A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. If the energy change during the charge redistribution is X/100J then value of X to the nearest integer is _____.

Solution

Physics sir · 3 steps

Charge is conserved, energy is not: share the charge and the drop in energy is your answer.

1. Before sharing. The 5 μF capacitor holds Q = CV = 5 × 10⁻⁶ × 220 = 1.1 × 10⁻³ C. Its energy is Uᵢ = 1/2CV² = 1/2 × 5 × 10⁻⁶ × 220² = 0.121 J.
2. Share the charge. Joining the two capacitors in a closed loop makes charge flow until both have the same voltage. Charge is conserved, so the common voltage is V = (1.1 × 10⁻³)/(7.5 × 10⁻⁶) ≈ 146.7 V. The final energy is U_f = Q²/2(C₁ + C₂) = (1.1 × 10⁻³)²/(2 × 7.5 × 10⁻⁶) ≈ 0.0807 J.

   Δ U = Uᵢ - U_f = (C₁ C₂)/2(C₁ + C₂)V² = 1/2 × (5 × 2.5)/7.5 × 10⁻⁶ × 220² ≈ 0.0403 J

3. So. Δ U = 0.0403 J = 4.03/100 J, so X ≈ 4.

The common slip: 8. Reporting the final stored energy, 0.0807 J, instead of the change gives 8. The question asks for the energy lost, Uᵢ - U_f.

The idea to keep. When a charged capacitor shares charge with another, the charge stays the same but energy is lost as heat. The loss is C₁C₂/2(C₁+C₂)(V₁-V₂)².

Answer from the NTA final answer key. Solution by PrepWiser.