Switch S shown in Fig. is closed for $t < 0$ and is opened at $t = 0$ . When currents through $L_{1}$ and $L_{2}$ are equal, their common value is

\frac{E}{R}

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\frac{E (L_{2} + L_{1})}{R L_{1}}

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\frac{E L_{1}}{R (L_{1} + L_{2})}

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\frac{E}{R} \frac{L_{1} + L_{2})}{L_{2}}

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Solution

The correct option is B $\frac{E L_{1}}{R (L_{1} + L_{2})}$
for $t < 0$ , no current flows through $L_{2}$ as this branch is shorted by switch S.,
for $t < 0$ steady state current $= \frac{E}{R}$ .
let the value of current when equal in both inductors be $i$
Constancy of flux implies that $\frac{E}{R} L_{1} = i (L_{1} + L_{2}) i . e ., i = \frac{E L_{1}}{R (L_{1} + L_{2})}$

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A source of constant voltage $V$ is connected to a resistance $R$ and two ideal inductors $L_{1}$ and $L_{2}$ through a switch $S$ as shown. There is no mutual inductance between the two inductors. The switch $S$ is initially open. At $t = 0$ , the switch is closed and current begins to flow. Which of the following options is/are correct?