---
title: "Voltage regulation of 2 MVA Δ-Y transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:delta-star-2mva-regulation"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/delta-star-2mva-regulation
site: BCT Engineering
---

# Voltage regulation of 2 MVA Δ-Y transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2078 Chaitra, 8 marks
- Video: [End Ch Q 2.11 (New) || 3-Phase Delta-Wye Transformer || Voltage Regulation || Efficiency || Chapman)](https://www.youtube.com/watch?v=2QN0r8JRKaI) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/delta-star-2mva-regulation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 2 MVA, Delta-star, 33/6.6 kV, three phase transformer has primary and secondary resistance of 8 ohm and 0.08 ohm per phase. The percentage impedance is 7 percent. Calculate the voltage regulation at full load at 0.75 pf lagging.

## Short answer

Refer secondary resistance to primary: R01=14 ohm. At full load, %R=0.857% and %X=6.947%, so regulation at 0.75 pf lagging is 0.857*0.75+6.947*0.6614=5.24%.

## Answer

Regulation is worked out per phase. The primary is delta (phase voltage = line voltage) and the secondary is star (phase voltage = line voltage/$\sqrt{3}$).

#### Step 1: Per-phase voltages and turns ratio
$$
\begin{aligned}
V_{1ph} &= 33000\ \text{V}, \quad V_{2ph} = \frac{6600}{\sqrt{3}} = 3810.5\ \text{V} \\
K &= \frac{V_{2ph}}{V_{1ph}} = \frac{3810.5}{33000} = 0.11547
\end{aligned}
$$

#### Step 2: Equivalent resistance referred to primary (per phase)
$$
\begin{aligned}
R_2' &= \frac{R_2}{K^2} = \frac{0.08}{0.11547^2} = 6.0\ \Omega \\
R_{01} &= R_1 + R_2' = 8 + 6 = 14\ \Omega
\end{aligned}
$$

#### Step 3: Percentage resistance and reactance
Full-load primary phase current:
$$
I_{1ph} = \frac{2 \times 10^6}{3 \times 33000} = 20.2\ \text{A}
$$
$$
\begin{aligned}
\%R &= \frac{I_{1ph} R_{01}}{V_{1ph}} \times 100 = \frac{20.2 \times 14}{33000} \times 100 = 0.857\% \\
\%Z &= 7\% \ (\text{given}) \\
\%X &= \sqrt{7^2 - 0.857^2} = 6.947\%
\end{aligned}
$$

#### Step 4: Regulation at 0.75 pf lagging
$\cos\phi = 0.75$, $\sin\phi = \sqrt{1 - 0.75^2} = 0.6614$.
$$
\begin{aligned}
\%\text{Reg} &\approx \%R\cos\phi + \%X\sin\phi \\
&= 0.857 \times 0.75 + 6.947 \times 0.6614 \\
&= 0.643 + 4.595 = 5.24\%
\end{aligned}
$$

If the more exact formula is used, adding the second-order term $\frac{(\%X\cos\phi - \%R\sin\phi)^2}{200}$ gives about 5.35%.

**Answer:** Full-load voltage regulation at 0.75 pf lagging is about **5.24%** (approximate formula).

## Check yourself

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

1. For the Delta-star 33/6.6 kV transformer, what are the per-phase voltages?
   A. 19053 V and 3810.5 V
   B. 33000 V and 6600/sqrt3 = 3810.5 V
   C. 33000/sqrt3 and 6600 V
   D. 33000 V and 6600 V
   Answer: B. Delta primary phase voltage equals line voltage; star secondary phase voltage is line/sqrt3. (Key point: Per-phase voltages)

2. How is R2' obtained?
   A. R2/K^2 = 6 ohm with K = 0.11547
   B. R2 x K^2
   C. R2 x K
   D. R2/K = 0.69 ohm
   Answer: A. Referring a secondary value to the primary divides by K^2, giving 0.08/0.01333 = 6 ohm. (Key point: Secondary resistance referred to primary)

3. How is %X found from %Z = 7% and %R = 0.857%?
   A. %X = 7 x 0.857
   B. %X = sqrt(7^2 - 0.857^2) = 6.947%
   C. %X = 7 + 0.857
   D. %X = 7 - 0.857
   Answer: B. Impedance is the phasor sum of R and X, so %X = sqrt(%Z^2 - %R^2). (Key point: Percentage reactance)

4. For a lagging pf of 0.75 which approximate formula gives the regulation?
   A. %R sin phi + %X cos phi
   B. %R cos phi - %X sin phi
   C. %R cos phi + %X sin phi
   D. %Z cos phi
   Answer: C. For lagging loads regulation = %R cos(phi) + %X sin(phi) = 0.643 + 4.595 = 5.24%. (Key point: Regulation formula)

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