### 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.

Letโs 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. Weโll label the two capacitances as ๐ถ one and ๐ถ two. Letโs say that ๐ถ one is the 135-microfarad capacitance and ๐ถ 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 theyโre connected in series. If we label the total capacitance in series as ๐ถ subscript S and the total capacitance in parallel as ๐ถ subscript P, then this ratio that weโre asked to calculate is equal to ๐ถ subscript P divided by ๐ถ subscript S. In order to find the values of ๐ถ subscript P and ๐ถ subscript S, weโll need to recall how we combine capacitances in series and in parallel.

Letโs 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 ๐ถ one, ๐ถ two, ๐ถ three, etcetera, then the total capacitance ๐ถ subscript T is equal to ๐ถ one plus ๐ถ two plus ๐ถ three and so on.

In our case, weโve just got two capacitors connected in parallel, and weโve labeled their total capacitance as ๐ถ subscript P. So we have that ๐ถ subscript P is equal to ๐ถ one plus ๐ถ two. When we substitute in our values for ๐ถ one and ๐ถ two, we find that ๐ถ 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, letโs clear some space and turn our attention to the case where the two capacitors are connected together in series.

In order to calculate ๐ถ 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, weโve labeled the total capacitance in series as ๐ถ subscript S. So we have that one over ๐ถ subscript S is equal to one over ๐ถ one plus one over ๐ถ 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 ๐ถ one by ๐ถ two over ๐ถ two and multiplying the second fraction one over ๐ถ two by ๐ถ one over ๐ถ one. We can then rewrite the right-hand side of the equation as ๐ถ one plus ๐ถ two divided by ๐ถ one time ๐ถ two.

Finally, to make ๐ถ subscript S the subject of the equation, we want to take the reciprocal of each side of the equation, so thatโs computing one divided by each side. When we do one divided by one over ๐ถ subscript S, this simply gives us ๐ถ 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 ๐ถ one times ๐ถ two divided by ๐ถ one plus ๐ถ two.

If we now substitute in that ๐ถ one is 135 microfarads and ๐ถ two is 264 microfarads, we end up with this expression for ๐ถ subscript S. In the numerator, weโve 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 ๐ถ subscript S of 89.3233 etcetera microfarads.

Now that weโve got values for both ๐ถ subscript P, the total capacitance in parallel, and ๐ถ subscript S, the total capacitance in series, letโs clear some space and calculate the ratio ๐ถ subscript P divided by ๐ถ Subscript S.

When we substitute our calculated values for ๐ถ subscript P and ๐ถ subscript S into this ratio of capacitances, we get this expression here. And since weโve 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 weโre calculating isnโt 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 ๐ถ subscript P is bigger than ๐ถ 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 ๐ถ one and ๐ถ two. Recall that ๐ถ subscript P is equal to ๐ถ one plus ๐ถ two. And we showed that ๐ถ subscript S is equal to ๐ถ one time ๐ถ two divided by ๐ถ one plus ๐ถ two. So using these expressions in place of ๐ถ subscript P and ๐ถ subscript S in this ratio, we have ๐ถ one plus ๐ถ two divided by the fraction ๐ถ one time ๐ถ two over ๐ถ one plus ๐ถ two.

We can then rewrite this ratio like this, with two factors of ๐ถ one plus ๐ถ two in the numerator divided by ๐ถ one time ๐ถ two in the denominator. Expanding out the brackets in the numerator, we have ๐ถ one squared plus ๐ถ two squared plus two factors of ๐ถ one times ๐ถ two. We can then split up this fraction into two individual fractions. In the first fraction, we can cancel the ๐ถ one in the numerator with the ๐ถ one in the denominator. And we can do the same thing with the factors of ๐ถ 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 ๐ถ subscript P over ๐ถ subscript S must be greater than one, making ๐ถ subscript P greater than ๐ถ subscript S, but actually weโve 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.

Letโs 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 ๐ถ subscript P to the total capacitance in series ๐ถ 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.