A coil of resistance 10Omega and an indu...

A coil of resistance `10Omega` and an inductance `5H` is connected to a `100` volt battery. Then energy stored in the coil is

125 erg

125 J

250 erg

250 J

Text Solution

Verified by Experts

The correct Answer is:

`U=1/2Li^(2)=1/2xx5xx(100/10)^(2)=1/2xx5xx100=250J`

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