CBSE Class 12 · 2014 · Delhi · CapacitanceShort Answer

A parallel plate capacitor of capacitance

C

is charged to a potential

V

. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Answer: $1 : 2$ (final energy is half the initial energy).

Solution

Initial energy:

U_{i} = \frac{1}{2} C V^{2}

On connecting an equal uncharged capacitor, charge is conserved, so the common potential is:

V^{'} = \frac{C V}{2 C} = \frac{V}{2}

Final energy:

U_{f} = \frac{1}{2} (2 C) (\frac{V}{2})^{2} = \frac{1}{4} C V^{2}

Ratio:

\frac{U _{f}}{U _{i}} = \frac{\frac{1}{4} C V ^{2}}{\frac{1}{2} C V ^{2}} = \frac{1}{2}

CBSE Class 12 Board · 2014 · Delhi · Subjective

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