---
title: "Coil on 240 V, power, current, inductance"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:coil-1-3kva-1-2kvar-240v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/coil-1-3kva-1-2kvar-240v
site: BCT Engineering
---

# Coil on 240 V, power, current, inductance

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

- Asked in: 2070 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/coil-1-3kva-1-2kvar-240v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

A coil takes 1.3 kVA and 1.2 kVAR when connected to a 240 V, 50 Hz sinusoidal supply. Calculate: (i) Power dissipated (ii) Current and (iii) Inductance of the coil.

## Short answer

Power dissipated is 500 W. Current is 5.42 A. Inductance of the coil is 0.130 H.

## Answer

Given S = 1.3 kVA = 1300 VA, Q = 1.2 kVAR = 1200 VAR, V = 240 V, f = 50 Hz.

#### (i) Power dissipated (active power)
From the power triangle S² = P² + Q²:

P = √(1300² − 1200²) = √(1 690 000 − 1 440 000) = √250 000 = **500 W**

(Power factor = P/S = 500/1300 = 0.385 lagging.)

#### (ii) Current
I = S/V = 1300/240 = **5.42 A**

#### (iii) Inductance of the coil
Q = I² X_L, so X_L = 1200 / 5.417² = 1200/29.34 = 40.9 Ω

L = X_L/(2πf) = 40.9/314.16 = **0.130 H** (≈ 130 mH)

The coil resistance is R = P/I² = 500/29.34 = 17.0 Ω. Check: Z = √(17.0² + 40.9²) = 44.3 Ω = 240/5.417.

## Check yourself

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

1. How is active power found from S and Q?
   A. P = √(S^2 − Q^2) = 500 W
   B. P = S + Q
   C. P = S − Q
   D. P = Q/S
   Answer: A. S^2 = P^2 + Q^2, giving P = √(1300^2 − 1200^2) = 500 W. (Key point: Power triangle)

2. How is the current found?
   A. I = Q/V
   B. I = V/S
   C. I = P/V
   D. I = S/V = 1300/240 = 5.42 A
   Answer: D. S = VI, so I = S/V. (Key point: Current)

3. How is L found?
   A. L = P/I^2
   B. X_L = Q/I^2 = 40.9 Ω, then L = X_L/(2πf) = 0.130 H
   C. L = Q/(2πf)
   D. L = V/I
   Answer: B. Reactive power is Q = I^2 X_L; then divide by ω. (Key point: Inductance)

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