An inductor of 10, text{mH} is connected to a | vedclass

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(C) The maximum current $I_{\max}$ in the $LR$ circuit is given by $I_{\max} = \frac{V}{R} = \frac{20\, 	ext{V}}{10 	imes 10^3\, \Omega} = 2 	imes

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(C) The maximum current $I_{\max}$ in the $LR$ circuit is given by $I_{\max} = \frac{V}{R} = \frac{20\, 	ext{V}}{10 	imes 10^3\, \Omega} = 2 	imes 10^{-3}\, 	ext{A} = 2\, 	ext{mA}$.When the switch is opened,the circuit acts as an $LR$ decay circuit. The current at time $t$ is given by $I(t) = I_{\max} e^{-Rt/L}$.Given $R = 10^4\, \Omega$,$L = 10 	imes 10^{-3}\, 	ext{H}$,and $t = 1 	imes 10^{-6}\, 	ext{s}$.The exponent is $-\frac{Rt}{L} = -\frac{10^4 	imes 10^{-6}}{10 	imes 10^{-3}} = -\frac{10^{-2}}{10^{-2}} = -1$.Thus,$I = 2 	imes e^{-1}\, 	ext{mA}$.Using $e^{-1} = 0.37$,we get $I = 2 	imes 0.37\, 	ext{mA} = 0.74\, 	ext{mA}$.Expressing this as $\frac{x}{100}\, 	ext{mA}$,we have $\frac{x}{100} = 0.74$,which implies $x = 74$.

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