Skip to main content

Three capacitors parallel and series

Short answer

Parallel charges: Q=CV gives 2.5 mC, 12.5 mC, 6.25 mC. Total parallel capacitance = 85 µF, series C_s = 6.25 µF. In series, common charge Q = C_s V = 1562.5 µC, so p.d. across each: 156.25 V, 31.25 V, 62.5 V.

Three capacitors A, B and C have capacitances 10, 50 and 25 μF respectively. Calculate: i) Charge on each when connected in parallel to a 250 V supply ii) Total capacitance and iii) p.d. across each when connected in series

Given: C_A = 10 μF, C_B = 50 μF, C_C = 25 μF, supply V = 250 V.

i) Parallel connection: charge on each

Each capacitor has the full 250 V across it, so Q = CV.

  • Q_A = 10 μF × 250 = 2500 μC = 2.5 mC
  • Q_B = 50 μF × 250 = 12500 μC = 12.5 mC
  • Q_C = 25 μF × 250 = 6250 μC = 6.25 mC

Total charge = 21250 μC (21.25 mC).

ii) Total capacitance

  • Parallel: C_p = 10 + 50 + 25 = 85 μF
  • Series: 1/C_s = 1/10 + 1/50 + 1/25 = 0.1 + 0.02 + 0.04 = 0.16, so C_s = 6.25 μF

iii) p.d. across each when connected in series (across 250 V)

In series every capacitor carries the same charge:

Q = C_s V = 6.25 μF × 250 = 1562.5 μC

  • V_A = Q/C_A = 1562.5 / 10 = 156.25 V
  • V_B = Q/C_B = 1562.5 / 50 = 31.25 V
  • V_C = Q/C_C = 1562.5 / 25 = 62.5 V

Check: 156.25 + 31.25 + 62.5 = 250 V.

Answer: Parallel charges 2.5 mC, 12.5 mC, 6.25 mC; C_parallel = 85 μF, C_series = 6.25 μF; series p.d. = 156.25 V, 31.25 V, 62.5 V (the smallest capacitor takes the largest voltage). The 250 V supply is assumed for the series case also.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 4 questions →