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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 30 / Problem 29E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Chapter 30, Problem 29E

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QUESTION:

Problem 29E

(a) What is the power input to the inductor from the battery as a function of time if the circuit is completed at t = 0? (b) What is the rate of dissipation of energy in the resistance of the inductor as a function of time? (c)What is the rate at which the energy of the magnetic field in the inductor is increasing, as a function of time? (d) Compare the results of parts (a), (b), and (c).

Questions & Answers

QUESTION:

Problem 29E

(a) What is the power input to the inductor from the battery as a function of time if the circuit is completed at t = 0? (b) What is the rate of dissipation of energy in the resistance of the inductor as a function of time? (c)What is the rate at which the energy of the magnetic field in the inductor is increasing, as a function of time? (d) Compare the results of parts (a), (b), and (c).

ANSWER:

Solution 29E

$The\ power\ input\ from\ the\ battery\ is\ P=εi.$

$The\ rate\ of\ dissipation\ of\ energy\ in\ the\ resistance\ is\ {i}^{2}\ R.$

$The\ voltage\ across\ the\ inductor\ has\ magnitude\ L=di\ /dt\ ,so\ the\ rate\ at\ which\ energy$

$is\ being\ stored\ in\ the\ inductor\ is\ iL=di/\ dt$

$a)P=\epsilon{}i$

$=\epsilon{}{i}_{0}(1-{e}^{-(R/L)t}$

$=\frac{{\epsilon{}}^{2}}{R}(1-{e}^{-(R/L)t}$

$=\frac{{(6\ V)}^{2}}{8\ \Omega{}}(1-{e}^{-(\frac{8\Omega{}}{2.5\ H})t}$

$=(4.5\ W)(1-{e}^{-(3.20\ {s}^{-1})t})$

$b)\ {P}_{R}={i}^{2}R$ $\$

$=\frac{{\epsilon{}}^{2}}{R}(1-{e}^{-(R/L)t}{)}^{2}$

$=\frac{{(6\ V)}^{2}}{8\ \Omega{}}(1-{e}^{-(\frac{8\Omega{}}{2.5\ H})t}{)}^{2}$

$=(4.5\ W)(1-{e}^{-(3.20\ {s}^{-1})t}{)}^{2}$

$c){P}_{L}=iL\frac{di}{dt}=\frac{\epsilon{}}{R}(1-{e}^{-(R/L)t}L\frac{\epsilon{}}{L}({e}^{-(R/L)t})$

$=\frac{{\epsilon{}}^{2}}{R}(1-{e}^{-(R/L)t}-{e}^{-2(R/L)t})$

&n

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