A $5\,\mu\text{F}$ capacitor is charged by a 100 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged $3\,\mu\text{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Energy lost $= 0.94 \times 10^{-2}\ \text{J}$.

CBSE Class 12 · 2005 · All India · CapacitanceShort Answer

5 μ F

capacitor is charged by a 100 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged

3 μ F

capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Answer: Energy lost $= 0.94 \times 1 0^{- 2} J$ .

Initial energy of the charged capacitor:

U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 5 \times 1 0^{- 6} \times (100)^{2} = 2.5 \times 1 0^{- 2} J

Initial charge:

q = 5 \times 1 0^{- 6} \times 100 = 5 \times 1 0^{- 4} C

On connecting, the common potential is:

V^{'} = \frac{5 \times 1 0 ^{- 4}}{( 5 + 3 ) \times 1 0 ^{- 6}} = 62.5 V

Final energy:

U_{f} = \frac{1}{2} (5 + 3) \times 1 0^{- 6} \times (62.5)^{2} = 1.56 \times 1 0^{- 2} J

Energy lost:

U_{i} - U_{f} = (2.5 - 1.56) \times 1 0^{- 2} = 0.94 \times 1 0^{- 2} J

dissipated as heat in the connecting wires.

CBSE Class 12 Board · 2005 · All India · Subjective