A $4\,\mu\text{F}$ capacitor is charged by a 200 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged $2\,\mu\text{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Energy lost $= 2.67 \times 10^{-2}\ \text{J}$.

CBSE Class 12 · 2005 · All India · CapacitanceShort Answer

4 μ F

capacitor is charged by a 200 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged

2 μ F

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

Answer: Energy lost $= 2.67 \times 1 0^{- 2} J$ .

Initial energy:

U_{i} = \frac{1}{2} \times 4 \times 1 0^{- 6} \times (200)^{2} = 8 \times 1 0^{- 2} J

Initial charge:

q = 4 \times 1 0^{- 6} \times 200 = 8 \times 1 0^{- 4} C

Common potential after connecting:

V^{'} = \frac{8 \times 1 0 ^{- 4}}{( 4 + 2 ) \times 1 0 ^{- 6}} = \frac{400}{3} V

Final energy:

U_{f} = \frac{1}{2} (4 + 2) \times 1 0^{- 6} \times (\frac{400}{3})^{2} = 5.33 \times 1 0^{- 2} J

Energy lost:

U_{i} - U_{f} = (8 - 5.33) \times 1 0^{- 2} = 2.67 \times 1 0^{- 2} J

CBSE Class 12 Board · 2005 · All India · Subjective