Capacitors in series
With capacitors, there is almost never any mutual interaction. This makes capacitors
somewhat easier to work with than inductors.
Capacitors in series add together like resistors in parallel. If you connect two capacitors
of the same value in series, the result will be half the capacitance of either
component alone. In general, if there are several capacitors in series, the composite
value will be less than any of the single components. It is important that you always use
the same size units when determining the capacitance of any combination. Don’t mix
microfarads with picofarads. The answer that you get will be in whichever size units you
use for the individual components.
Suppose you have several capacitors with values C1, C2, C3, . . . Cn all connected in
series. Then 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 total capacitance, 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 extremely tiny compared with the values of all the others, the composite capacitance
can be taken as the value of the smallest component
Problem 11-1
Two capacitors, with values of C1 = 0.10 μF and C2 = 0.050μF, are connected in series
(Fig. 11-3). What is the total capacitance?
Using the above 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.
Problem 11-2
Two capacitors with values of 0.0010 μF and 100 pF are connected in series. What is the
total capacitance?
Convert to the same size units. A value of 100 pF represents 0. 000100 μF. Then you
can say that C1 = 0.0010μF and C2 = 0.00010μF. The reciprocals are 1/C1 = 1000 and 1/C2 = 10,000. Therefore, 1/C = 1,000 + 10,000 = 11,000, and C = 0. 000091 μF. This
number is a little awkward, and you might rather say it’s 91 pF.
In the above problem, you could have chosen pF to work with, rather than μF. In
either case, there is some tricky decimal placement involved. It’s important to doublecheck
calculations when numbers get like this. Calculators will take care of the decimal
placement problem, sometimes using exponent notation and sometimes not, but a calculator
can only work with what you put into it! If you put a wrong number in, you will
get a wrong answer, perhaps off by a factor of 10, 100, or even 1,000.
Problem 11-3
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 total value C is just 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 pF.
