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(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-t/	au})$,where $I_0 = E/R$ and $	au = L/R$.Given $L = 20 \, H$ and $R =

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$2 \ln 2 \, s$

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(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-t/	au})$,where $I_0 = E/R$ and $	au = L/R$.Given $L = 20 \, H$ and $R = 10 \, \Omega$,the time constant $	au = 20/10 = 2 \, s$.So,$I = \frac{E}{10}(1 - e^{-t/2})$.The rate of dissipation of energy (Joule's heat) across the resistance is $P_R = I^2 R = I^2 	imes 10$.The rate at which magnetic energy is stored in the inductor is $P_L = \frac{d}{dt}(\frac{1}{2} L I^2) = L I \frac{dI}{dt}$.Equating $P_R = P_L$,we get $I^2 R = L I \frac{dI}{dt}$,which simplifies to $I R = L \frac{dI}{dt}$.Substituting $I = \frac{E}{10}(1 - e^{-t/2})$ and $\frac{dI}{dt} = \frac{E}{10} 	imes \frac{1}{2} e^{-t/2}$:$\frac{E}{10}(1 - e^{-t/2}) 	imes 10 = 20 	imes \frac{E}{20} e^{-t/2}$.$1 - e^{-t/2} = e^{-t/2}$.$1 = 2 e^{-t/2} \implies e^{-t/2} = 1/2$.Taking the natural logarithm on both sides: $-t/2 = \ln(1/2) = -\ln 2$.Therefore,$t = 2 \ln 2 \, s$.

$A$ $20 \, H$ inductor coil is connected to a $10 \, \Omega$ resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor is:

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