Find V_A - V_B in steady state. (in , V) | vedclass

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(C) In steady state,the inductor acts as a short circuit (zero resistance).The upper branch has a $6 \, \Omega$ resistor in series with a parallel com

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$24$

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(C) In steady state,the inductor acts as a short circuit (zero resistance).The upper branch has a $6 \, \Omega$ resistor in series with a parallel combination of a $4 \, H$ inductor and a $2 \, \Omega$ resistor. Since the inductor is a short circuit,the $2 \, \Omega$ resistor is bypassed. Thus,the resistance of the upper branch is $6 \, \Omega$.The lower branch has a $2 \, H$ inductor in series with a $2 \, \Omega$ resistor and a $4 \, \Omega$ resistor. The inductor acts as a short circuit,so the resistance of the lower branch is $2 \, \Omega + 4 \, \Omega = 6 \, \Omega$.The total current entering the parallel combination is $2 \, A$ in the upper branch. Since both branches have equal resistance $(6 \, \Omega)$,the current in the lower branch is also $2 \, A$.The total current flowing through the $3 \, \Omega$ resistor is $I_{total} = 2 \, A + 2 \, A = 4 \, A$.The potential difference $V_A - V_B$ is the sum of the potential drop across the upper branch and the $3 \, \Omega$ resistor: $V_A - V_B = (I_{upper} 	imes R_{upper}) + (I_{total} 	imes R_{series}) = (2 \, A 	imes 6 \, \Omega) + (4 \, A 	imes 3 \, \Omega) = 12 \, V + 12 \, V = 24 \, V$.

Find $V_A - V_B$ in steady state. (in $, V$)

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