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

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 $1 + 3 = 4 Ω$ resistance = $3 \times 4 = 12 V$ .Hence $I_{1} = \frac{12}{6} = 2 A$ . Therefore $I = I_{1} + I_{2} = 2 + 3 = 5 A$ , 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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In the circuit shown in figure, I2=3A in the steady state. The potential difference across the 4Ω resistor

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