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A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be

1. 4.0 A
2. 8.0 A
3. 20 / √13 A
4. 2.0 A

Correct answer: 2.0 A

Solution

The inductive reactance is proportional to frequency, so at 100 Hz, the reactance doubles to 40 ohms. The impedance is Z = √(R² + X_L²) = √(30² + 40²) = 50 ohms. The current is I = V/Z = 200/50 = 4.0 A.

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