Question

A variable condenser is permanently connected to a 100 V battery. If capacitor is changed from 2μF to 10μF. then energy changes is equal to 

(A) 2 × 10^–2J 

(B) 2.5 × 10^–2J 

(C) 6.5 × 10^–2J 

(D) 4 × 10^–2J

Answer 1

The correct option (D) 4 × 10^–2J  

Explanation:

Energy change = U₁ – U₂

= (1/2)C₁V² – (1/2)C₂V²

= (1/2)V²(C₁ – C₂)

= (1/2) × (100)²(2 – 10) × 10^–6

= – 4 × 10^–2 J