---
title: "Star load (8+j6) line, phase current"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:star-load-8j6-400v-ryb"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-load-8j6-400v-ryb
site: BCT Engineering
---

# Star load (8+j6) line, phase current

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

- Asked in: 2073 Chaitra, 6 marks
- Video: [Three Phase Balanced Star Connected Load | Balanced Star Connected Load | Problem 1](https://www.youtube.com/watch?v=JYwPvszUkGs) — Aryabhat Institute
- 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-load-8j6-400v-ryb
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A balanced star-connected load of (8 + j6)Ω per phase is connected to a balanced 3 phase 400V, 50Hz supply. Find the line current, phase current and total power consumed. Take RYB phase sequence.

## Short answer

Impedance 8+j6 Ω = 10∠36.87° Ω, phase voltage 230.9 V. Phase current = line current = 23.09 A. Total power = 3 × 23.09² × 8 = 12.8 kW.

## Answer

Given a balanced star load Z_ph = 8 + j6 = 10∠36.87° Ω, V_L = 400 V, 50 Hz, RYB sequence.

- V_ph = 400/√3 = 230.9 V
- Phase voltages: V_R = 230.9∠0°, V_Y = 230.9∠−120°, V_B = 230.9∠+120° V

#### Currents
I_R = V_R/Z = 23.09∠−36.87° A

I_Y = 23.09∠−156.87° A

I_B = 23.09∠+83.13° A

- **Phase current = line current = 23.09 A** (for star)

#### Power
- Power factor = 8/10 = 0.8 lagging
- **Total power P = 3 I_ph² R = 3 × 23.09² × 8 = 12 800 W = 12.8 kW**

Check: √3 V_L I_L cos φ = 1.732 × 400 × 23.09 × 0.8 = 12.8 kW. Also Q = 9.6 kVAR, S = 16 kVA.

## 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 star load on a 400 V supply?
   A. 400 V
   B. 230.9 V
   C. 693 V
   D. 133 V
   Answer: B. V_ph = 400/√3 = 230.9 V. (Key point: Phase voltage)

2. How do phase and line current compare in star?
   A. Line is √3 times phase
   B. They are equal, 23.09 A
   C. Phase is √3 times line
   D. Line is 3 times phase
   Answer: B. In star the line carries the phase current. (Key point: Phase vs line current)

3. Which gives the total power?
   A. 3 I² R = 12.8 kW
   B. 3 I² Z = 16 kW
   C. 3 I² X = 9.6 kW
   D. I² R = 4.27 kW
   Answer: A. Active power uses R: 3 x 23.09² x 8 = 12 800 W (pf = 0.8). (Key point: Power)

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