The energy stored in the capacitor is. (A) 4.8 × 10^–6 J (B) 9.6 × 10^–6 J

Question

The circuit shown in fig consists of the following

E₁ = 6, E₂ = 2, E₃ = 3 Volt

R₁ = 6, R₄ = 3 Ohm

R₃ = 4, R₂ = 2 Ohm

C = 5μF

E₁ = 6V E₂ = 2V   E₃ = 3V  R₁ = 6Ω

R₂ = 2Ω    R₃ = 4Ω R₄ = 3Ω

The energy stored in the capacitor is.

(A) 4.8 × 10^–6 J

(B) 9.6 × 10^–6 J

(C) 1.44 × 10^–5 J 

(D) 1.92 × 10^–5 J

Answer 1

 The correct option (C) 1.44 × 10^–5 J

Explanation:

P.D between F & E is V = E₂ + I₃R₂

= 2 + (0.2 × 2) = 2.4V.

Points E & A are at the same potential

∴ p.d betⁿ F & A is also 2.4V

Energy stored = (1/α) CV²

= (1/2) × 5 × 10^–6 × (2.4)²

= 14.4 × 10^–6 J

= 1.44 × 10^–5 J