitors in parallel with capacitances of 𝐶 one, 𝐶 two, 𝐶 three, etcetera, then the total capacitance 𝐶 subscript T is equal to 𝐶 one plus 𝐶 two plus 𝐶 three and so on.

In our case, we’ve just got two capacitors connected in parallel, and we’ve labeled their total capacitance as 𝐶 subscript P. So we have that 𝐶 subscript P is equal to 𝐶 one plus 𝐶 two. When we substitute in our values for 𝐶 one and 𝐶 two, we find that 𝐶 subscript P is equal to 135 microfarads plus 264 microfarads, which works out as 399 microfarads.

Now that we found the total capacitance in parallel, let’s clear some space and turn our attention to the case where the two capacitors are connected together in series.

In order to calculate 𝐶 subscript S, the total capacitance in series, we need to recall that when we combine multiple capacitors together in series, we add the reciprocals of the individual capacitances, and this gives us the reciprocal of the total capacitance. In our case, we’ve labeled the total capacitance in series as 𝐶 subscript S. So we have that one over 𝐶 subscript S is equal to one over 𝐶 one plus one over 𝐶 two.

We can add the fractions on the right-hand side by getting them over a common denominator. And we can achieve 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 of the equation as 𝐶 one plus 𝐶 two divided by 𝐶 one time 𝐶 two.

Finally, to make 𝐶 subscript S the subject of the equation, we want to take the reciprocal of each side of the equation, so that’s computing one divided by each side. When we do one divided by one over 𝐶 subscript S, this simply gives us 𝐶 subscript S. Then on the right, when we calculate one divided by a fraction, the effect is to flip the numerator and denominator of the fraction around. So we end up with 𝐶 one times 𝐶 two divided by 𝐶 one plus 𝐶 two.

If we now substitute in that 𝐶 one is 135 microfarads and 𝐶 two is 264 microfarads, we end up with this expression for 𝐶 subscript S. In the numerator, we’ve got 135 microfarads multiplied by 264 microfarads. And that works out as 35,640 with units of microfarads squared. Then in the denominator, 135 microfarads plus 264 microfarads works out as 399 microfarads. We can cancel one factor of microfarads from the numerator and denominator, which leaves us with units of microfarads. Evaluating the expression gives a result for 𝐶 subscript S of 89.3233 etcetera microfarads.

Now that we’ve got values for both 𝐶 subscript P, the total capacitance in parallel, and 𝐶 subscript S, the total capacitance in series, let’s clear some space and calculate the ratio 𝐶 subscript P divided by 𝐶 Subscript S.

When we substitute our calculated values for 𝐶 subscript P and 𝐶 subscript S into this ratio of capacitances, we get this expression here. And since we’ve got units of microfarads in both the numerator and the denominator, then these units cancel out leaving us with a dimensionless quantity. Now this makes sense because what we’re calculating isn’t a capacitance but rather a ratio between two capacitances.

Doing the division on the right-hand side, we get a result of 4.4669 etcetera. Since this ratio is greater than one, that means that 𝐶 subscript P is bigger than 𝐶 subscript S, which we can also see of course by looking at our individually calculated capacitances. Expressed in words, this statement is saying that the total capacitance of these two capacitors connected in parallel is greater than the total capacitance of these same two capacitors connected in series.

In fact, as a quick aside, we can show that this statement would always hold true no matter what values we have for the individual capacitances 𝐶 one and 𝐶 two. Recall that 𝐶 subscript P is equal to 𝐶 one plus 𝐶 two. And we showed that 𝐶 subscript S is equal to 𝐶 one time 𝐶 two divided by 𝐶 one plus 𝐶 two. So using these expressions in place of 𝐶 subscript P and 𝐶 subscript S in this ratio, we have 𝐶 one plus 𝐶 two divided by the fraction 𝐶 one time 𝐶 two over 𝐶 one plus 𝐶 two.

We can then rewrite this ratio like this, with two factors of 𝐶 one plus 𝐶 two in the numerator divided by 𝐶 one time 𝐶 two in the denominator. Expanding out the brackets in the numerator, we have 𝐶 one squared plus 𝐶 two squared plus two factors of 𝐶 one times 𝐶 two. We can then split up this fraction into two individual fractions. In the first fraction, we can cancel the 𝐶 one in the numerator with the 𝐶 one in the denominator. And we can do the same thing with the factors of 𝐶 two. Then this first term is simply equal to two.

The second term must be positive because capacitors can only have positive values. That means that the sum of these two terms must be greater than this value of two. So not only have we shown that the ratio 𝐶 subscript P over 𝐶 subscript S must be greater than one, making 𝐶 subscript P greater than 𝐶 subscript S, but actually we’ve shown that this ratio must be greater than two.

Putting this statement into words, we can say that for any two capacitors, the total capacitance when we combine them in parallel must be at least twice the total capacitance when we combine them in series.

Let’s now turn our attention back to the question at hand. We found that for the two capacitances given to us in this question, the ratio of the total capacitance in parallel 𝐶 subscript P to the total capacitance in series 𝐶 subscript S is equal to 4.4669 etcetera. This ratio is exactly what the question was asking us to find, but it tells us to give our answer to two decimal places. So rounding our result to two decimal places gives us our final answer of 4.47.

The ratio of the total capacitance in parallel to the total capacitance in series, to two decimal places, is 4.47.