1.

Initially switch S is opened and energy stored between the plates of capacitor is E. Now the switch S is closed. Work done by battery after the switch S is closed is W. Find (2E)/W.

Answer»


Solution :`E=1/2(C/2)V^(2) Q_(i)=(CV)/2 Q_(F)=(2CV)/3`
`Q_(f)-Q_(i)=(2CV)/3-(CV)/2=(CV)/6`
`E/W=(CV^(2)//4)/(CV^(2)//5)=3/2 implies(2E)/W=3`


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