---
title: "Delta load line currents and power, ABC sequence"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:delta-unbalanced-52-45-380v-abc"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-unbalanced-52-45-380v-abc
site: BCT Engineering
---

# Delta load line currents and power, ABC sequence

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

- Asked in: 2068 Chaitra, 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-unbalanced-52-45-380v-abc
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A delta connected load of ZAB = 52∠45°Ω, ZBC = 52∠-30°Ω and ZCA = 10∠0°Ω are connected to a 380V, 3 phase ac source. Find the magnitude of the line currents and total power absorbed by loads, when phase sequence is ABC.

## Short answer

Compute phase currents from Vph/Zph for each delta branch, then line currents via difference. |IA|=45.1 A, |IB|=5.59 A, |IC|=44.5 A. Total power = sum of Iph²×R = 18.8 kW.

## Answer

Delta load: Z_AB = 52∠45° Ω, Z_BC = 52∠−30° Ω, Z_CA = 10∠0° Ω. V_L = 380 V, sequence ABC.

V_AB = 380∠0°, V_BC = 380∠−120°, V_CA = 380∠120° V (phase voltage = line voltage in delta).

#### Phase currents
- I_AB = 380∠0° / 52∠45° = 7.31∠−45° = 5.17 − j5.17 A
- I_BC = 380∠−120° / 52∠−30° = 7.31∠−90° = 0 − j7.31 A
- I_CA = 380∠120° / 10∠0° = 38.0∠120° = −19.0 + j32.91 A

#### Line currents
- I_A = I_AB − I_CA = 24.17 − j38.08 = **45.1 A** (∠−57.6°)
- I_B = I_BC − I_AB = −5.17 − j2.14 = **5.59 A** (∠−157.5°)
- I_C = I_CA − I_BC = −19.0 + j40.22 = **44.5 A** (∠115.3°)

(Sum = 0.)

#### Total power
- P_AB = 7.31² × 52 cos 45° = 1964 W
- P_BC = 7.31² × 52 cos 30° = 2405 W
- P_CA = 38² × 10 × cos 0° = 14 440 W

**Total power = 18 809 W ≈ 18.8 kW**

## Check yourself

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

1. In delta, what voltage is across each phase impedance?
   A. 219.4 V
   B. 380 V
   C. 658 V
   D. 127 V
   Answer: B. Phase voltage equals line voltage in delta. (Key point: Phase voltages)

2. What is the phase current of Z_CA = 10∠0°?
   A. 38.0 A
   B. 7.31 A
   C. 65.8 A
   D. 21.9 A
   Answer: A. I_CA = 380/10 = 38.0 A. (Key point: Phase current)

3. How is the line current I_A found?
   A. I_AB + I_CA
   B. I_AB - I_CA
   C. I_AB x √3
   D. I_AB
   Answer: B. By KCL at junction A, I_A = I_AB - I_CA = 45.1 A. (Key point: Line current)

4. How is total power calculated?
   A. Sum of I² Z cos θ for each phase: about 18.8 kW
   B. √3 V_L I_L cos φ
   C. Sum of I² Z
   D. V_L x I_A
   Answer: A. P = 1964 + 2405 + 14 440 = 18 809 W. (Key point: Total power)

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