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(D) The energy stored in an inductor when the current reaches its steady state value $I_0$ is given by $U = \frac{1}{2} L I_0^2$.Here,the steady state

Vedclass · Accepted Answer

$1000$

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(D) The energy stored in an inductor when the current reaches its steady state value $I_0$ is given by $U = \frac{1}{2} L I_0^2$.Here,the steady state current $I_0$ is determined by the Ohm's law for the circuit: $I_0 = \frac{V}{R}$.Since both circuits have the same e.m.f. $V = 10 \ V$ and the same resistance $R = 40 \ \Omega$,the steady state current $I_0$ is the same for both circuits.Therefore,the ratio of energy stored in circuit $A$ to circuit $B$ is $\frac{U_A}{U_B} = \frac{\frac{1}{2} L_A I_0^2}{\frac{1}{2} L_B I_0^2} = \frac{L_A}{L_B}$.Given $L_A = 10 \ H$ and $L_B = 10 \ mH = 10 	imes 10^{-3} \ H = 0.01 \ H$.Thus,the ratio is $\frac{10}{0.01} = 1000$.

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