---
title: "Currents and efficiency of 132/11 kV transformer at 4 MW"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:star-delta-132-11kv-4mw-currents"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/star-delta-132-11kv-4mw-currents
site: BCT Engineering
---

# Currents and efficiency of 132/11 kV transformer at 4 MW

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2072 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/star-delta-132-11kv-4mw-currents
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 132 kV/11 kV Star/Delta, 3-phase transformer has balanced star connected 3-phase load of 4 MW at p.f of 0.8 lagging per phase. The primary winding has resistance and leakage reactance of 10 Ω and 30 Ω respectively. The secondary winding has resistance and leakage reactance of 0.02 Ω and 0.06 Ω respectively. Given that the iron loss of transformer is 20 kW. Calculate: i) Secondary phase and line currents ii) Primary line current iii) Efficiency of transformer

## Short answer

Use S = P/pf = 5 MVA. Secondary line current = 5 MVA/(sqrt3 times 11 kV) = 262.43 A, phase current = 151.52 A. Primary line current = 21.87 A; total copper loss 15.73 kW, iron loss 20 kW, efficiency = 4000/(4035.73) = 99.11%.

## Answer

Primary is **star** (line current = phase current), secondary is **delta** (phase voltage = line voltage = 11 kV, line current = $\sqrt3$ × phase current). Winding resistances are taken as per-phase values.

Load apparent power:
$$
S = \frac{P}{\cos\phi} = \frac{4 \times 10^6}{0.8} = 5\ \text{MVA}
$$

#### (i) Secondary line and phase currents
$$
\begin{aligned}
I_{2L} &= \frac{S}{\sqrt3\,V_{2L}} = \frac{5\times10^6}{\sqrt3 \times 11000} = 262.43\ \text{A} \\
I_{2ph} &= \frac{I_{2L}}{\sqrt3} = \frac{262.43}{\sqrt3} = 151.52\ \text{A}
\end{aligned}
$$

#### (ii) Primary line current
Neglecting no-load current, input VA = output VA:
$$
I_{1L} = I_{1ph} = \frac{5\times10^6}{\sqrt3 \times 132000} = 21.87\ \text{A}
$$
Check through the turns ratio: $V_{1ph} = 132000/\sqrt3 = 76210$ V, $K = 11000/76210 = 0.1443$, $I_{1ph} = K I_{2ph} = 0.1443 \times 151.52 = 21.87$ A.

#### (iii) Efficiency
Copper losses (three phases):
$$
\begin{aligned}
P_{cu1} &= 3I_{1ph}^2R_1 = 3 \times 21.87^2 \times 10 = 14348\ \text{W} \\
P_{cu2} &= 3I_{2ph}^2R_2 = 3 \times 151.52^2 \times 0.02 = 1377\ \text{W} \\
P_{cu} &= 14348 + 1377 = 15725\ \text{W} = 15.73\ \text{kW}
\end{aligned}
$$
Total losses $= P_i + P_{cu} = 20 + 15.73 = 35.73$ kW.
$$
\eta = \frac{4000}{4000 + 35.73}\times 100 = 99.11\%
$$
(The leakage reactances do not cause power loss, so they are not needed for efficiency.)

| Quantity | Value |
|---|---|
| Secondary line current | 262.43 A |
| Secondary phase current | 151.52 A |
| Primary line current | 21.87 A |
| Copper loss | 15.73 kW |
| Efficiency | 99.11% |

**Answer:** (i) $I_{2L} = 262.43$ A, $I_{2ph} = 151.52$ A; (ii) $I_{1L} = 21.87$ A; (iii) $\eta = 99.11\%$.

## Check yourself

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

1. With 4 MW at 0.8 pf, the apparent power is:
   A. 3.2 MVA
   B. 4 MVA
   C. 5 MVA
   D. 6.4 MVA
   Answer: C. S = P/cos(phi) = 4/0.8 = 5 MVA. (Key point: Load VA)

2. In the delta secondary, the phase current relates to the line current by:
   A. Iph = IL
   B. Iph = IL / sqrt(3)
   C. Iph = sqrt(3) IL
   D. Iph = IL / 3
   Answer: B. In delta, IL = sqrt3 Iph, so Iph = 262.43/sqrt3 = 151.52 A. (Key point: Delta secondary)

3. In the star primary, the line current is found as:
   A. S/(sqrt3 x 132 kV) = 21.87 A
   B. S/(3 x 132 kV)
   C. S/(sqrt3 x 11 kV)
   D. S/(132 kV)
   Answer: A. Input VA equals output VA, and star has line current equal to phase current. (Key point: Primary line current)

4. Why are leakage reactances not needed in the efficiency calculation?
   A. They are very large
   B. They cause no power loss
   C. They are already in the iron loss
   D. They only affect the secondary
   Answer: B. Reactances do not dissipate power; only resistances (copper loss) and iron loss count. Efficiency is 99.11%. (Key point: Efficiency)

## Close variants set in other exams

- [Currents and efficiency of 132/11 kV transformer at 5 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-5mw-currents) — Electrical Machines I, chapter 2, asked 3×
- [Primary current and efficiency of 11 kV/380 V Δ/Y transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Adelta-star-11kv-380v-40kw-efficiency) — Electrical Machines I, chapter 2, asked 3×

## Related in Electrical Machines I

- [Primary current and efficiency of 100 kVA Δ/Y transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Adelta-star-100kva-11kv-100a-efficiency) — Electrical Machines I, chapter 2

## Related in other subjects

- [Design: 100 kVA 11/132 kV transformer with winding data](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-11-132kv) — Electric Machine Design, chapter 3, asked 3×
- [500 kVA delta/star transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-500kva-33-11-efficiency) — Electrical Machine, chapter 2
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- [Delta/star 120 kVA transformer values](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Adelta-star-120kva-line-phase-values) — Electrical Machine, chapter 2