Capacitance Help (page 3)

By — McGraw-Hill Professional
Updated on Sep 9, 2011

Problem 2

Two capacitors with values of 0.0010 μF and 100 pF are connected in series. What is the total capacitance?

Solution 2

Convert to the same size units. A value of 100 pF represents 0.000100 μF. Then you can say that C 1 = 0.0010 μF and C 2 = 0.000100 μF. The reciprocals are 1/ C 1 = 1000 μF −1 and 1/ C 2 = 10,000 μF −1 . Therefore:

1/ C = 1000 μF −1 + 10,000 μF −1 = 11,000 μF −1

C = 0.000091 μF

This number is a little awkward, and you might rather say that it is 91 pF.

In the preceding problem, you can choose picofarads to work with rather than microfarads. In either case, there is some tricky decimal placement involved. It’s important to double-check 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 enter a wrong number, you will get a wrong answer, and if you miss a digit, you’ll be off by a factor of 10 (an order of magnitude).

Capacitors In Parallel

Capacitances in parallel add like resistances or inductances in series (Fig. 15-7). That is, the total capacitance is the sum of the individual component values. Again, you need to be sure that you use the same size units all the way through.


More About Alternating Current Capacitance Capacitors In Parallel

Fig. 15-7 . Capacitances in parallel add like resistances or inductances in series.

Capacitors In Parallel Practice Problem


Three capacitors are set up in parallel, having values of C 1 = 0.100 μF, C 2 = 0.0100 μF, and C 3 = 0.00100 μF. What is C , the total capacitance?


Add them up: C = 0.100 μF + 0.0100 μF + μF 0.00100 = 0.11100 μF. Because the values are given to three significant figures, the final answer should be stated as C = 0.111 μF.

Practice problems of these concepts can be found at: Alternating Current Practice Test

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