Question

A capacitor of 20 μF and charged to 500 V is connected in parallel with another capacitor of 10 μF charged to 200 V. Find the common potential.

Answer 1

Charge on one capacitor, q₁ = C₁ V₁ = (20 × 10^–6) × 500 = 0.01 C

Charge on second capacitor, q₂ = C₂ V₂ = (10 × 10^–6) × 200 = 0.002 C

Total charge on capacitors, q = q₁ + q₂ = 0.01 + 0.002 = 0.012 C

Total capacitance, C = C₁ + C₂ = (20 × 10^–6) + (10 × 10^–6) = 30 × 10^–6 F

∴ Common potential,