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A coil has \(L=0.04\) H and \(R=12\ \Omega\). When it is connected to a 220 V, 50 Hz supply, the current flowing through the coil, in amperes, is

1. 10.7
2. 11.7
3. 14.7
4. 12.7

Correct answer: 12.7

Solution

The coil behaves like a series RL circuit. Compute \(X_L=2\pi fL=2\pi\times 50\times 0.04\approx 12.57\,\Omega\), so \(Z=\sqrt{12^2+12.57^2}\approx 17.4\,\Omega\). Hence \(I=V/Z\approx 220/17.4\approx 12.7\) A.

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