---
title: "Delta connected coils line current and power"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:delta-coils-7ohm-0-03h-400v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-coils-7ohm-0-03h-400v
site: BCT Engineering
---

# Delta connected coils line current and power

Basic Electrical Engineering · Chapter 8: Three-Phase Circuit Analysis

- Asked in: 2073 Shrawan, 8 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/delta-coils-7ohm-0-03h-400v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

Three similar coils each of resistance 7Ω and inductance of 0.03 H are connected in Delta to a 400 V, 3 phase, 50 Hz supply. Calculate the line current and the total power consumed.

## Short answer

Coil impedance Z = 7 + j9.425 = 11.74∠53.4° Ω. Phase current = 400/11.74 = 34.07 A. Line current = √3 × 34.07 = 59.0 A. Total power = 3 × 34.07² × 7 = 24.4 kW.

## Answer

Each coil: R = 7 Ω, L = 0.03 H, delta connected to V_L = 400 V, 50 Hz.

- X_L = 2πfL = 314.16 × 0.03 = 9.425 Ω
- Z_ph = 7 + j9.425 = 11.74∠53.4° Ω
- Power factor = R/Z = 7/11.74 = 0.596 lagging

#### Phase current
In delta, V_ph = V_L = 400 V.

I_ph = 400/11.74 = 34.07 A

#### Line current
I_L = √3 I_ph = 1.732 × 34.07 = **59.0 A**

#### Total power
P = 3 I_ph² R = 3 × 34.07² × 7 = **24 380 W ≈ 24.4 kW**

Check: √3 V_L I_L cos φ = 1.732 × 400 × 59.01 × 0.596 = 24.4 kW.

## Check yourself

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

1. What is the coil reactance?
   A. X_L = 0.03 Ω
   B. X_L = 2πfL = 9.425 Ω
   C. X_L = 1/(2πfL)
   D. X_L = 7 Ω
   Answer: B. X_L = 2π x 50 x 0.03 = 9.425 Ω. (Key point: Reactance)

2. What voltage is across each coil?
   A. 230.9 V
   B. 400 V
   C. 692 V
   D. 133 V
   Answer: B. In delta V_ph = V_L = 400 V. (Key point: Phase voltage in delta)

3. How is the line current found from I_ph = 34.07 A?
   A. I_L = I_ph
   B. I_L = √3 I_ph = 59.0 A
   C. I_L = I_ph/√3
   D. I_L = 3 I_ph
   Answer: B. In delta I_L = √3 I_ph = 59.0 A. (Key point: Line current)

4. What is the total power?
   A. 3 I_ph² R = 24.4 kW
   B. I_ph² R = 8.1 kW
   C. √3 x 400 x 59 = 40.9 kW
   D. 3 I_ph² Z = 40.9 kW
   Answer: A. Power is dissipated only in the resistance: 3 x 34.07² x 7 = 24 380 W. (Key point: Total power)

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