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: 28


$C = 25~\mu F$

Work Step by Step

We can find the equivalent capacitance $C_s$ of the two capacitors in series: $\frac{1}{C} = \frac{1}{20\times 10^{-6}~F}+\frac{1}{30\times 10^{-6}~F}$ $\frac{1}{C} = \frac{3}{60\times 10^{-6}~F}+\frac{2}{60\times 10^{-6}~F}$ $\frac{1}{C} = \frac{5}{60\times 10^{-6}~F}$ $C = \frac{60\times 10^{-6}~F}{5}$ $C = 12~\mu F$ We can find the equivalent capacitance $C$: $C = 12~\mu F + 13~\mu F$ $C = 25~\mu F$
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