A Mixture of Series and Parallel Capacitance

Strategy

To find the total capacitance, we first identify which capacitors are in series and which are in parallel. Capacitors $C_{1}$ and $C_{2}$ are in series. Their combination, labeled $C_{S}$ in the figure, is in parallel with $C_{3}$ .

Solution

Since $C_{1}$ and $C_{2}$ are in series, their total capacitance is given by $\frac{1}{C_{S}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$ . Entering their values into the equation gives

\frac{1}{C_{S}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{1.000 μF} + \frac{1}{5.000 μF} = \frac{1.200}{μF} .

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Inverting gives

C_{S} = 0 . 833 µF .

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

\begin{array}{l} C_{tot} & = & C_{S} + C_{S} \\ = & 0.833 μF + 8 . 000 μF \\ = & 8.833 μF . \end{array}

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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.