Two Capacitors, One Switch, and a Stored-Energy Puzzle
Capacitor C₁ = 6.0 μF is charged to 12 V by a battery that is then removed. C₁ is then connected across an initially uncharged capacitor C₂ = 3.0 μF through a switch. After the switch is closed, the system reaches equilibrium. A student computes the final energy by taking the initial energy and dividing by the ratio of capacitances, obtaining a value that does not match energy conservation. The instructor notes that charge is conserved but energy is not, because the connecting wires have resistance and the redistribution process dissipates energy as heat (and possibly radiation).
1. Find the final common voltage and the final stored energy, and compare with the initial energy.
Read the worked answer
Initial charge Q₀ = C₁V₀ = (6.0 μF)(12 V) = 72 μC; initial energy U₀ = ½C₁V₀² = ½(6.0×10⁻⁶)(144) = 4.32×10⁻⁴ J. Charge is conserved: Q₀ = (C₁+C₂)V_f → V_f = 72 μC/9.0 μF = 8.0 V. Final energy U_f = ½(C₁+C₂)V_f² = ½(9.0×10⁻⁶)(64) = 2.88×10⁻⁴ J. The difference, 1.44×10⁻⁴ J, is dissipated in the wires regardless of the resistance value (as R→0 the dissipation rate increases but the total stays the same).
2. Explain why the final energy is independent of the wire resistance, and predict what happens if a dielectric of κ = 2 is inserted into C₂ before closing the switch.
Read the worked answer
The energy lost equals the integral of I²R dt. Since the charge-transfer process is a simple RC transient, the total dissipated energy equals ½C_eq(ΔV)² for the equivalent redistribution and is fixed by the initial and final charge configurations, not by R. With a dielectric in C₂, C₂' = 6.0 μF, so C_eq = 12 μF. Charge conserved: V_f' = 72 μC/12 μF = 6.0 V. U_f' = ½(12 μF)(36) = 2.16×10⁻⁴ J, so more energy is lost (2.16×10⁻⁴ J dissipated) - the dielectric increases the capacitance and lowers the final voltage, increasing dissipation.