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A 400 pF capacitor charged by a 100 V dc...

A 400 pF capacitor charged by a 100 V dc supply is disconnected from the supply and connected to another uncharged 400 pF capacitor calculate the loss of energy

Text Solution

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`U_i = 1/2 CV^2 = 1/2 qV = 2 xx 10^(-6) J`
After connecting charged capacitor to uncharged capacitor of same capacity 1/2 of the charge will flow
` therefore U_f = 1/2 q. V = 1/2 (q/2) V = (qV)/(4) = 1 xx 10^(-6) J`
Loss in energy ,` U_i - U_f = 1 xx 10^(-6) J`
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