---
title: "Combined balanced three-phase loads"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:three-phase-loads-415v-four-wire"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/three-phase-loads-415v-four-wire
site: BCT Engineering
---

# Combined balanced three-phase loads

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

- Asked in: 2074 Asoj, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/three-phase-loads-415v-four-wire
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

The following balanced three phase loads are connected to a 415 V, three phase, four wire supply. (i) 160 kVA at 0.7 power factor lagging (ii) 50 kVA at 0.65 power factor leading (iii) 50 kW at unity power factor. Calculate (a) the total load in kVA (b) the line current (c) the combined power factor

## Short answer

Sum active and reactive powers: 194.5 kW, 76.3 kVAR lag. Total apparent power is 208.9 kVA. Line current at 415 V is 290.6 A. Combined power factor is 0.931 lagging.

## Answer

Given a balanced supply of 415 V (line), three loads in parallel.

#### Each load as complex power S = P + jQ
1. 160 kVA at 0.7 lag: P₁ = 160 × 0.7 = 112 kW, Q₁ = 160 × sin(45.57°) = +114.3 kVAR
2. 50 kVA at 0.65 lead: P₂ = 50 × 0.65 = 32.5 kW, Q₂ = −50 × sin(49.46°) = −38.0 kVAR
3. 50 kW at unity: P₃ = 50 kW, Q₃ = 0

#### Totals
- P = 112 + 32.5 + 50 = 194.5 kW
- Q = 114.3 − 38.0 = 76.3 kVAR (lagging)

#### (a) Total load in kVA
S = √(194.5² + 76.3²) = **208.9 kVA**

#### (b) Line current
I_L = S/(√3 V_L) = 208 900/(1.732 × 415) = **290.6 A**

#### (c) Combined power factor
cos φ = P/S = 194.5/208.9 = **0.931 lagging**

## Check yourself

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

1. How should the loads be combined to find the total?
   A. Add the kVA values arithmetically
   B. Add the P values and Q values separately (with leading Q negative), then combine
   C. Add the power factors
   D. Average the power factors
   Answer: B. kVA values have different angles, so add real and reactive parts as complex power. (Key point: Method)

2. How is the 50 kVA, 0.65 leading load treated?
   A. Q is positive
   B. Q is negative (-38.0 kVAR)
   C. Q is zero
   D. P is negative
   Answer: B. A leading load supplies reactive power, so Q is negative. (Key point: Leading load)

3. What is the total load?
   A. About 208.9 kVA
   B. About 260 kVA
   C. About 194.5 kVA
   D. About 76.3 kVA
   Answer: A. S = √(194.5² + 76.3²) = 208.9 kVA. (Key point: Total kVA)

4. Which formula gives the line current?
   A. I_L = S/V_L
   B. I_L = S/(√3 V_L)
   C. I_L = P/V_L
   D. I_L = S/(3 V_L)
   Answer: B. For three-phase S = √3 V_L I_L, so I_L = 208 900/(1.732 x 415) = 290.6 A. (Key point: Line current)

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