Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 27 - Circuits - Problems - Page 799: 61b

Answer

$C = 1.61\times 10^{-10}~F$

Work Step by Step

We can find the time constant $\tau$: $\frac{q}{C} = \mathscr{E}~(1-e^{-t/\tau})$ $\frac{q}{C} = 12.0~V~(1-e^{-1.30~\mu s/\tau}) = 5.00~V$ $1-e^{-1.30~\mu s/\tau} = \frac{5.00}{12.0}$ $e^{-1.30~\mu s/\tau} = 1-\frac{5.00}{12.0}$ $e^{-1.30~\mu s/\tau} = \frac{7.00}{12.0}$ $e^{1.30~\mu s/\tau} = \frac{12.0}{7.00}$ $\frac{1.30~\mu s}{\tau} = ln(\frac{12.0}{7.00})$ $\tau = \frac{1.30~\mu s}{ln(\frac{12.0}{7.00})}$ $\tau = 2.41~\mu s$ We can find the capacitance $C$: $RC = \tau$ $C = \frac{\tau}{R}$ $C = \frac{2.41~\mu s}{15.0~k \Omega}$ $C = \frac{2.41\times 10^{-6}~s}{1.50\times 10^4~\Omega}$ $C = 1.61\times 10^{-10}~F$
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