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Q.

When an ideal battery of emf of 10 volt is connected across an inductor, rate of energy dissipated in it is 10 Watt. When a 10 volt 50 Hz alternating source is connected across the same inductor, power consumed by the inductor is 2.5 Watt. Then reactance of the inductor is

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a

20Ω

b

103Ω

c

52Ω

d

5Ω

answer is B.

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Detailed Solution

102R=10⇒R=10Ω = Resistance of the inductor.When A.C. source is connected, power consumption =V2RZ2∴102×10Z2=2.5⇒Z2=400⇒XL2+102=400⇒XL=103Ω
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