- 2074 Asoj · 5 marks
Energy stored in parallel capacitors
Short answer
Equivalent capacitance is 30 μF. Energy U = ½CV² = ½ × 30×10⁻⁶ × 100² = 0.15 J.
Two capacitors having capacitance 25μF and 5μF are connected in parallel and charged with a 100V power supply. Calculate the total energy stored in the two capacitors.
Capacitors in parallel have the same potential difference, and their capacitances add.
Given: C₁ = 25 μF, C₂ = 5 μF, V = 100 V.
- Equivalent capacitance: C = C₁ + C₂ = 25 + 5 = 30 μF = 30 × 10⁻⁶ F
- Energy stored: U = ½ C V²
- U = ½ × 30 × 10⁻⁶ × (100)² = ½ × 30 × 10⁻⁶ × 10⁴ = 0.15 J
Check by adding each capacitor
- U₁ = ½ × 25 × 10⁻⁶ × 10⁴ = 0.125 J
- U₂ = ½ × 5 × 10⁻⁶ × 10⁴ = 0.025 J
- U₁ + U₂ = 0.15 J
Answer: total energy stored = 0.15 J (125 mJ in the 25 μF and 25 mJ in the 5 μF capacitor).
✓ Answer checked by Claude Opus 5.5 (high effort)
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