Friday, October 11, 2013

**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 C_{2}. Let’s say C_{2} is K times bigger than C_{1}. (K could be 2, 3, 5, 10 or whatever value you want.) Then we could write C_{2} = KC_{1}. 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 C_{1} (which we’ll call Q_{1}) is the same as the charge on C_{2} (which we’ll call Q_{2}). So Q_{1} = Q_{2}. 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 V_{1} (the voltage across C_{1}) = Q_{1} / C_{1} and similarly V_{2} = Q_{2} / C_{2}. Since C_{2} = KC_{1} and Q_{2} = Q_{1}, we can substitute these values on the right side of the equation for V_{2} and write V_{2} = Q_{1}/ KC_{1}.

Now we wish to see how many times larger V_{1} is than V_{2}. We can do this by dividing V_{1} / V_{2}. Since V_{1} = Q_{1} / C_{1} and V_{2} = Q_{1} / KC_{1}, we can write the equation V_{1} / V_{2} = Q_{1} / C_{1} ÷ Q_{1} / KC_{1}. To divide by a fraction, we invert the fraction and multiply. Our equation then becomes V_{1 }/ V_{2} = (Q_{1} / C_{1}) x (KC_{1} / Q_{1}). On the right side, the Q_{1}’s cancel and so do the C_{1}’s. We are just left with V_{1} / V_{2} = K. This is another way of saying that V_{1} is K times larger than V_{2}. And when we started, we said that C_{2} was K times larger than C_{1}.

In other words, the voltages compare by the same factor as the capacitance values, but the higher voltage is across the smaller capacitor.

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