---
title: "Neutral current and power, 380 V unbalanced load"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:unbalanced-star-neutral-current-380v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/unbalanced-star-neutral-current-380v
site: BCT Engineering
---

# Neutral current and power, 380 V unbalanced load

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

- Asked in: 2068 Shrawan, 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-380v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

For the circuit shown below, calculate the current through the neutral and total power of the circuit. [Figure: 380 V balanced source, star connected; load impedances 10 Ω, 30 Ω and 15 Ω in the three phases; neutral current IN.]

## Short answer

Phase voltage 219.4 V; phase currents 21.94 A, 14.63 A, 7.31 A. Neutral current magnitude is 12.67 A. Total power consumed is 9.63 kW.

## Answer

From the figure: a 380 V balanced three-phase star source feeds a star load with resistors of 10 Ω, 30 Ω and 15 Ω, and a neutral wire carrying I_N. Assume R–Y–B sequence with R = 10 Ω, Y = 15 Ω, B = 30 Ω. (The magnitude of I_N and the total power are the same whichever way the three resistors are assigned to the phases.)

V_ph = 380/√3 = 219.4 V. V_R = 219.4∠0°, V_Y = 219.4∠−120°, V_B = 219.4∠120° V.

#### Currents
- I_R = 219.4/10 = 21.94∠0° A = 21.94 + j0
- I_Y = 219.4/15 = 14.63∠−120° A = −7.31 − j12.67
- I_B = 219.4/30 = 7.31∠120° A = −3.66 + j6.33

#### Neutral current
I_N = I_R + I_Y + I_B = (21.94 − 7.31 − 3.66) + j(−12.67 + 6.33) = 10.97 − j6.33 A

**|I_N| = 12.67 A** (∠−30°). This magnitude is the same whichever order the phases are assumed in.

#### Total power
The load is purely resistive, so P = Σ I²R:

- 21.94² × 10 = 4813 W
- 14.63² × 15 = 3209 W
- 7.31² × 30 = 1604 W

**Total power = 9627 W ≈ 9.63 kW**

## Check yourself

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

1. What is the phase voltage of the 380 V star source?
   A. 380 V
   B. 219.4 V
   C. 658 V
   D. 127 V
   Answer: B. V_ph = 380/√3 = 219.4 V. (Key point: Phase voltage)

2. What is the current in the 30 Ω phase?
   A. 21.94 A
   B. 14.63 A
   C. 7.31 A
   D. 12.67 A
   Answer: C. I = 219.4/30 = 7.31 A. (Key point: Current in 30 Ω)

3. How is the neutral current found?
   A. Algebraic sum of magnitudes
   B. Phasor sum of the three phase currents: 12.67 A
   C. V_L / total R
   D. Average of currents
   Answer: B. The currents are 120° apart with unequal magnitudes, so their phasor sum is 12.67 A. (Key point: Neutral current)

4. What is the total power?
   A. About 9.63 kW from the sum of I²R
   B. About 12.67 kW
   C. About 3.2 kW
   D. About 28.9 kW
   Answer: A. 4813 + 3209 + 1604 = 9627 W. (Key point: Total power)

## Related in Basic Electrical Engineering

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