---
title: "Star system with three loads, neutral current"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:star-three-loads-400v-neutral-current"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-three-loads-400v-neutral-current
site: BCT Engineering
---

# Star system with three loads, neutral current

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

- Asked in: 2072 Kartik, 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/star-three-loads-400v-neutral-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A three phase star connected system with line voltage 400 V is connected to three loads: 25∠0°, 11∠-20° and 15∠10° (also connected in star). Find the line to line current, total power and current in the neutral of the system.

## Short answer

Line currents: I_R = 9.24 A ∠0°, I_Y = 20.99 A ∠-100°, I_B = 15.40 A ∠110°. Total power = 10.2 kW. Neutral current = 6.2 A (phasor sum of three line currents).

## Answer

Taking the three loads Z_R = 25∠0° Ω, Z_Y = 11∠−20° Ω, Z_B = 15∠10° Ω as star connected with a neutral wire, in RYB sequence. The loads are unbalanced.

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

#### Line currents
- I_R = 230.9∠0° / 25∠0° = **9.24 A ∠0°**
- I_Y = 230.9∠−120° / 11∠−20° = **20.99 A ∠−100°**
- I_B = 230.9∠120° / 15∠10° = **15.40 A ∠+110°**

In rectangular form: I_R = 9.24 + j0; I_Y = −3.65 − j20.67; I_B = −5.27 + j14.47 A.

#### Neutral current
I_N = I_R + I_Y + I_B = (9.24 − 3.65 − 5.27) + j(0 − 20.67 + 14.47) = 0.32 − j6.20 A

**|I_N| = 6.2 A** (at about −87°)

#### Total power
P = Σ I²R_load, with R = Z cos θ of each load:

- Phase R: 9.238² × 25 × cos 0° = 2133 W
- Phase Y: 20.995² × 11 × cos(−20°) = 4556 W
- Phase B: 15.396² × 15 × cos 10° = 3502 W

**Total P = 10 191 W ≈ 10.2 kW**

**Answer:** line currents 9.24 A, 20.99 A, 15.40 A; total power about 10.19 kW; neutral current about 6.2 A.

## Check yourself

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

1. How is each line current found?
   A. Phase voltage divided by that phase impedance
   B. Line voltage divided by that impedance
   C. Sum of all impedances
   D. Average of the impedances
   Answer: A. In star each phase has V_ph = 230.9 V across its own impedance. (Key point: Line currents)

2. How is the neutral current found?
   A. I_N = I_R + I_Y + I_B as phasors
   B. I_N = I_R + I_Y + I_B as magnitudes
   C. I_N = 0 always
   D. I_N = V/Z
   Answer: A. The neutral carries the phasor sum; the loads are unbalanced, so it is not zero (about 6.2 A). (Key point: Neutral current)

3. How is total power found for the unbalanced loads?
   A. √3 V_L I_L cos φ
   B. Sum of I² R_load over the phases, with R = Z cos θ
   C. 3 x the power of one phase
   D. V_L x I_N
   Answer: B. The load is unbalanced, so each phase is computed separately and added: about 10.2 kW. (Key point: Power)

4. Why is the neutral current non-zero here?
   A. The supply voltage is unbalanced
   B. The loads are unequal, so the line currents do not sum to zero
   C. The frequency differs
   D. The neutral has resistance
   Answer: B. Unequal impedances make unequal currents with unequal angles, so their sum is not zero. (Key point: Unbalanced effect)

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