---
title: "Energy stored in parallel capacitors"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:parallel-capacitors-energy-stored"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/parallel-capacitors-energy-stored
site: BCT Engineering
---

# Energy stored in parallel capacitors

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2074 Asoj, 5 marks
- Video: [How To Calculate The Energy Stored In a Capacitor](https://www.youtube.com/watch?v=PTyB6_Kt_5A) — 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/engineering_physics/parallel-capacitors-energy-stored
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

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.

## Short answer

Equivalent capacitance is 30 μF. Energy U = ½CV² = ½ × 30×10⁻⁶ × 100² = 0.15 J.

## Answer

Capacitors in parallel have the same potential difference, and their capacitances add.

Given: C₁ = 25 μF, C₂ = 5 μF, V = 100 V.

1. Equivalent capacitance: C = C₁ + C₂ = 25 + 5 = 30 μF = 30 × 10⁻⁶ F
2. Energy stored: U = ½ C V²
3. 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).**

## Check yourself

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

1. For capacitors in parallel, which statement is correct?
   A. They share the same charge and 1/C values add
   B. They share the same potential difference and 1/C values add
   C. They share the same charge and capacitances add
   D. They share the same potential difference and capacitances add
   Answer: D. C = C1 + C2 = 30 μF for a common V. (Key point: Parallel rule)

2. What is the total stored energy?
   A. 0.30 J (forgetting the ½)
   B. 0.015 J
   C. 0.15 J
   D. 1.5 x 10^-3 J
   Answer: C. U = ½ C V^2 = ½ x 30 x 10^-6 x 10^4 = 0.15 J. (Key point: Energy)

3. How is the energy shared between the 25 μF and 5 μF capacitors at the same V?
   A. Equally: 0.075 J each
   B. In proportion to capacitance: 0.125 J and 0.025 J
   C. In inverse proportion to capacitance
   D. Entirely in the smaller capacitor
   Answer: B. U = ½CV^2 at the same V is proportional to C. (Key point: Energy split)

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