An Uncharged Capacitor Is Connected To A Battery. Show That Half The Energy Supplied By Battery Is Lost As Heat While Charging The Conductor

The charge stored on capacitor = Q

If Q charge has passed through battery work done by battery = QV

Energy stored in capacitor = \(frac{1}{2} CV^{2} = frac{1}{2} QV\)

Since Q = CV

Therefore,

The energy lost as heat = Work done by battery − Energy store in a capacitor

= \(QV – frac{1}{2} QV

= \(frac{1}{2} QV\)

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