College Physics

Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “...” indicates the expression is valid for any number of

capacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be

Cp= 1.000 μF+5.000 μF+8.000 μF = 14.000 μF. (19.70)

The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated inFigure

19.22(b).

Total Capacitance in Parallel,Cp

Total capacitance in parallelCp=C 1 +C 2 +C 3 + ...

More complicated connections of capacitors can sometimes be combinations of series and parallel. (SeeFigure 19.23.) To find the total capacitance

of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

Figure 19.23(a) This circuit contains both series and parallel connections of capacitors. SeeExample 19.10for the calculation of the overall capacitance of the circuit. (b)

C 1 andC 2 are in series; their equivalent capacitanceCSis less than either of them. (c) Note thatCSis in parallel withC 3. The total capacitance is, thus, the sum of

CSandC 3.

Example 19.10 A Mixture of Series and Parallel Capacitance

Find the total capacitance of the combination of capacitors shown inFigure 19.23. Assume the capacitances inFigure 19.23are known to three

decimal places (C 1 = 1.000 μF,C 2 = 3.000 μF, andC 3 = 8.000 μF), and round your answer to three decimal places.

Strategy

To find the total capacitance, we first identify which capacitors are in series and which are in parallel. CapacitorsC 1 andC 2 are in series.

Their combination, labeledCSin the figure, is in parallel withC 3.

Solution

SinceC 1 andC 2 are in series, their total capacitance is given by^1

CS

=^1

C 1

+^1

C 2

+^1

C 3

. Entering their values into the equation gives

1 (19.71)

CS

=^1

C 1

+^1

C 2

=^1

1.000 μF

+^1

5.000 μF

=1.200

μF

.

Inverting gives

CS= 0.833 μF. (19.72)

This equivalent series capacitance is in parallel with the third capacitor; thus, the total is the sum

Ctot = CS+CS (19.73)

= 0.833 μF + 8.000 μF

= 8.833μF.

Discussion

This technique of analyzing the combinations of capacitors piece by piece until a total is obtained can be applied to larger combinations of

capacitors.

19.7 Energy Stored in Capacitors

Most of us have seen dramatizations in which medical personnel use adefibrillatorto pass an electric current through a patient’s heart to get it to

beat normally. (ReviewFigure 19.24.) Often realistic in detail, the person applying the shock directs another person to “make it 400 joules this time.”

The energy delivered by the defibrillator is stored in a capacitor and can be adjusted to fit the situation. SI units of joules are often employed. Less

686 CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD

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