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In the circuit shown in figure, switch S is shifted to position 2 from position 1 at t = 0, having been in position 1 for a long time. The current in the circuit just after shifting of switch will be (battery and both the inductors are ideal)

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$\frac{5}{4} \frac{E}{R}$

$\frac{5}{9} \frac{E}{R}$

$\frac{4}{5} \frac{E}{R}$

$\frac{E}{R}$

answer is C.

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Detailed Solution

Let i₁ be the current in the circuit before shifting

$i_{1} = \frac{E}{R} . . . . . (i)$

Since the flux associated with the inductors will be same just before and just after shifting, therefore

$i_{1} 5 L = i_{2} 9 L$

$i_{2} = \frac{5}{9} \frac{E}{R}$

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