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 762: 131

Answer

(a) $273\mu C$ (b) decrease (c) $58.7ms$

Work Step by Step

(a) We know that $q(t)=C\epsilon$ We plug in the known values to obtain: $q(t)=(91\mu F)(3.00V)$ $q(t)=273\mu C$ (b) We know that $t=RC$. As the capacitance value has decreased, the time to charge will also decrease because there is not much charge which will eventually be added to the capacitor. (c) We can determine the required time as follows: $t=(1.6084)(126+275)(91\times 10^{-6})$ $\implies t=(1.6094)(401)(91\times 10^{-6})$ $\implies t=58.7ms$
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