---
title: "Star three impedances four-wire system"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:star-three-impedances-4-wire-400v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-three-impedances-4-wire-400v
site: BCT Engineering
---

# Star three impedances four-wire system

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

- Asked in: 2069 Asar, 8 marks
- Video: [How to Analyze a 3-Phase, 4-Wire System with an Unbalanced Star Load](https://www.youtube.com/watch?v=lE0RvSU_fkM) — Atiigah Tech Academy
- 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-three-impedances-4-wire-400v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

Three impedances of (6+j8)Ω, (8+j6)Ω, (3+j4)Ω, are connected in star to a 3-phase, 4 wire system for which the line voltage is 400V. Find the line currents, and active, reactive and apparent power per phase. Also find the current through neutral wire.

## Short answer

Line currents: 23.09 A (−53.13°), 23.09 A (−156.87°), 46.19 A (66.87°). Active, reactive powers: 13.87 kW, 16.0 kVAR. Neutral current magnitude 18.4 A.

## Answer

Star connected load Z_R = 6 + j8 = 10∠53.13° Ω, Z_Y = 8 + j6 = 10∠36.87° Ω, Z_B = 3 + j4 = 5∠53.13° Ω. V_L = 400 V, four-wire, RYB sequence assumed.

V_ph = 400/√3 = 230.9 V. V_R = 230.9∠0°, V_Y = 230.9∠−120°, V_B = 230.9∠120° V.

#### Line currents
- I_R = 230.9∠0° / 10∠53.13° = **23.09 A** ∠−53.13° = 13.86 − j18.48
- I_Y = 230.9∠−120° / 10∠36.87° = **23.09 A** ∠−156.87° = −21.24 − j9.07
- I_B = 230.9∠120° / 5∠53.13° = **46.19 A** ∠66.87° = 18.14 + j42.48

#### Neutral current
I_N = I_R + I_Y + I_B = (13.86 − 21.24 + 18.14) + j(−18.48 − 9.07 + 42.48) = 10.76 + j14.93 A

**|I_N| = 18.4 A** (∠54.2°)

#### Power per phase

| Phase | I (A) | P = I²R (W) | Q = I²X (VAR) | S = I²Z (VA) |
|---|---|---|---|---|
| R | 23.09 | 3200 | 4267 | 5333 |
| Y | 23.09 | 4267 | 3200 | 5333 |
| B | 46.19 | 6400 | 8533 | 10 667 |
| Total | | 13 867 | 16 000 | 21 333 (sum) |

Total active power 13.87 kW, total reactive power 16.0 kVAR (all lagging), total apparent power √(13 867² + 16 000²) = 21.17 kVA (the arithmetic sum is 21.33 kVA).

## Check yourself

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

1. What is the line current in the R phase (6+j8 Ω)?
   A. 40 A
   B. 23.09 A
   C. 46.19 A
   D. 30 A
   Answer: B. I_R = 230.9/10 = 23.09 A. (Key point: Line current in R)

2. Which phase carries the largest current and why?
   A. R, as it has the largest reactance
   B. Y, as it has the smallest pf
   C. B, as it has the smallest impedance (5 Ω)
   D. All the same
   Answer: C. With equal phase voltage, the 5 Ω phase gives 46.19 A. (Key point: Largest current)

3. How is the neutral current obtained?
   A. Arithmetic sum of the line currents
   B. Phasor sum I_R + I_Y + I_B, about 18.4 A
   C. Zero because the supply is balanced
   D. Difference of the largest and smallest
   Answer: B. The neutral carries the phasor sum of the unbalanced line currents. (Key point: Neutral current)

4. Active power per phase is:
   A. I² Z
   B. I² R
   C. I² X
   D. V I
   Answer: B. P = I²R, Q = I²X and S = I²Z. (Key point: Power per phase)

## Related in Basic Electrical Engineering

- [Star load (8+j6) line, phase current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Astar-load-8j6-400v-ryb) — Basic Electrical Engineering, chapter 8
- [Star system with three loads, neutral current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Astar-three-loads-400v-neutral-current) — Basic Electrical Engineering, chapter 8
- [Phase order and star load (6+j8)](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aphase-order-star-load-6j8) — Basic Electrical Engineering, chapter 8
- [Neutral current and power, unbalanced star](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aunbalanced-star-neutral-current-415v) — Basic Electrical Engineering, chapter 8
- [Star alternator supplying delta load (8+j6)](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Astar-alternator-delta-load-8j6) — Basic Electrical Engineering, chapter 8
- [Phase and line voltage in star](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aphase-voltage-line-voltage-star) — Basic Electrical Engineering, chapter 8

## Related in other subjects

- [Line currents with isolated and grounded neutral](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-impedance-unbalanced-voltages-neutral) — Power System Analysis II, chapter 4
- [Sequence impedances of star load with Zn](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-load-sequence-zn) — Power System Analysis II, chapter 4, asked 2×
- [Direction of active and reactive power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adirection-active-reactive-flow) — Power System Analysis I, chapter 6
- [Fault currents of isolated-neutral 380 V generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-faults-380v-isolated-neutral) — Power System Analysis II, chapter 5