Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 26 - Potential and Field - Exercises and Problems - Page 738: 25


$C = 3.0~\mu F$

Work Step by Step

We can find the equivalent capacitance $C$: $\frac{1}{C} = \frac{1}{6.0\times 10^{-6}~F}+\frac{1}{10\times 10^{-6}~F}+\frac{1}{16\times 10^{-6}~F}$ $\frac{1}{C} = \frac{(10)(16)}{(10)(16)(6.0\times 10^{-6}~F)}+\frac{(6.0)(16)}{(6.0)(16)(10\times 10^{-6}~F)}+\frac{(6.0)(10)}{(6.0)(10)(16\times 10^{-6}~F)}$ $\frac{1}{C} = \frac{316}{(10)(16)(6.0\times 10^{-6}~F)}$ $C = \frac{(10)(16)(6.0\times 10^{-6}~F)}{316}$ $C = 3.0~\mu F$
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