---
title: "Star alternator supplying delta load (8+j6)"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:star-alternator-delta-load-8j6"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-alternator-delta-load-8j6
site: BCT Engineering
---

# Star alternator supplying delta load (8+j6)

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

- Asked in: 2068 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/star-alternator-delta-load-8j6
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A star connected alternator supplies a delta connected load. The impedance of the load branch is (8 + J6)Ω. The line voltage is 230 volt. Determine: i) Current in the load branch ii) Power consumed by the load iii) Power factor of the load iv) Reactive power of the load

## Short answer

Load branch current = 23 A. Power consumed = 12.7 kW. Power factor = 0.8 lagging. Reactive power = 9.52 kVAR.

## Answer

The alternator is star connected, but the load is delta connected, so the line voltage across each load branch is V_L = 230 V (phase voltage of load = line voltage).

Z_ph = 8 + j6 = 10∠36.87° Ω

#### i) Current in the load branch
I_ph = V_L/Z = 230/10 = **23 A** (line current = √3 × 23 = 39.8 A)

#### iii) Power factor
cos φ = R/Z = 8/10 = **0.8 lagging**

#### ii) Power consumed
P = 3 I_ph² R = 3 × 23² × 8 = **12 696 W ≈ 12.7 kW**

Check: √3 V_L I_L cos φ = 1.732 × 230 × 39.84 × 0.8 = 12.7 kW.

#### iv) Reactive power
Q = 3 I_ph² X = 3 × 23² × 6 = **9522 VAR ≈ 9.52 kVAR** (lagging)

Apparent power S = √(12 696² + 9522²) = 15.87 kVA.

## Check yourself

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

1. What voltage is across each branch of the delta load?
   A. 230/√3 = 132.8 V
   B. 230 V
   C. 398 V
   D. 115 V
   Answer: B. In delta the branch voltage equals the line voltage; the alternator connection does not change this. (Key point: Load voltage)

2. What is the load branch current?
   A. 23 A
   B. 39.8 A
   C. 13.3 A
   D. 46 A
   Answer: A. I_ph = 230/10 = 23 A. (Key point: Branch current)

3. What is the load power factor?
   A. 0.6 lagging
   B. 0.8 lagging (R/Z)
   C. 0.8 leading
   D. 1.0
   Answer: B. cos φ = R/Z = 8/10. (Key point: Power factor)

4. Reactive power is:
   A. 3 I² R = 12.7 kVAR
   B. 3 I² X = 9.52 kVAR
   C. 3 I² Z = 15.9 kVAR
   D. I² X = 3.17 kVAR
   Answer: B. Q = 3 x 23² x 6 = 9522 VAR. (Key point: Reactive power)

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