Correct Answer: Option B

Explanation

V = 220V, F = 60Hz and L = 3.5H

I = \(\frac{V}{X_L}\)

but \(X_L = 2 \pi FL\) = 2 x 3.142 x 60 x 3.5 = 1319.64Ω

I = \(\frac{220}{1319.6}\) = 0.1667A OR 0.1670A

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