---
title: "Regulation and zero-regulation power factor of 100 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:regulation-100kva-zero-regulation-pf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/regulation-100kva-zero-regulation-pf
site: BCT Engineering
---

# Regulation and zero-regulation power factor of 100 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2081 Chaitra (new course), 3+3 marks
- Video: [Solved Example Problem 8 – Transformer Efficiency and Regulation #bee](https://www.youtube.com/watch?v=_8orv_GeRT0) — Power Systemz
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/regulation-100kva-zero-regulation-pf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A single phase, 100 kVA, 2000/200 V, 50 Hz transformer has an impedance drop of 10% and resistance drop of 5%. Calculate: a) The regulation at full load 0.8 power factor lagging. b) The value of power factor at which regulation is zero.

## Short answer

Use percent regulation = epsilon_r cos phi + epsilon_x sin phi with epsilon_x = sqrt(10^2 - 5^2) = 8.66%. At 0.8 lagging regulation is 9.20%; zero regulation occurs at 0.866 leading pf.

## Answer

Per-unit resistance and impedance drops are given: $\varepsilon_r = 5\%$, $\varepsilon_z = 10\%$, so the reactance drop is

$$
\varepsilon_x = \sqrt{10^2 - 5^2} = 8.66\%
$$

#### a) Regulation at full load, 0.8 pf lagging

$$
\begin{aligned}
\%\text{Reg} &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi \\
&= 5\times0.8 + 8.66\times0.6 \\
&= 4.0 + 5.196 = 9.20\%
\end{aligned}
$$

#### b) Power factor for zero regulation

Regulation is zero for a **leading** pf, when

$$
\begin{aligned}
\varepsilon_r\cos\phi - \varepsilon_x\sin\phi &= 0 \\
\tan\phi &= \frac{\varepsilon_r}{\varepsilon_x} = \frac{5}{8.66} = 0.577 \Rightarrow \phi = 30^\circ \\
\cos\phi &= \frac{\varepsilon_x}{\varepsilon_z} = \frac{8.66}{10} = 0.866
\end{aligned}
$$

**Answer:** (a) regulation = **9.20%**; (b) zero regulation at **pf = 0.866 leading**.

## Check yourself

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

1. How is the reactance drop found from εz = 10% and εr = 5%?
   A. εx = 10 − 5 = 5%
   B. εx = √(10² − 5²) = 8.66%
   C. εx = 10 + 5 = 15%
   D. εx = 5/10 = 0.5%
   Answer: B. Drops form a right triangle: εz² = εr² + εx². (Key point: Reactance drop)

2. What is the regulation at full load 0.8 pf lag?
   A. εr cos φ − εx sin φ = −1.2%
   B. εr cos φ + εx sin φ = 5×0.8 + 8.66×0.6 = 9.20%
   C. εr sin φ + εx cos φ = 10.9%
   D. εz = 10%
   Answer: B. For lagging pf the resistive and reactive parts add. (Key point: Regulation formula)

3. At what pf is regulation zero?
   A. 0.866 leading
   B. 0.866 lagging
   C. 0.5 leading
   D. Unity
   Answer: A. Zero needs εr cos φ = εx sin φ with leading current, tan φ = 5/8.66, φ = 30°. (Key point: Zero regulation)

4. Which expression gives cos φ at zero regulation?
   A. cos φ = εr/εz
   B. cos φ = εr/εx
   C. cos φ = εx/εz
   D. cos φ = εz/εx
   Answer: C. tan φ = εr/εx, so cos φ = εx/εz = 0.866. (Key point: Cos formula)

## Related in Electrical Machines I

- [Full-load and half-load efficiency of 100 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A100kva-1200w-960w-efficiency) — Electrical Machines I, chapter 2
- [Equivalent circuit, efficiency and regulation of 5 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-5kva-220-400-lv) — Electrical Machines I, chapter 2
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- [Voltage regulation of transformer for lagging load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Avoltage-regulation-define-lagging-expression) — Electrical Machines I, chapter 2
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