Question

(1 point) A resistor R = 15 N is in series with an inductor L = 5 mH and a 110 V battery. The switch is closed at t = 0. At what time will the power loss in R equal the rate at which energy is being stored in L? ms

Answer 1

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When the switch is closed at time t=0 ,

The current through the circuit

$I=I_0(1-e^{-Rt/L})$ --------------------(1)

Power loss in Resistance

$P=I^2R$ ---------------(2)

Rate at which energy is stored in inductor

$\frac{dE}{dt}=\frac{d}{dt}\left ( \frac{1}{2}LI^2 \right )=LI\frac{dI}{dt}$ ---------------(3)

$I^2R=LI\frac{dI}{dt}$

$IR=L\frac{dI}{dt}$

this happens when

potential across the resistance = potential across inductor = $\frac{\varepsilon}{2}$

$IR=L\frac{dI}{dt}=\frac{\varepsilon}{2}$

$I=\frac{\varepsilon}{2R}$ -------------------(4)

from (1) and (4)

$\frac{\varepsilon}{2R}=I_0(1-e^{-Rt/L})$

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

$\frac{1}{2}=(1-e^{-Rt/L})$

$e^{-Rt/L}=\frac{1}{2}$

$e^{Rt/L}={2}$

${Rt/L}=\ln{2}$

L t= In 2 R

t = ln (2)*5x10^-3/15) = 0.2311 x 10^-3 s = 0.231 ms

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