Find the steady state current through $L_{1}$ in the Fig. -

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None of these

$\frac{V_{0} L_{1}}{R (L_{1} + L_{2})}$

$\frac{V_{0} L_{2}}{R (L_{1} + L_{2})}$

$\frac{V_{0}}{R}$

answer is C.

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Detailed Solution

After a long time steady current through the resistor , $I_{0} = \frac{V_{0}}{R}$ .

At time t , $L_{1} \frac{d i_{1}}{d t} = L_{2} \frac{d i_{2}}{d t} \Rightarrow I n t e g r a t i n g, L_{1} I_{1} = I_{2} L_{2} . . . . . . (1)$

Also $I_{1} + I_{2} = I_{0} . . . . . . (2)$

Solving $I_{1} = \frac{V_{0} L_{2}}{R (L_{1} + L_{2})}$

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