---
title: "Three coils in delta, 415 V"
subject: "Basic Electrical and Electronics Engineering"
chapter: 1
key: "semester_1/basic_electrical_and_electronics_engineering:delta-coils-415v-numerical"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/delta-coils-415v-numerical
site: BCT Engineering
---

# Three coils in delta, 415 V

Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits

- Asked in: 2081 Kartik (new course), 4 marks
- Video: [Delta and Wye - Volts, Amps, & VA](https://www.youtube.com/watch?v=dLqz_zPpgHI) — Dave Gordon
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/delta-coils-415v-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1

## Question

Three identical coils, each having resistance of 20 Ω and inductance of 0.08 H are connected in delta across a three-phase, 415 V, 50 Hz supply. Calculate: i) The phase current, ii) The line current, iii) The total power consumed, and iv) power factor.

## Short answer

Each coil impedance Z = sqrt(20² + (2π×50×0.08)²) = 32.12 Ω. In delta, phase voltage = 415 V; phase current = 415/32.12 = 12.92 A. Line current = √3 × 12.92 = 22.38 A. Total power = 3 × 12.92² × 20 = 10.02 kW. Power factor = R/Z = 20/32.12 = 0.623 lagging.

## Answer

**Given:** each coil $R=20\ \Omega$, $L=0.08$ H, delta connection, $V_L=415$ V, $f=50$ Hz. In delta, $V_{ph}=V_L=415$ V.

**Coil impedance:**

$$
X_L=2\pi\times50\times0.08=25.13\ \Omega, \qquad Z_{ph}=\sqrt{20^2+25.13^2}=32.12\ \Omega
$$

**(i) Phase current:**

$$
I_{ph}=\frac{415}{32.12}=12.92\ \text{A}
$$

**(ii) Line current:**

$$
I_L=\sqrt3\,I_{ph}=\sqrt3\times12.92=22.38\ \text{A}
$$

**(iii) Total power:**

$$
P=3I_{ph}^2R=3\times(12.92)^2\times20=10016\ \text{W}
$$

**(iv) Power factor:**

$$
\cos\phi=\frac{R}{Z}=\frac{20}{32.12}=0.623\ \text{lagging}\ (\phi=51.5^\circ)
$$

(Check: $P=\sqrt3V_LI_L\cos\phi=\sqrt3\times415\times22.38\times0.6227=10016$ W.)

**Answer: (i) 12.92 A, (ii) 22.38 A, (iii) 10.02 kW, (iv) 0.623 lagging.**

## Check yourself

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

1. In a delta connection what voltage is across each coil?
   A. 415/sqrt(3) = 240 V
   B. 415 x sqrt(3) = 719 V
   C. Full line voltage, 415 V
   D. 415/3 = 138 V
   Answer: C. In delta, the phase voltage equals the line voltage. (Key point: Delta voltage)

2. With XL = 25.13 Ω and R = 20 Ω, what is the phase current?
   A. 12.92 A
   B. 16.5 A
   C. 20.75 A
   D. 7.46 A
   Answer: A. Z = sqrt(20^2 + 25.13^2) = 32.12 Ω, so Iph = 415/32.12 = 12.92 A. (Key point: Phase current)

3. How are line and phase currents related in delta, and what is IL here?
   A. IL = Iph = 12.92 A
   B. IL = Iph/sqrt(3) = 7.46 A
   C. IL = 3 Iph = 38.8 A
   D. IL = sqrt(3) Iph = 22.38 A
   Answer: D. In delta, IL = sqrt(3) Iph = 1.732 x 12.92 = 22.38 A. (Key point: Line current)

4. Which gives total power and power factor?
   A. P = Iph^2 R = 3.34 kW; pf = 0.623
   B. P = 3 Iph^2 R = 10.02 kW; pf = R/Z = 0.623 lagging
   C. P = 3 Iph^2 XL; pf = XL/Z
   D. P = 3 V Iph = 16 kW; pf = 1
   Answer: B. Only R dissipates power: 3 x 12.92^2 x 20 = 10016 W, and cos(phi) = R/Z = 0.623 lagging. (Key point: Power and power factor)

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