In the circuit shown below, the current through the inductor is

\frac{2}{1 + j} A

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\frac{- 1}{1 + j} A

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\frac{1}{1 + j} A

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0 A

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Solution

The correct option is C $\frac{1}{1 + j} A$

Apply KCL node at 'A',
so, current flowing through $1 Ω$ is
$(1 - I_{2})$
Applying KVL in ABCD Loop,

$1 ∠ 0 - 1 ∠ 0 + 1 (1 - I_{2}) - j I_{2} = 0$

$∴ I_{2} = \frac{1}{1 + j}$

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