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Q.

A coil of inductance 0.1 H is connected to 50 V 100 Hz generator and current is found to be 0.5 A. The potential difference across resistance of coil is

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a

15 V

b

20 V

c

25 V

d

39 V

answer is D.

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Detailed Solution

I=EZ⇒0.5=50Z⇒Z=100 ΩZ2=R2+ω2L2,  then  R=78 ΩNow, VR=VLR2−VL2=39 V∴VR2+VL2=VLR2
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A coil of inductance 0.1 H is connected to 50 V 100 Hz generator and current is found to be 0.5 A. The potential difference across resistance of coil is