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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 |
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Answer» SOLUTION :`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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