Capacitors in Series

With capacitors, there is rarely any mutual interaction. This makes capacitors easier to work with than inductors. We don’t have to worry about mutual capacitance very often, the way we have to be concerned about mutual inductance when working with wire coils.

Capacitors in series add together like resistors or inductors in parallel. Suppose you have several capacitors with values C1, C2, C3, . . . , Cn connected in series. You can find the reciprocal of the total capacitance, 1/C, using the following formula:

1/C = 1/C1 + 1/C2 + 1/C3 + . . . + 1/Cn

The net capacitance of the series combination, C, is found by taking the reciprocal of the number you get for 1/C.

If two or more capacitors are connected in series, and one of them has a value that is tiny compared with the values of all the others, the net capacitance is roughly equal to the smallest capacitance.

Problem 1

Suppose two capacitors, with values of C1 = 0.10 μF and C2 = 0.050 μF, are connected in series (below figure). What is the net capacitance?

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Capacitors in parallel.

Using the preceding formula, first find the reciprocals of the values. They are 1/C1 = 10 and 1/C2 = 20. Then 1/C = 10 + 20 = 30, and C = 1/30 = 0.033 μF. Note that we can work with reciprocal capacitances in this calculation only because the values of the components are specified in the same units.

Problem 2

Suppose two capacitors with values of 0.0010 μF and 100 pF are connected in series. What is the net capacitance?
In this case, you must convert to the same size units before doing any calculations. A value of 100 pF represents 0.000100 μF. Thus, C1 = 0.0010 μF and C2 = 0.000100 μF. The reciprocals are 1/C1 = 1000 and 1/C2 = 10,000. Therefore, 1/C = 1000 + 10,000 = 11,000, so C = 1/11,000 = 0.000091 μF. (You might rather say it’s 91 pF.)

Problem 3

Suppose five capacitors, each of 100 pF, are in series. What is the total capacitance?
If there are n capacitors in series, all of the same value so that C1 = C2 = C3 = . . . = Cn, the net capacitance C is equal to 1/n of the capacitance of any of the components alone. Because there are five 100-pF capacitors here, the total is C = 100/5 = 20.0 pF.