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(C) In steady state,an inductor behaves as a conducting wire (short circuit).Therefore,the two inductors in the circuit act as simple wires.The circui

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$10$

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(C) In steady state,an inductor behaves as a conducting wire (short circuit).Therefore,the two inductors in the circuit act as simple wires.The circuit now consists of three resistors of $3 \, \Omega$ each,connected in parallel.The equivalent resistance of these three parallel resistors is given by:$\frac{1}{R_{p}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \, \Omega^{-1}$$\Rightarrow R_{p} = 1 \, \Omega$This parallel combination is in series with the $2 \, \Omega$ resistor.Total equivalent resistance of the circuit is $R_{eq} = 2 \, \Omega + R_{p} = 2 \, \Omega + 1 \, \Omega = 3 \, \Omega$.The current $i$ through the battery is given by Ohm's law:$i = \frac{V}{R_{eq}} = \frac{30 \, V}{3 \, \Omega} = 10 \, A$.

For the given circuit,the current $i$ through the battery when the key is closed and the steady state has been reached is .....$A$.

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