In the circuit shown in Fig., $I_{2} = 3 A$ in the steady state. The potential difference across the $4 Ω$ resistor is

12 V

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18 V

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20 V

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24 V

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Solution

The correct option is C
20 V

In the steady state, the resistance of the inductor is zero, i.e. it behaves as a short circuit. Hence the circuit can be drawn as follows.

$I_{2} = 3 A$ .

Therefore, p.d. across the branch in which $I_{2}$ is flowing = $3 \times 4 = 12 V .$

Hence $I_{1} = \frac{12}{6} = 2 A .$

$I = I_{1} + I_{2} = 2 + 3 = 5 A,$ which is the current flowing through $4 Ω$ resistance.

So p.d across $4 Ω$ resistance = $5 \times 4 = 20 V .$

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