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Question

When a 100 V DC is applied across a solenoid, a current of 1.0 A flows in it. When a 100 V AC is applied across the same coil, the current drops to 0.5 A. If the frequency of the AC source is 50 Hz, the impedance and inductance of the solenoid are

A
200 Ω and 0.55 H
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B
100 Ω and 0.86 H
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C
200 Ω and 1.0 H
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D
100 Ω and 0.93 H
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Solution

The correct option is A 200 Ω and 0.55 H
From given, Z=VrmsIrms=100 V0.5 A=200 Ω
Now from DC circuit,
R=100V1A=100 Ω
For detailed solution watch the next video
Now in AC circuit
Z=R2+X2L ; XL=2πfL
Z=(100)2+(2π×50×L)2
(200)2=(2π×50×L)2+(100)2L=0.55 H

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