In the circuit shown, one of the inductors used is ideal. The potential difference across the batteries is10 V. The current drawn from the battery at steady state, after the switch is closed is

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0.4A

0.3A

0.2A

answer is C.

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Detailed Solution

An ideal inductor acts like a short circuit for dc, Therefore, no current flows through 10 $Ω$
resistance, The effective resistance of the circuit
R =5 + 5 = 10 $Ω$
Therefore, Current
$I = \frac{E}{R} = \frac{10}{10} = 1 A$

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