A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage `V=10 sin (100t)`. The peak current in the circuit will be:
A. `2` amp
B. `1` amp
C. `10` amp
D. `20` amp

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1 Answers

Correct Answer - D
`I_(omega)=V_(0)/(omegaL)=10/(100xx5xx10^(-3))`

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