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(A) At $t = 0$,the switch $S$ is open. The inductor acts as an open circuit,but the current flows through the $20 \, V$ battery,the $5 \, \Omega$ resi

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(A) At $t = 0$,the switch $S$ is open. The inductor acts as an open circuit,but the current flows through the $20 \, V$ battery,the $5 \, \Omega$ resistor,the $5 \, \Omega$ resistor,and the $10 \, V$ battery. The current $I_0$ through the inductor branch is $0 \, A$ because the switch is open.At steady state $(t 	o \infty)$,the inductor acts as a short circuit (zero resistance). The capacitor acts as an open circuit. The circuit simplifies to a loop with the $20 \, V$ battery,the $5 \, \Omega$ resistor,and the inductor branch (with $5 \, \Omega$ resistance). The $10 \, V$ battery branch is in parallel with the inductor branch. The current $I$ through the inductor is determined by the $20 \, V$ source and the $5 \, \Omega$ resistor in series with the inductor branch: $I = \frac{20 \, V}{5 \, \Omega} = 4 \, A$.The change in flux is $\Delta \phi = L \Delta I = L(I_{final} - I_{initial})$.Given $L = 500 \, mH = 0.5 \, H$,$I_{initial} = 0 \, A$,and $I_{final} = 4 \, A$.$\Delta \phi = 0.5 	imes (4 - 0) = 2 \, Wb$.

Switch $S$ is closed at $t = 0$ in the circuit shown. The change in magnetic flux in the inductor $(L = 500 \, mH)$ from $t = 0$ to an instant when it reaches steady state is ...... $Wb$.

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