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(B) In a steady state, an inductor acts as a short circuit (zero resistance).The total resistance in the circuit is $R = 5 \, \Omega$.The total curren

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$8/3 \, A$

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(B) In a steady state, an inductor acts as a short circuit (zero resistance).The total resistance in the circuit is $R = 5 \, \Omega$.The total current $I$ flowing from the battery is $I = V/R = 20 \, V / 5 \, \Omega = 4 \, A$.This current $I$ splits into two parallel branches containing inductors $L_1 = 5 \, mH$ and $L_2 = 10 \, mH$.For parallel inductors, the current divides in the inverse ratio of their inductances: $I_1 / I_2 = L_2 / L_1$.Using the current divider rule for inductors: $I_{L1} = I 	imes (L_2 / (L_1 + L_2))$.Substituting the values: $I_{L1} = 4 	imes (10 / (5 + 10)) = 4 	imes (10 / 15) = 4 	imes (2 / 3) = 8/3 \, A$.

In the given circuit, the current through the $5 \, mH$ inductor in steady state is

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