Two capacitors, one with a capacitance of 12 F and the other with an unknown capacitance C (where C > 12 F ), are connected in parallel to a DC supply. The energy stored in this combination is U_p . When the same two capacitors are connected in series to the same DC supply, the energy stored is U_s . If U_p = 4.5 U_s , the value of C (in F ) is _____.

Step-by-step solution

Let the supply voltage be V . In parallel combination, the equivalent capacitance is: C_p = 12 + C The energy stored is: U_p = 1 2 C_p V^2 = 1 2 (12 + C) V^2 In series combination, the equivalent capacitance is: C_s = 12C 12 + C The energy stored is: U_s = 1 2 C_s V^2 = 1 2 ( 12C 12 + C ) V^2 Given that U_p = 4.5 U_s , we have: 1 2 (12 + C) V^2 = 4.5 1 2 ( 12C 12 + C ) V^2 Dividing both sides by 1 2 V^2 : 12 + C = 4.5 ( 12C 12 + C ) (12 + C)^2 = 54C 144 + 24C + C^2 = 54C C^2 - 30C + 144 = 0 Solving the quadratic eq