---
title: "Unbalanced delta loads currents and power"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:delta-unbalanced-loads-31j59-200v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-unbalanced-loads-31j59-200v
site: BCT Engineering
---

# Unbalanced delta loads currents and power

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

- Asked in: 2071 Shrawan, 8 marks
- Video: [Delta Unbalanced Resistive Load](https://www.youtube.com/watch?v=N-JdRfJdFkE) — Zack Hartle
- 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-loads-31j59-200v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

Three loads (31+j59) Ω, (30-j40)Ω and (80+j60)Ω are connected in delta to a 3 phase, 200 V supply. Find the phase currents, line currents and total power absorbed.

## Short answer

Phase currents: I_AB=3.00 A, I_BC=4.00 A, I_CA=2.00 A. Line currents: 4.78 A, 1.04 A, 5.82 A. Total active power absorbed is 1.08 kW.

## Answer

Loads connected in delta: Z_AB = 31 + j59 Ω, Z_BC = 30 − j40 Ω, Z_CA = 80 + j60 Ω. V_L = 200 V, sequence assumed ABC, with the loads in the order AB, BC, CA.

Phase voltages equal line voltages: V_AB = 200∠0°, V_BC = 200∠−120°, V_CA = 200∠120° V.

#### Phase currents
- |Z_AB| = 66.65 Ω, so I_AB = 200/66.65 = **3.00 A** ∠−62.3° = 1.40 − j2.66
- |Z_BC| = 50 Ω, so I_BC = 200/50 = **4.00 A** ∠−66.9° = 1.57 − j3.68
- |Z_CA| = 100 Ω, so I_CA = 200/100 = **2.00 A** ∠83.1° = 0.24 + j1.99

#### Line currents (by KCL at the junctions)
- I_A = I_AB − I_CA = 1.16 − j4.65 = **4.78 A** ∠−76.0°
- I_B = I_BC − I_AB = 0.17 − j1.02 = **1.04 A** ∠−80.3°
- I_C = I_CA − I_BC = −1.33 + j5.67 = **5.82 A** ∠103.2°

(Their sum is zero.)

#### Total power
- P_AB = 3.00² × 31 = 279.2 W
- P_BC = 4.00² × 30 = 480 W
- P_CA = 2.00² × 80 = 320 W

**Total power = 1079 W ≈ 1.08 kW**

## Check yourself

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

1. What voltage appears across each load in a delta connection on 200 V?
   A. 115.5 V
   B. 200 V
   C. 346 V
   D. 66.7 V
   Answer: B. In delta the phase voltage equals the line voltage. (Key point: Phase voltage)

2. What is the phase current of the (30 - j40) Ω load?
   A. 4.00 A
   B. 6.67 A
   C. 2.00 A
   D. 3.00 A
   Answer: A. |Z| = 50 Ω, so I = 200/50 = 4.00 A. (Key point: Phase current)

3. How are line currents found for an unbalanced delta?
   A. I_L = √3 I_ph for each line
   B. By KCL at the junctions, e.g. I_A = I_AB - I_CA
   C. I_L = I_ph
   D. By adding the phase current magnitudes
   Answer: B. The load is unbalanced, so the √3 rule does not apply; phasor KCL is used. (Key point: Line currents)

4. How is total power found?
   A. Sum of I²R of each load: about 1.08 kW
   B. √3 V_L I_L cos φ using one line
   C. Sum of I²Z of each load
   D. 200 x sum of currents
   Answer: A. P = 279.2 + 480 + 320 = 1079 W. (Key point: Total power)

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