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
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