Prove that in charging a capacitor half of the energy supplied by the battery is stored in the capacitor and remaining half is lost during charging.
Text Solution
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When switch `S` is closed `q=CV` chare is stored in the capacitor. Charge transferred from the bttery is also `q`. Hence, energy supplied by the battery `=qV=(CV)(V)=CV^2` Half of its energy i.e. `1/2 CV^2` is stoed in the capacitor and the remaining `50%` or `1/2CV^2` is lost.
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