Three capacitors of capacitance 2μF, 4μFand 8μF are connected in parallel and a P.D of 6V is maintained across each capacitor, the total energy stored is?

Analyze the question: Three capacitors in parallel, each with 6 V across it. We need the total energy stored.

Concept:

• In parallel, capacitances add: C_total = C₁ + C₂ + C₃
• Energy stored in a capacitor: E = ½CV²

Solution:

1. Total capacitance: C = 2 + 4 + 8 = 14 μF = 14 × 10⁻⁶ F

2. Energy stored: E = ½ × C × V² E = ½ × (14 × 10⁻⁶) × (6)² E = ½ × 14 × 10⁻⁶ × 36 E = 7 × 36 × 10⁻⁶ E = 252 × 10⁻⁶ J E = 2.52 × 10⁻⁴ J

Check: Individual energies are ½(2)(36) = 36 μJ, ½(4)(36) = 72 μJ and ½(8)(36) = 144 μJ. These add up to 252 μJ = 2.52 × 10⁻⁴ J. ✓

Tip: Watch the powers of 10. Option A (2.52 × 10⁻⁶) is a trap for students who forget that 252 × 10⁻⁶ must be converted to standard form: 252 × 10⁻⁶ = 2.52 × 10⁻⁴. For parallel capacitors, add the capacitances first, then use ½CV².