A 2 F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 , is

Step-by-step solution

q_i=C_i V=2 V=q (say) This charge will remain constant after switch is shifted from position 1 to position 2. aligned U_i & = 1 2 q^2 C_i = q^2 2 2 = q^2 4 U_f & = 1 2 q^2 C_f = q^2 2 10 = q^2 20 aligned Energy dissipated =U_i-U_f= q^2 5 This energy dissipated (= q^2 5 ) is 80 % of the initial stored energy (= q^2 4 ) . Analysis of Question (i) This question is moderately tough. (ii) In a capacitor circuit, redistribution of charge takes place under following three conditions. (a) A switch is closed. (b) A closed s