Two capacitors of capacities 1 F and C F are connected in series and the combination is charged to a potential difference of 120 ~V . If the charge on the combination is 80 C , the energy stored in the capacitor of capacity C in J is

Step-by-step solution

Capacitances 1 F and C F are connected in series, then C_ eq = C 1+C Given, V=120 ~V and q=80 C q=C_ eq V80= C C+1 120 or C=2 F The energy stored in the capacitor of capacity CU= 1 2 q^2 C = 1 2 (80 10⁻⁶ )^2 2 10⁻⁶ = 1 2 80 10⁻⁶ 80 10⁻⁶ 2 10⁻⁶ U=1600 J