## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 537: 27

#### Answer

a) 95% of the steady state value $\dfrac{V}{R}$ b) 86% of the steady state value $\dfrac{V}{R}$

#### Work Step by Step

(a) Since, $L\dfrac{di}{dt}+Ri=0$ Also, $i=(\dfrac{V}{R})(1-e^{-(\frac{R}{L})t})$ $\implies i=(\dfrac{V}{R}) (1-e^{-3})\approx (0.9502) \dfrac{V}{R}$ Hence, 95% of the steady state value $\dfrac{V}{R}$ (b)Consider $L\dfrac{di}{dt}+Ri=0$ Also, $i=(\dfrac{V}{R})(1-e^{-(\frac{R}{L})t})$ Suppose $t=\dfrac{2L}{R}$, then we have $i=(\dfrac{V}{R})(1-e^{-(2)})$ $\implies i=(\dfrac{V}{R}) (1-e^{-3})\approx (0.8647) (\dfrac{V}{R})$ Hence, 86% of the steady state value $\dfrac{V}{R}$

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