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

# Capacitance and series capacitors

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

- Asked in: 2073 Chaitra, 4 marks; 2071 Chaitra, 4 marks (asked 2 times)
- Video: [Capacitors in Series and Parallel Explained!](https://www.youtube.com/watch?v=zaT4JorVUz0) — 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/capacitors-in-series-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4

## Question

Define capacitance and derive relation for connection of capacitors in series.

## Short answer

Capacitance C = Q/V. In series, each capacitor stores the same charge Q, so total voltage V = Q/C1 + Q/C2 + Q/C3. Hence 1/Ceq = 1/C1 + 1/C2 + 1/C3; for two, Ceq = C1C2/(C1+C2). Equivalent capacitance is less than the smallest individual capacitance.

## Answer

#### Capacitance
**Capacitance** is the ability of a capacitor to store charge. It is the ratio of the charge stored to the potential difference across it:

C = Q/V (farad, F)

1 farad is the capacitance that stores 1 coulomb of charge at 1 volt. For a parallel plate capacitor, C = ε A/d.

#### Capacitors in series
```text
 o--| |--| |--| |--o
    C1   C2   C3
  +V1-  +V2- +V3-
 <------- V ------->
```

In a series connection the same charge *Q* is stored on every capacitor (charge cannot go anywhere else), and the supply voltage divides:

V = V₁ + V₂ + V₃

Using V = Q/C for each capacitor:

V₁ = Q/C₁, V₂ = Q/C₂, V₃ = Q/C₃

For the equivalent capacitor C_eq, V = Q/C_eq. Substituting:

Q/C_eq = Q/C₁ + Q/C₂ + Q/C₃

Dividing by Q:

**1/C_eq = 1/C₁ + 1/C₂ + 1/C₃**

For two capacitors: C_eq = C₁C₂/(C₁ + C₂).

#### Notes
- C_eq is less than the smallest capacitor.
- The voltage divides in the inverse ratio of the capacitances: V₁ = V × C_eq/C₁ (the smaller capacitor takes the larger voltage).
- Example: 4 μF and 12 μF in series give 3 μF; at 100 V, Q = 300 μC, V₁ = 75 V, V₂ = 25 V.

## Check yourself

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

1. How is capacitance defined?
   A. C = Q/I
   B. C = V/Q
   C. C = Q/V, the charge stored per volt
   D. C = QV
   Answer: C. One farad stores one coulomb at one volt. (Key point: Capacitance)

2. What is the same for all capacitors in series?
   A. Capacitance
   B. Charge Q
   C. Voltage
   D. Energy
   Answer: B. Charge cannot go elsewhere, so each capacitor stores the same Q while the voltage divides. (Key point: Series principle)

3. What is the series relation?
   A. C_eq = C1 + C2 + C3
   B. 1/C_eq = 1/C1 + 1/C2 + 1/C3
   C. 1/C_eq = C1 + C2 + C3
   D. C_eq = C1C2C3
   Answer: B. From V = V1 + V2 + V3 with V = Q/C and dividing by Q. (Key point: Result)

4. In series, which capacitor takes the largest voltage?
   A. The largest capacitor
   B. The one nearest the supply
   C. The smallest capacitor
   D. All equal
   Answer: C. V = Q/C with common Q, so voltage is inversely proportional to C. (Key point: Voltage division)

## Related in Basic Electrical Engineering

- [Three capacitors parallel and series](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Athree-capacitors-parallel-series) — Basic Electrical Engineering, chapter 4

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