A 600pF capacitor is charged with battery 100V.calculate the electrostatic potential energy stored and charge on capacitor
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Sol. C1 = C2 = 600 pF
V1 = 100 V , V2 = 0
Loss of Energy = C1 C2(V1 -V2)2 /2(C1+C2)
By putting the given values we get ,
Loss of energy = 6 x 10–5 J
V1 = 100 V , V2 = 0
Loss of Energy = C1 C2(V1 -V2)2 /2(C1+C2)
By putting the given values we get ,
Loss of energy = 6 x 10–5 J
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