---
title: "kW, kVA, kVAR in parallel impedances"
subject: "Basic Electrical Engineering"
chapter: 7
key: "semester_1/basic_electrical_engineering:parallel-impedances-12j16-10-j20-kw-kva"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-impedances-12j16-10-j20-kw-kva
site: BCT Engineering
---

# kW, kVA, kVAR in parallel impedances

Basic Electrical Engineering · Chapter 7: Power in AC Circuits

- Asked in: 2068 Shrawan, 8 marks; 2068 Chaitra, 8 marks (asked 2 times)
- Video: [53. kVA, kVAR, kW, & Power factor Calculation Of Parallel Combination of Impedance In A.C. Circuit.](https://www.youtube.com/watch?v=ybxkuI3lPog) — Electrical Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/parallel-impedances-12j16-10-j20-kw-kva
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-7

## Question

A voltage of 200∠53° V is applied across impedances in parallel. The value of impedances are (12 + j16)Ω and (10–j20)Ω. Determine the kW, kVA and kVAR in each branch and the power factor of the whole circuit.

## Short answer

Branch 1: I₁=10A, 1.2kW, 1.6kVAR lag, 2kVA. Branch 2: I₂=8.94A, 0.8kW, 1.6kVAR lead, 1.789kVA. Total P=2kW, Q=0, S=2kVA, power factor=1 unity.

## Answer

The magnitude of the voltage is 200 V. Its phase angle (53°) does not affect the powers, which depend only on the magnitudes and the impedances. (The same problem in another paper gives 200∠0° V, with identical results.)

Z₁ = 12 + j16 = 20∠53.13° Ω, Z₂ = 10 − j20 = 22.36∠−63.43° Ω

#### Branch 1
- I₁ = 200/20 = 10 A (lagging 53.13°)
- kVA: S₁ = VI₁ = 200 × 10 = 2000 VA = **2.0 kVA**
- kW: P₁ = I₁²R = 100 × 12 = 1200 W = **1.2 kW**
- kVAR: Q₁ = I₁²X = 100 × 16 = 1600 VAR = **1.6 kVAR** (lagging)
- pf₁ = 12/20 = 0.6 lagging

#### Branch 2
- I₂ = 200/22.36 = 8.944 A (leading 63.43°)
- kVA: S₂ = 200 × 8.944 = 1789 VA = **1.789 kVA**
- kW: P₂ = I₂²R = 80 × 10 = 800 W = **0.8 kW**
- kVAR: Q₂ = I₂²X = 80 × 20 = 1600 VAR = **1.6 kVAR** (leading)
- pf₂ = 10/22.36 = 0.447 leading

#### Whole circuit
- P = 1200 + 800 = 2000 W
- Q = 1600 (lag) − 1600 (lead) = 0 VAR
- S = √(2000² + 0²) = 2000 VA = 2 kVA

Total current I = S/V = 10 A.

**Power factor = P/S = 2000/2000 = 1 (unity)**

The lagging reactive power of the first branch is cancelled exactly by the leading reactive power of the second.

## Check yourself

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

1. Why does the 53° phase of the voltage not matter?
   A. It is the pf
   B. Powers depend only on voltage magnitude and impedances
   C. It changes the power
   D. It is ignored by mistake
   Answer: B. Phase angle is a reference only. (Key point: Which values matter)

2. Branch 1 (12 + j16) has:
   A. I = 10 A, P = 1.2 kW, Q = 1.6 kVAR lagging
   B. I = 8.94 A, Q leading
   C. I = 10 A, P = 2 kW
   D. P = 1.6 kW
   Answer: A. Z = 20 Ω; P = I^2R, Q = I^2X. (Key point: Branch 1)

3. Branch 2 (10 − j20) has:
   A. P = 1.6 kW
   B. I = 10 A
   C. Q lagging
   D. I = 8.944 A, P = 0.8 kW, Q = 1.6 kVAR leading
   Answer: D. Z = 22.36 Ω; capacitive reactance gives leading Q. (Key point: Branch 2)

4. Total P = 2000 W and net Q is:
   A. -1.6 kVAR
   B. 1.6 kVAR
   C. 3.2 kVAR
   D. 0 VAR because lag and lead cancel, pf = 1
   Answer: D. Q = 1600 lag − 1600 lead = 0, so S = P and pf = 1. (Key point: Whole circuit)

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