---
title: "Three capacitors parallel and series"
subject: "Basic Electrical Engineering"
chapter: 4
key: "semester_1/basic_electrical_engineering:three-capacitors-parallel-series"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/three-capacitors-parallel-series
site: BCT Engineering
---

# Three capacitors parallel and series

Basic Electrical Engineering · Chapter 4: Inductance & Capacitance in Electric Circuits

- Asked in: 2072 Kartik, 6 marks
- Video: [Equivalent Capacitance - Capacitors In Series and Parallel](https://www.youtube.com/watch?v=PPhSmZ6w9WY) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/three-capacitors-parallel-series
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4

## Question

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

## 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.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why Q = CV for each in parallel?
   A. Charges are equal
   B. Capacitance equal
   C. Each has full 250 V
   D. Voltage divides
   Answer: C. Parallel capacitors have the same voltage so Q_A = 2.5 mC, Q_B = 12.5 mC, Q_C = 6.25 mC. (Key point: Parallel charge)

2. What are C_parallel and C_series?
   A. 6.25 uF and 85 uF
   B. 85 uF and 6.25 uF
   C. 85 uF and 85 uF
   D. 60 uF and 8 uF
   Answer: B. Parallel adds; series 1/C = 0.16. (Key point: Totals)

3. Why does the 10 uF take the largest p.d.?
   A. It is nearest the supply
   B. Same Q, V = Q/C, so smallest C gets largest V (156.25 V)
   C. Voltage splits equally
   D. It has the largest charge
   Answer: B. Q = 1562.5 uC; V_A = 156.25 V, V_B = 31.25 V, V_C = 62.5 V. (Key point: Series voltages)

## Related in Basic Electrical Engineering

- [Capacitance and series capacitors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitors-in-series-derivation) — Basic Electrical Engineering, chapter 4, asked 2×
- [Total power of series-parallel circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-total-power-265uf-19mh) — Basic Electrical Engineering, chapter 6
- [Parallel RL and RC branches, total current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-rl-rc-total-current-pf-power) — Basic Electrical Engineering, chapter 6, asked 2×
- [Resistors A, B in parallel with C in series](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aresistors-a-b-c-4320w) — Basic Electrical Engineering, chapter 1
- [Series-parallel AC circuit analysis](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-r1-15-l1-0-05-200v) — Basic Electrical Engineering, chapter 6, asked 2×
- [Series-parallel circuit with I1, I2, I3 phasors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-v100-z1-z2-z3) — Basic Electrical Engineering, chapter 6, asked 2×

## Related in other subjects

- [Energy stored in parallel capacitors](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aparallel-capacitors-energy-stored) — Engineering Physics, chapter 7
- [Load sharing of 50 kVA and 25 kVA transformers in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aparallel-50kva-25kva-60kva-load) — Electrical Machines I, chapter 2
- [Capacitance and charge matrices, transposed and untransposed](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acapacitance-charge-matrix-midspan) — High Voltage Engineering, chapter 5
- [Capacitance and charge matrices (9.81 m clearance)](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acapacitance-charge-matrix-9-81m) — High Voltage Engineering, chapter 5, asked 3×