A coil of inductance 5.0 mH and negligib...

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:

`2` amp

`1` amp

`10` amp

`20` amp

Text Solution

Verified by Experts

The correct Answer is:

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

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