A coil of inductance 1, H and resistance 10, | vedclass

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(C) The rate at which magnetic energy is stored in the coil is given by $P_m = \frac{d}{dt}(\frac{1}{2}Li^2) = Li \frac{di}{dt}$.The rate at which ene

Vedclass · Accepted Answer

$0.36$

Vedclass · Answer

(C) The rate at which magnetic energy is stored in the coil is given by $P_m = \frac{d}{dt}(\frac{1}{2}Li^2) = Li \frac{di}{dt}$.The rate at which energy is supplied by the battery is $P_b = Ei$.The ratio of these rates is $\frac{P_m}{P_b} = \frac{Li \frac{di}{dt}}{Ei} = \frac{L}{E} \frac{di}{dt}$.For an $RL$ circuit,the current at time $t$ is $i = \frac{E}{R}(1 - e^{-Rt/L})$.The rate of change of current is $\frac{di}{dt} = \frac{E}{R} \cdot \frac{R}{L} e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.Substituting this into the ratio expression:$\frac{P_m}{P_b} = \frac{L}{E} \left( \frac{E}{L} e^{-Rt/L} ight) = e^{-Rt/L}$.Given $R = 10\, \Omega$,$L = 1\, H$,and $t = 0.1\, s$:$\frac{P_m}{P_b} = e^{-(10 	imes 0.1) / 1} = e^{-1} \approx \frac{1}{2.718} \approx 0.368 \approx 0.36$.

$A$ coil of inductance $1\, H$ and resistance $10\, \Omega$ is connected to a battery of $emf$ $50\, V$ and negligible internal resistance at time $t = 0.$ The ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at $t = 0.1\, s$ is:

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