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A coil has a resistance of 30 ohm and an inductive reactance of 20 ohm when the frequency is 50 Hz. If a 200 V AC supply of frequency 100 Hz is applied across the coil, the current through it will be

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

Correct answer: 4.0 A

Solution

At 100 Hz, XL = 20*(100/50) = 40 ohm; R = 30 ohm unchanged. Z = sqrt(30^2 + 40^2) = 50 ohm. I = 200/50 = 4.0 A.

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