An uncharged capacitor is connected to a battery. Show that half the energy supplied by battery is lost as heat while charging the conductor.

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Solution

Step 1: Given data
An uncharged capacitor is connected to a battery.
Step 2: To show
Half the energy supplied by the battery is lost as heat while charging the conductor.
Step 3: Proof
Work done by the battery, $W = Q V$
where Q is charge and V is voltage.
Energy stored in the capacitor, $E = \frac{1}{2} C V^{2} = \frac{1}{2} Q V (C = Q / V)$
The remaining energy is converted into heat energy,
Heat energy,
$E_{h e a t} = W - E \Rightarrow E_{h e a t} = Q V - \frac{1}{2} Q V \Rightarrow E_{h e a t} = \frac{1}{2} Q V \Rightarrow E_{h e a t} = \frac{1}{2} W$
Therefore, it is proved that half the energy supplied by the battery is lost as heat while charging the conductor.

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