---
title: "4.2 kV/120 V transformer voltage regulation"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-4-2kv-120v-regulation-unity"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-4-2kv-120v-regulation-unity
site: BCT Engineering
---

# 4.2 kV/120 V transformer voltage regulation

Electrical Machines · Chapter 2: Transformer

- Asked in: 2066 Magh (old course), 8 marks
- Video: [Voltage Regulation of Transformer Example 2 : Numerical on voltage regulation of transformer](https://www.youtube.com/watch?v=QFp-dGgGVQ0) — E Gyan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-4-2kv-120v-regulation-unity
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 4.2 kV/120 V, 50 Hz, 1-phase transformer has following series parameters: $R_1 = 1.4\ \Omega$, $X_1 = 3.5\ \Omega$, $R_2 = 0.04\ \Omega$, $X_2 = 0.1\ \Omega$. If the transformer draws 500 A current on the secondary at rated terminal voltage, calculate voltage regulation at unity power factor.

## Short answer

Referred to secondary, equivalent resistance and reactance are 0.04114 Ω and 0.10286 Ω. No-load voltage E2 = √((120+500×0.04114)² + (500×0.10286)²) = 149.684 V. Voltage regulation = (149.684-120)/120×100 = 24.74%.

## Answer

Voltage regulation is the change in secondary terminal voltage from no load to load, as a fraction of the rated terminal voltage:

$$
\%\text{VR} = \frac{E_2 - V_2}{V_2}\times 100
$$

where $E_2$ is the no-load secondary voltage and $V_2$ is the load terminal voltage.

#### Equivalent parameters referred to secondary

$$
a = \frac{4200}{120} = 35, \qquad a^2 = 1225
$$

$$
\begin{aligned}
R_{02} &= R_2 + \frac{R_1}{a^2} = 0.04 + \frac{1.4}{1225} = 0.04114\ \Omega \\
X_{02} &= X_2 + \frac{X_1}{a^2} = 0.1 + \frac{3.5}{1225} = 0.10286\ \Omega
\end{aligned}
$$

#### No-load voltage at unity p.f.

With $V_2 = 120$ V, $I_2 = 500$ A and $\cos\phi = 1$:

$$
\begin{aligned}
E_2 &= \sqrt{(V_2 + I_2R_{02})^2 + (I_2X_{02})^2} \\
&= \sqrt{(120 + 20.571)^2 + (51.429)^2} \\
&= 149.684\ \text{V}
\end{aligned}
$$

#### Regulation

$$
\%\text{VR} = \frac{149.684 - 120}{120}\times 100 = 24.74\%
$$

**Answer:** voltage regulation at unity power factor $= 24.74\%$ (no-load secondary voltage $= 149.684$ V).

## Check yourself

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

1. In this answer, the percent voltage regulation is defined as
   A. (E2 - V2)/V2 x 100, with E2 the no-load voltage
   B. (V2 - E2)/E2 x 100
   C. (E2 + V2)/2 x 100
   D. I2 Z02/E2 only at lagging pf
   Answer: A. Regulation is the rise in voltage from load to no load as a fraction of the rated terminal voltage V2. (Key point: Regulation definition used)

2. With a = 35, R02 is
   A. 0.04 + 1.4 x 1225 = 1715 Ω
   B. 0.04 + 1.4/35 = 0.08 Ω
   C. 0.04 + 1.4/1225 = 0.04114 Ω
   D. 1.4 + 0.04/1225 = 1.4 Ω
   Answer: C. Primary resistance is divided by a^2 = 1225 when referred to the secondary. (Key point: Equivalent resistance referred to secondary)

3. With V2 = 120 V, I2 = 500 A, R02 = 0.04114 Ω, X02 = 0.10286 Ω at unity pf, E2 is
   A. 120 + 20.57 + 51.43 = 192 V
   B. sqrt((V2 + I2R02)^2 + (I2X02)^2) = 149.68 V
   C. sqrt(120^2 + 51.43^2) = 130.6 V
   D. 120 V
   Answer: B. At unity pf the resistive drop adds in phase with V2 and the reactive drop is at 90°, so the phasor sum gives 149.68 V. (Key point: No-load voltage at unity pf)

4. The resulting voltage regulation is
   A. 12.4%
   B. 24.74%
   C. 49.7%
   D. 2.47%
   Answer: B. (149.684 - 120)/120 x 100 = 24.74%. (Key point: Result)

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