Q.
The capacities of two capacitors are C1 and C2 and their respective potentials are V1 and V2. If they are connected by a thin wire, then the loss of energy will be given by
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Electrostatic Potential and Capacitance
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Solution:
Initial energy, Ui=21C1V12+21C2V22.
Final energy Uf=21(C1+C2)V2( where V=C1C2C1V1+C2V2)
Hence, energy loss, ΔU=Ui−Uf=2(C1+C2)C1C2(V1−V2)2