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A coil has a resistance of 30 ohm and an inductive reactance of 20 ohm at 50 Hz. If an AC source of 200 V (rms) at 100 Hz is connected across this coil, what is the current through the coil?

1. 2 A
2. 4 A
3. 5 A
4. 10 A

Correct answer: 4 A

Solution

At 50 Hz, XL = 20 ohm. Since XL = 2*pi*f*L, L = 20/(100*pi) H. At 100 Hz, XL = 2*pi*100*L = 2*20 = 40 ohm. Impedance Z = sqrt(R² + XL²) = sqrt(900 + 1600) = sqrt(2500) = 50 ohm. Current I = V/Z = 200/50 = 4 A.

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