1.

A capacitor of capacitnace C is charged to a potential difference V and then disconnected from the battery. Now it is connected to an inductor of inductance L at t=0. Then

Answer»

Energy stored in capacitor and INDUCTOR will be equal at TIME `t=(pi)/(2)sqrtLC`
POTENTIAL difference across inductor will be `(V)/(2)` at time`t=(pi)/(3)sqrtLC`
The RATE of increase of energy in magnetic field will be maximum at `(pi)/(4)sqrtLC`
When the potentail difference across the capacitor `SQRT(((3C)/(L)))`

Answer :B::C::D


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