---
title: "Pure capacitor circuit, zero real power"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:pure-capacitor-instantaneous-current"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/pure-capacitor-instantaneous-current
site: BCT Engineering
---

# Pure capacitor circuit, zero real power

Basic Electrical Engineering · Chapter 6: Single-phase AC Circuits

- Asked in: 2071 Chaitra, 6 marks; 2070 Chaitra, 4 marks (asked 2 times)
- Video: [AC voltage across a capacitor (derivation) | Alternating currents | Physics | Khan Academy](https://www.youtube.com/watch?v=79Q_SU4BDZI) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/pure-capacitor-instantaneous-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Derive the equation for instantaneous current flowing through a pure capacitor when excited by AC sinusoidal voltage V = Vm Sinωt. Draw the waveform of voltage and current and phasor diagram of the circuit. Show analytically and graphically that it does not consume real power.

## Short answer

i = C dV/dt = (Vm/Xc) sin(ωt+90°), current leads voltage by 90°. Instantaneous power p = (VmIm/2) sin 2ωt averages to zero over a cycle. A pure capacitor exchanges reactive power, consuming no real power.

## Answer

A pure capacitor has capacitance C only (no resistance, no leakage).

#### Derivation of current

Applied voltage: v = Vm sin ωt. Charge on the plates: q = C v. Current is the rate of flow of charge:

i = dq/dt = C dv/dt = C Vm ω cos ωt

**i = (Vm / X_C) sin(ωt + 90°) = Im sin(ωt + 90°)**

where X_C = 1/(ωC) = 1/(2πfC) is the capacitive reactance (ohm) and Im = Vm/X_C. So the current leads the voltage by 90°.

#### Waveforms and phasor diagram

```text
       i        v
     .--.     .--.
 ___/    \___/    \___ ωt
  i peaks a quarter cycle before v
```

```text
        I (leads by 90°)
        ▲
        │
 ───────┴──────────► V (reference)
```

#### Power

p = v i = Vm sin ωt × Im cos ωt = (Vm Im / 2) sin 2ωt

Average power over one cycle:

P = (1/2π) ∫₀^{2π} (Vm Im/2) sin 2ωt d(ωt) = 0

Also P = V I cos 90° = 0.

#### Graphical proof

```text
  p
   │  .-.      .-.
 ──┼─/───\────/───\── ωt
   │       \_/      \_/
 positive loops = negative loops
```

The power wave has twice the supply frequency. During the quarter-cycles when |v| rises, p is positive and the capacitor charges, storing energy ½CV². In the next quarter-cycles, p is negative and the capacitor returns the same energy to the source. The positive and negative areas are equal, so net energy used is zero. A pure capacitor consumes no real power; it only exchanges reactive power.

## Check yourself

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

1. Which relation gives the current in a capacitor?
   A. i = C dv/dt
   B. i = C ∫v dt
   C. i = L dv/dt
   D. i = v/C
   Answer: A. Current is the rate of change of charge, i = dq/dt = C dv/dt. (Key point: Basic relation)

2. For v = Vm sin ωt, what is the current in a pure capacitor?
   A. Im sin(ωt − 90°), lagging
   B. Im cos(ωt − 90°) lagging
   C. Im sin ωt, in phase
   D. Im sin(ωt + 90°), leading by 90°
   Answer: D. i = C Vm ω cos ωt = Im sin(ωt + 90°), so current leads voltage by 90°. (Key point: Phase of current)

3. What is X_C and how is Im related?
   A. X_C = ωC, Im = Vm/X_C
   B. X_C = 1/(ωC), Im = Vm/X_C
   C. X_C = 1/(ωC), Im = Vm X_C
   D. X_C = ωC, Im = Vm ωC
   Answer: B. X_C = 1/(2πfC) ohm and Im = Vm/X_C. (Key point: Capacitive reactance)

4. What shows the average power in a pure capacitor is zero?
   A. Voltage is constant
   B. The capacitor has high resistance
   C. The current is in phase with voltage
   D. p = (VmIm/2) sin 2ωt has equal positive and negative loops
   Answer: D. The capacitor charges (stores ½CV^2) and then returns it; P = VI cos 90° = 0. (Key point: Zero real power)

## Related in Basic Electrical Engineering

- [Pure inductive circuit, zero real power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Apure-inductive-circuit-power-derivation) — Basic Electrical Engineering, chapter 6, asked 2×
- [Phasor relation in a pure capacitor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Apure-capacitor-phasor-relationship) — Basic Electrical Engineering, chapter 4
- [Reactive power and power phasor diagram](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Areactive-power-phasor-diagram) — Basic Electrical Engineering, chapter 7
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