Q. I have read that when two capacitors of different values are connected in series, the one with the higher value has less voltage across it than the smaller one. This seems backward to me. If they were resistors, the one with the larger value would have more voltage across it than the smaller one. Why is this so?
A. The easiest way to explain this is mathematically. Suppose we have two capacitors, C1 and C2. Let’s say C2 is K times bigger than C1. (K could be 2, 3, 5, 10 or whatever value you want.) Then we could write C2 = KC1. The capacitors are in a series, so the charging current has to be the same in both, and it has to flow for the same amount of time. Charge is current multiplied by the time the current flows, so we can write Q = It, where Q is the charge in Coulombs, I is the current in Amps, and t is the time in seconds.
Since the current is the same for both capacitors, and the time the current flows is also the same, the charge on C1 (which we’ll call Q1) is the same as the charge on C2 (which we’ll call Q2). So Q1 = Q2. Now we can use the equation V = Q / C for each capacitor, where V is the voltage in Volts, Q is the charge in Coulombs, and C is capacitance in Farads. This means that V1 (the voltage across C1) = Q1 / C1 and similarly V2 = Q2 / C2. Since C2 = KC1 and Q2 = Q1, we can substitute these values on the right side of the equation for V2 and write V2 = Q1/ KC1.
Now we wish to see how many times larger V1 is than V2. We can do this by dividing V1 / V2. Since V1 = Q1 / C1 and V2 = Q1 / KC1, we can write the equation V1 / V2 = Q1 / C1 ÷ Q1 / KC1. To divide by a fraction, we invert the fraction and multiply. Our equation then becomes V1 / V2 = (Q1 / C1) x (KC1 / Q1). On the right side, the Q1’s cancel and so do the C1’s. We are just left with V1 / V2 = K. This is another way of saying that V1 is K times larger than V2. And when we started, we said that C2 was K times larger than C1.
In other words, the voltages compare by the same factor as the capacitance values, but the higher voltage is across the smaller capacitor.