A capacitor of 5.0 F is initially charged to 12 V . Following this, its positive plate is attached to the negative terminal of a 12 V battery, and its negative plate is attached to the positive terminal. Determine the heat developed in the connecting wires.

Step-by-step solution

Initial charge on the capacitor: q_i = CV = 5.0 10⁻⁶ 12 = 60 C Initial energy of the capacitor: U_i = 1 2 CV^2 = 1 2 5.0 10⁻⁶ (12)^2 = 360 J When the capacitor is connected to the battery with reversed polarity, the final charge on the originally positive plate becomes q_f = -60 C . Charge that flows through the battery: q = |q_f - q_i| = |-60 - 60| = 120 C Work done by the battery: W = q V = 120 C 12 V = 1440 J Final energy of the capacitor: U_f = 1 2 CV^2 = 1 2 5.0 10⁻⁶ (12)^2 = 360 J Heat developed in the connec