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(D) After a long time,the inductor acts as a short circuit (a wire with zero resistance) because the current becomes steady.The circuit now consists o

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$6$

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(D) After a long time,the inductor acts as a short circuit (a wire with zero resistance) because the current becomes steady.The circuit now consists of two resistors of $5\,\Omega$ each,connected in parallel across the $15\, V$ battery.The equivalent resistance $R_{eq}$ of the two parallel resistors is given by:$R_{eq} = \frac{R 	imes R}{R + R} = \frac{5 	imes 5}{5 + 5} = \frac{25}{10} = 2.5\,\Omega$The current $i$ through the battery is given by Ohm's law:$i = \frac{V}{R_{eq}} = \frac{15}{2.5} = 6\, A$

In the figure shown,a circuit contains two identical resistors with resistance $R = 5\,\Omega$ and an inductance with $L = 2\, mH$. An ideal battery of $15\, V$ is connected in the circuit. What will be the current through the battery long after the switch $S$ is closed (in $, A$)?

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