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: 27


$C = 7.5~\mu F$

Work Step by Step

We can find the equivalent capacitance $C_p$ of the two capacitors in parallel: $C_p = 20~\mu F + 10~\mu F$ $C_p = 30~\mu F$ We can find the equivalent capacitance $C$: $\frac{1}{C} = \frac{1}{10\times 10^{-6}~F}+\frac{1}{30\times 10^{-6}~F}$ $\frac{1}{C} = \frac{3}{30\times 10^{-6}~F}+\frac{1}{30\times 10^{-6}~F}$ $\frac{1}{C} = \frac{4}{30\times 10^{-6}~F}$ $C = \frac{30\times 10^{-6}~F}{4}$ $C = 7.5~\mu F$
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