Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 21 - Electric Current and Direct-Current Circuits - Problems and Conceptual Exercises - Page 759: 81

Answer

(a) $9.75ms$ (b) $668\mu C$ (c) $68.6mA$

Work Step by Step

(a) We can find the time constant as $t=RC$ We plug in the known values to obtain: $t=(175\Omega)(55.7\times 10^{-6}F)$ $t=9.75\times 10^{-3}s$ $t=9.75ms$ (b) We can find the maximum charge as $q_{max}=C\mathcal{E}$ We plug in the known values to obtain: $q_{max}=(55.7\times 10^{-6}F)(12.0V)$ $q_{max}=6.68\times 10^{-4}C$ $q_{max}=668\mu C$ (c) The initial current in the circuit can be determined as $I_{t}=(\frac{\mathcal{E}}{R})e^{-t/r}$ For initial current $t=0$ $\implies I(0)=(\frac{\mathcal{E}}{R})e^{0}$ $I=\frac{\mathcal{E}}{R}$ We plug in the known values to obtain: $I=\frac{12.0V}{175\Omega}$ $I=0.06857A$ $I=68.6mA$
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