Question

Two identical capacitors have the same capacitance C. one of them is charged to a potential V₁ and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

(A) (1/4)C (V₁² – V₂²)

(B) (1/4)C (V₁² + V₂²)

(C) (1/4)C (V₁ – V₂)²

(D) (1/4)C (V₁ + V₂)²

Answer 1

The correct option (C) (1/4)C (V₁ – V₂)²

Explanation:

initial energy = U_i = U₁ + U₂

U_i = (1/2)CV₁² + (1/2)CV₂²    (1)

when they are connected, the potential across each is

V = (1/2) (V₁ + V₂)

Final energy = (1/2)CV² + (1/2)CV² = CV²

U_f = C[(V₁ + V₂)/2]²   (2)

∴ Decrease in energy = U_i – U_f

= (1/2)(CV₁² + CV₂²) – C[(V₁ + V₂) / 2]² from (1) & (2)

∴ Decrease in energy = (1/2) C(V₁² + V₂²) – (1/4) C(V₁² + 2V₁V₂ + V₂²)

= (1/4) C(V₁ – V₂)²