Capacitors: Numerical Problems Set (6)

hard • 3 min read
Tags Capacitors

Question

A capacitor C1=4 μC_1 = 4~\muF is charged to V0=100V_0 = 100 V and then disconnected from the source. It is then connected in parallel with an uncharged capacitor C2=6 μC_2 = 6~\muF. Find (a) the common voltage after connection, (b) the final charge on each capacitor and (c) the energy lost during the process. Where does the lost energy go?

Solution — Step by Step

Q0=C1V0=4×10−6×100=4×10−4 C=400 μCQ_0 = C_1 V_0 = 4 \times 10^{-6} \times 100 = 4 \times 10^{-4} \text{ C} = 400~\mu\text{C}

Charge is conserved (the system is isolated after disconnection). When connected in parallel, the total charge Q0Q_0 redistributes across C1+C2C_1 + C_2 at a common voltage VV.

V=Q0C1+C2=400×10−610×10−6=40 VV = \frac{Q_0}{C_1 + C_2} = \frac{400 \times 10^{-6}}{10 \times 10^{-6}} = 40 \text{ V} Q1′=C1V=4×40=160 μCQ_1' = C_1 V = 4 \times 40 = 160~\mu\text{C} Q2′=C2V=6×40=240 μCQ_2' = C_2 V = 6 \times 40 = 240~\mu\text{C}

Check: Q1′+Q2′=400 μQ_1' + Q_2' = 400~\muC =Q0= Q_0. Conservation of charge confirmed.

Initial energy: Ui=12C1V02=12×4×10−6×104=0.02U_i = \tfrac{1}{2} C_1 V_0^2 = \tfrac{1}{2} \times 4 \times 10^{-6} \times 10^4 = 0.02 J =20= 20 mJ.

Final energy: Uf=12(C1+C2)V2=12×10×10−6×1600=8U_f = \tfrac{1}{2}(C_1 + C_2) V^2 = \tfrac{1}{2} \times 10 \times 10^{-6} \times 1600 = 8 mJ.

Energy lost: ΔU=20−8=12\Delta U = 20 - 8 = 12 mJ.

V=40V = 40 V, Q1′=160 μQ_1' = 160~\muC, Q2′=240 μQ_2' = 240~\muC, energy lost =12= 12 mJ.

Why This Works

When two capacitors at different potentials are connected, charge flows momentarily until potentials equalise. This transient current passes through the connecting wires, which always have some resistance — and resistance dissipates energy as heat. Even with “ideal” wires, the energy is radiated as electromagnetic waves. The 60% loss here is independent of resistance value, which is the surprising part.

The general formula for energy lost: ΔU=12C1C2C1+C2(V1−V2)2\Delta U = \tfrac{1}{2}\frac{C_1 C_2}{C_1 + C_2}(V_1 - V_2)^2. With V2=0V_2 = 0 here, that gives 12⋅2410⋅104×10−6=12\tfrac{1}{2} \cdot \tfrac{24}{10} \cdot 10^4 \times 10^{-6} = 12 mJ — matches.

Alternative Method

Use the energy-loss formula directly: ΔU=12C1C2C1+C2(ΔV)2=12(2.4×10−6)(100)2=12\Delta U = \tfrac{1}{2}\frac{C_1 C_2}{C_1 + C_2}(\Delta V)^2 = \tfrac{1}{2}(2.4 \times 10^{-6})(100)^2 = 12 mJ. Skips the explicit calculation of UiU_i and UfU_f.

Common Mistake

Students conserve energy instead of charge. They write 12C1V02=12(C1+C2)V2\tfrac{1}{2}C_1 V_0^2 = \tfrac{1}{2}(C_1+C_2)V^2 and solve for V≈63V \approx 63 V — wrong. Energy is not conserved when capacitors share charge through real (or even ideal) wires. Always conserve charge, then compute energies separately.

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