A 4 F capacitor is charged by a 200 ~V battery. It is then disconnected from the supply and is connected to another uncharged 2 F capacitor. During the process, loss of energy (in J ) is

Step-by-step solution

Charge stored at the capacitor q=C₁ V₁=4 200=800 C When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors V= C₁ V₁+C₂ V₂ C₁+C₂ = 800+0 4+2 = 800 6 ~V Loss in energy = Initial energy - Final energy aligned = & 1 2 C₁ V₁^2- 1 2 (C₁+C₂ ) V^2 = & 1 2 4 10⁻⁶ (200)^2 & - 1 2 (4+2) 10⁻⁶ ( 800 6 )^2 = & 2 10⁻⁶ 4 10^4- 3 10⁻⁶ 64 10^4 36 = & 8 10⁻²- 64 12 10⁻² = & 8 10⁻²-5.33 10⁻² = & 2.67 10⁻² ~J aligned