A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2 . 5 μF capacitor. If the energy change during the charge redistribution is X 100 J then value of X to the nearest integer is :

Step-by-step solution

C 1 = 5 μ F    V 1 = 220 Volt C 2 = 2 . 5 μ F    V 2 = 0 Heat loss; Δ H = U 1 - U t = 1 2 C 1 C 2 C 1 + C 2 v 1 - v 2 2 = 1 2 × 5 × 2 . 5 ( 5 + 2 . 5 ) ( 220 - 0 ) 2 μ J = 5 2 × 3 × 22 × 22 × 100 × 10 - 6 J = 5 × 11 × 22 3 × 10 - 4 J = 55 × 22 3 × 10 - 4 J = 1210 3 × 10 - 4 J = 1210 3 × 10 - 3 J = 4 × 10 - 2 According to questions x 100 = 4 × 10 - 2 so, x = 4