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

# Pure inductive circuit, zero real power

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

- Asked in: 2072 Kartik, 6 marks; 2068 Shrawan, 6 marks (asked 2 times)
- Video: [AC voltage across pure inductor (derivation) | Alternating current | Physics | Khan Academy](https://www.youtube.com/watch?v=vOJxWKOMCPM) — 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-inductive-circuit-power-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Derive the expression for electrical current in a pure inductive circuit when input power is Vm Sinωt. Draw the wave form of voltage and current and phasor diagram of the circuit. Show analytically and graphically that it does not consume real power.

## Short answer

i = (Vm/ωL) sin(ωt−90°), current lags voltage by 90°. Instantaneous power p = −(VmIm/2) sin 2ωt; its average over a cycle is zero. The inductor stores and returns energy without consuming real power.

## Answer

A pure inductive circuit has inductance L only (R = 0).

#### Derivation of current

Applied voltage v = Vm sin ωt. The induced emf is e = −L di/dt and it balances the applied voltage:

v = L di/dt  ⟹  di = (Vm/L) sin ωt dt

Integrating: i = −(Vm/ωL) cos ωt = (Vm/ωL) sin(ωt − 90°)

**i = Im sin(ωt − 90°), where Im = Vm/(ωL) = Vm/X_L**

X_L = ωL = 2πfL (ohm) is the inductive reactance. The current lags the voltage by 90°.

#### Waveforms and phasor diagram

```text
       v        i (lags 90°)
        .--.     .--.
   ____/    \___/    \___  ωt
  i reaches peak a quarter cycle after v.
```

```text
   ─────────────► V (reference)
   │
   │
   ▼
   I (lags V by 90°)
```

#### Power

Instantaneous power:

p = v i = Vm sin ωt × Im sin(ωt − 90°) = −Vm Im sin ωt cos ωt

**p = −(Vm Im / 2) sin 2ωt**

Average power over a cycle:

P = (1/2π) ∫₀^{2π} p d(ωt) = 0, since the mean of sin 2ωt is zero.

Also P = V I cos φ = V I cos 90° = 0.

#### Graphical proof

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

The power wave has twice the supply frequency. In the quarter-cycles when v and i have the same sign, p is positive and the inductor stores energy (½Li²) in its magnetic field. In the next quarter-cycles p is negative and it returns the same energy to the source. The positive and negative areas are equal, so the net energy consumed is zero. The inductor consumes no real power; it only exchanges reactive power Q = V I VAR.

## Check yourself

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

1. Which relation links applied voltage and current in a pure inductor?
   A. v = i/(L)
   B. v = L di/dt
   C. v = L ∫i dt
   D. v = R i
   Answer: B. The applied voltage balances the induced emf: v = L di/dt. (Key point: Starting equation)

2. What current results when v = Vm sin ωt is applied to a pure inductor?
   A. i = (Vm/ωL) sin ωt
   B. i = (Vm/ωL) sin(ωt + 90°)
   C. i = (Vm ωL) sin(ωt − 90°)
   D. i = (Vm/ωL) sin(ωt − 90°)
   Answer: D. Integrating gives −(Vm/ωL) cos ωt = (Vm/ωL) sin(ωt − 90°); current lags by 90°. (Key point: Current phase)

3. What is the instantaneous power in the pure inductor?
   A. p = (Vm Im/2) sin 2ωt, always positive
   B. p = −(Vm Im/2) sin 2ωt, alternating at twice supply frequency
   C. p = Vm Im sin^2 ωt
   D. p = Vm Im cos ωt
   Answer: B. p = vi = −Vm Im sin ωt cos ωt = −(VmIm/2) sin 2ωt, which has equal positive and negative loops. (Key point: Instantaneous power)

4. Why does a pure inductor consume no real power?
   A. The current is zero
   B. It stores energy in the magnetic field and returns the same energy, so average power is zero
   C. Its resistance is zero so energy is wasted as heat only
   D. The voltage is zero on average
   Answer: B. Positive and negative power loops are equal: stored energy ½Li^2 is returned to the source, so P = VI cos 90° = 0. (Key point: Why no real power)

## Related in Basic Electrical Engineering

- [Pure capacitor circuit, zero real power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Apure-capacitor-instantaneous-current) — Basic Electrical Engineering, chapter 6, asked 2×
- [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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