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(C) Before the switch $S$ is closed,the current $i$ flows through the single inductor $L$ and the resistor $R$. The steady-state current is $i = \frac

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$\frac{E}{2}$

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(C) Before the switch $S$ is closed,the current $i$ flows through the single inductor $L$ and the resistor $R$. The steady-state current is $i = \frac{\varepsilon}{R}$.The energy stored in the inductor is $E = \frac{1}{2} L i^2 = \frac{1}{2} L \left( \frac{\varepsilon}{R} ight)^2$.After the switch $S$ is closed,the two inductors $L$ and $L$ are connected in parallel. The equivalent inductance is $L_{eq} = \frac{L 	imes L}{L + L} = \frac{L}{2}$.After a long time,the inductors act as short circuits (ideal inductors). The total current flowing through the circuit remains $i = \frac{\varepsilon}{R}$.This current $i$ splits equally into the two parallel branches,so the current in each inductor is $i' = \frac{i}{2} = \frac{\varepsilon}{2R}$.The total energy stored in both inductors is $E' = \frac{1}{2} L (i')^2 + \frac{1}{2} L (i')^2 = L (i')^2$.Substituting $i' = \frac{i}{2}$,we get $E' = L \left( \frac{i}{2} ight)^2 = \frac{1}{4} L i^2 = \frac{1}{2} \left( \frac{1}{2} L i^2 ight) = \frac{E}{2}$.

Initially,the switch $S$ is open and the energy stored in the inductance $L$ connected in series with the battery is $E$. Now,the switch $S$ is closed. The energy stored in both the inductors after a long time is

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