---
title: "Phasor relation in a pure capacitor"
subject: "Basic Electrical Engineering"
chapter: 4
key: "semester_1/basic_electrical_engineering:pure-capacitor-phasor-relationship"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/pure-capacitor-phasor-relationship
site: BCT Engineering
---

# Phasor relation in a pure capacitor

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

- Asked in: 2068 Chaitra, 5 marks
- Video: [Phasor Relationship for Capacitor](https://www.youtube.com/watch?v=AejbzRS_MQs) — Neso Academy
- 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-phasor-relationship
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4

## Question

Describe and illustrate the phasor relationship that exist between the voltage that appears across the terminals of a pure capacitor and the current that flows through it in steady state when the capacitor is excited by a sinusoidal source.

## Short answer

In a pure capacitor, the current leads the voltage by 90° (π/2). The phasor diagram shows voltage I lagging, with magnitude V/X_C where X_C=1/(ωC). In complex form, V = V∠0°, I = (V/X_C)∠90°.

## Answer

**Statement:** In a pure capacitor the current leads the voltage by 90° (π/2). Equivalently, the voltage lags the current by 90°.

#### Description

Let the applied voltage be v = Vm sin ωt. The charge is q = C v, so the current is

i = dq/dt = C dv/dt = ωC Vm cos ωt = (Vm / X_C) sin(ωt + 90°)

where X_C = 1/(ωC) is the capacitive reactance in ohm. So Im = Vm / X_C and in rms terms I = V / X_C.

- The current is maximum when the voltage passes through zero (rate of change of v is highest).
- The current is zero when the voltage is at its peak.

#### Phasor diagram (V as reference)

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

In complex form: V = V∠0°, I = (V / X_C)∠90° = V / (−jX_C).

#### Waveforms

```text
      v        i
        .--.  .--.
   ____/    \/    \____ ωt
        i leads v by 90°
```

The current wave reaches its peak a quarter-cycle before the voltage wave does.

#### Power
p = v i = (Vm Im / 2) sin 2ωt. Its average over a cycle is zero, so a pure capacitor consumes no real power. It takes energy in one quarter-cycle and returns it in the next.

## Check yourself

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

1. Phase relationship?
   A. Current leads voltage by 90 degrees
   B. Current leads by 180
   C. Current lags by 90
   D. In phase
   Answer: A. i = C dv/dt. (Key point: Phase)

2. X_C is?
   A. omega/C
   B. 1/(omega C)
   C. omega C
   D. C/omega
   Answer: B. Ohm; I = V/X_C. (Key point: Reactance)

3. Average power?
   A. Zero
   B. Vm Im
   C. VI
   D. Vm Im/2
   Answer: A. p = (VmIm/2) sin 2wt averages zero. (Key point: Power)

4. When is current max?
   A. When voltage passes through zero
   B. Never
   C. When voltage is max
   D. When q is max
   Answer: A. Rate of change of v highest at zero crossing. (Key point: Instants)

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