A capacitor of capacitance $C_1 = 1.0 \mu F$, initially charged to a voltage $V_0 = 300$ V, is conne

Question

A capacitor of capacitance $C_1 = 1.0 \mu F$, initially charged to a voltage $V_0 = 300$ V, is connected in parallel with an uncharged capacitor of capacitance $C_2 = 2.0 \mu F$ through a switch. All connecting wires have small but finite resistance. What is the total amount of heat generated in the circuit from the moment the switch is closed until the system reaches a steady electrostatic equilibrium (in mJ)?

Professor PC Thomas Classes · Answer

The initial energy of the system is entirely stored in $C_1$: $U_i = \frac{1}{2} C_1 V_0^2 = \frac{1}{2} (1 	imes 10^{-6} 	ext{ F}) (300 	ext{ V})^2 = 45 	imes 10^{-3} 	ext{ J} = 45 	ext{ mJ}$. By conservation of charge, the initial charge $Q = C_1 V_0 = 300 \mu C$ is redistributed over the equivalent parallel capacitance $C_{eq} = C_1 + C_2 = 3.0 \mu F$. The common steady-state voltage is $V_f = \frac{Q}{C_1 + C_2} = \frac{300 \mu C}{3.0 \mu F} = 100 	ext{ V}$. The final energy stored in the capacitors is $U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} (3.0 	imes 10^{-6} 	ext{ F}) (100 	ext{ V})^2 = 15 	imes 10^{-3} 	ext{ J} = 15 	ext{ mJ}$. The heat generated in the connecting wires is the difference between the initial and final stored energies: $H = U_i - U_f = 45 	ext{ mJ} - 15 	ext{ mJ} = 30 	ext{ mJ}$.