---
title: "Neutral current and power, unbalanced star"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:unbalanced-star-neutral-current-415v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/unbalanced-star-neutral-current-415v
site: BCT Engineering
---

# Neutral current and power, unbalanced star

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

- Asked in: 2072 Chaitra, 8 marks
- Video: [The Easiest Most Accurate Way to Calculate Three Phase Neutral Current in an Unbalanced Load](https://www.youtube.com/watch?v=ExHJGSJSE6o) — Joe Robinson Training
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/unbalanced-star-neutral-current-415v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

For the circuit shown in figure below, calculate the current through the neutral and the total power consumed in the load. [Figure: 415 V balanced supply feeding a four-wire star load with 10 Ω in one phase, 15 Ω in another phase and 20 Ω in the third phase; neutral current IN.]

## Short answer

Phase voltage 239.6 V; phase currents 23.96 A (0°), 15.97 A (−120°), 11.98 A (120°). Neutral current magnitude is 10.57 A. Total power is 12.44 kW.

## Answer

From the figure: a 415 V balanced supply feeds a four-wire star load with 10 Ω, 15 Ω and 20 Ω resistors in the three phases and a neutral wire carrying I_N. The sequence is not shown, so assume R–Y–B with 10 Ω in R, 15 Ω in Y and 20 Ω in B.

V_ph = 415/√3 = 239.6 V. Take V_R = 239.6∠0°, V_Y = 239.6∠−120°, V_B = 239.6∠+120° V.

#### Line (phase) currents
- I_R = 239.6/10 = 23.96∠0° A = 23.96 + j0
- I_Y = 239.6/15 = 15.97∠−120° A = −7.99 − j13.83
- I_B = 239.6/20 = 11.98∠+120° A = −5.99 + j10.38

#### Neutral current
I_N = I_R + I_Y + I_B = (23.96 − 7.99 − 5.99) + j(0 − 13.83 + 10.38) = 9.98 − j3.46 A

**|I_N| = 10.57 A** (at −19.1°). The neutral current flows because the load is unbalanced.

#### Total power
All loads are resistive, so P = Σ I²R:

- Phase R: 23.96² × 10 = 5741 W
- Phase Y: 15.97² × 15 = 3827 W
- Phase B: 11.98² × 20 = 2870 W

**Total power = 12 438 W ≈ 12.44 kW**

Interchanging the phases of the load changes the angle of I_N but not its magnitude, and the total power stays the same, so the answer does not depend on the assumed order.

## Check yourself

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

1. What is the phase voltage for a 415 V star supply?
   A. 415 V
   B. 239.6 V
   C. 718 V
   D. 138 V
   Answer: B. V_ph = 415/√3 = 239.6 V. (Key point: Phase voltage)

2. What is the current in the 10 Ω phase?
   A. 41.5 A
   B. 23.96 A
   C. 15.97 A
   D. 11.98 A
   Answer: B. I = 239.6/10 = 23.96 A. (Key point: Currents)

3. Why does current flow in the neutral?
   A. The load is unbalanced
   B. The supply frequency is high
   C. The loads are resistive
   D. The supply is balanced
   Answer: A. For a balanced load the three currents sum to zero; the unequal resistors leave a net 10.57 A. (Key point: Neutral current)

4. How is total power found for this resistive load?
   A. √3 V_L I_N
   B. Sum of I²R over the three phases
   C. 3 x V_ph x I_N
   D. V_L²/R
   Answer: B. Total = 5741 + 3827 + 2870 = 12.44 kW. (Key point: Power)

## Related in Basic Electrical Engineering

- [Neutral current and power, 380 V unbalanced load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aunbalanced-star-neutral-current-380v) — Basic Electrical Engineering, chapter 8
- [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
- [Star three impedances four-wire system](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Astar-three-impedances-4-wire-400v) — Basic Electrical Engineering, chapter 8
- [Combined balanced three-phase loads](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Athree-phase-loads-415v-four-wire) — Basic Electrical Engineering, chapter 8
- [Delta load (15+j20) phase and line currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-15j20-220v-phasors) — Basic Electrical Engineering, chapter 8

## Related in other subjects

- [Unbalanced source feeding star load](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aunbalanced-source-star-load-components) — Power System Analysis II, chapter 4
- [Sequence networks of Y load, neutral current](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-load-sequence-networks-neutral) — Power System Analysis II, chapter 4, asked 2×
- [Balanced circulating current: 10 kV alternator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acirculating-current-10kv-10-ohm) — Switchgear and Protection, chapter 7
- [Talker and listener echo in four-wire circuit](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Afour-wire-echo-numerical) — Telecommunications, chapter 2