1.

A coil of resistance 20Omegaand self-inductance 5 H connected with battery 100 V. What will be the value of energy stored ?

Answer»

31.25 J
62.5 J
125 J
250 J

Solution :The ENERGY STORED in the COIL
`=1/2LI^2`
`=1/2L[E/R]^2 "" [because I=E/R]`
`=1/2xx5xx[100/20]^2=1/2xx5xx25`=62.5 J


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