Capacitors: Edge Cases and Subtle Traps (2)

medium • 2 min read
Tags Capacitors

Question

A capacitor of capacitance C1=4 μC_1 = 4\,\muF is charged to a potential of V1=100V_1 = 100 V and then disconnected from the source. It is then connected in parallel to an uncharged capacitor C2=6 μC_2 = 6\,\muF. Find the common potential after sharing, and the energy lost in the process. Where does this lost energy go?

Solution — Step by Step

Q1=C1V1=4×100=400 μCQ_1 = C_1 V_1 = 4 \times 100 = 400 \, \mu\text{C}

Q2=0Q_2 = 0 initially.

Total charge stays 400 μ400\,\muC and is now shared between the two capacitors at the same potential VV.

V=QtotalC1+C2=40010=40 VV = \frac{Q_{total}}{C_1 + C_2} = \frac{400}{10} = 40 \, \text{V}

Ui=12C1V12=12(4)(104)=2×104 μU_i = \tfrac{1}{2} C_1 V_1^2 = \tfrac{1}{2}(4)(10^4) = 2 \times 10^4\,\muJ =0.02= 0.02 J.

Uf=12(C1+C2)V2=12(10)(1600)=8000 μU_f = \tfrac{1}{2}(C_1 + C_2) V^2 = \tfrac{1}{2}(10)(1600) = 8000\,\muJ =0.008= 0.008 J.

 ΔU=Ui−Uf=0.012 J=60%\Delta U = U_i - U_f = 0.012 \, \text{J} = 60\%

The charge transfer involves a current through the connecting wires. Even with “ideal” zero-resistance wires, electromagnetic radiation and transient oscillations carry away exactly this energy. Real wires dissipate it as heat.

Final answer: common potential =40= 40 V, energy lost =0.012= 0.012 J (60% of original), dissipated as heat/radiation in the wires.

Why This Works

Charge is conserved (no external source). Energy is not, because the charge transfer happens against varying potential difference, and the work done crosses through the wire’s resistance no matter how small.

The 60% loss is independent of wire resistance — it is set by the capacitor values alone. This is one of those non-intuitive results that JEE loves to test.

Alternative Method

Compute lost energy directly:

ΔU=C1C2(V1−V2)22(C1+C2)=4⋅6⋅10022⋅10=12000 μJ=0.012 J\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} = \frac{4 \cdot 6 \cdot 100^2}{2 \cdot 10} = 12000\,\mu\text{J} = 0.012 \, \text{J}

Same answer in one line.

Students often write “energy is conserved” since “wires have no resistance”. Wrong — charge redistribution through any connecting medium dissipates energy. This is why high-frequency power transfer always uses inductors, not direct capacitor sharing.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Capacitor problems — charging, discharging, energy stored, combinations

medium

Capacitor problems — charging, discharging, energy stored, combinations

medium

Capacitor with dielectric inserted — how does capacitance change

hard

Capacitors: Diagram-Based Questions (4)

easy

Capacitors: Exam-Pattern Drill (1)

easy