---
title: "Currents and efficiency of 132/11 kV transformer at 5 MW"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:star-delta-132-11kv-5mw-currents"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/star-delta-132-11kv-5mw-currents
site: BCT Engineering
---

# Currents and efficiency of 132/11 kV transformer at 5 MW

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2079 Chaitra, 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-5mw-currents
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 132/11 kV, star/delta, 3-phase, 50 Hz transformer has balanced star connected three-phase 5 MW at 0.8 pf lagging per phase. The primary winding has a resistance and leakage reactance of 10 Ω and 30 Ω per phase respectively and the secondary winding has a resistance and leakage reactance of 0.02 Ω and 0.06 Ω per phase respectively. Given that the iron loss of transformer is 20 kW. Calculate secondary phase and line currents, primary line current and efficiency of transformer.

## Short answer

Secondary line current is 328.0 A and phase current is 189.4 A (delta); primary line current is 27.34 A (star). Copper loss is 24.57 kW plus iron loss 20 kW gives efficiency 99.12%.

## Answer

**Reading of the data:** the load is 5 MW in total (balanced, three-phase) at 0.8 pf lag, supplied at 11 kV line voltage from the delta secondary; the 132 kV primary is star-connected. Winding R and X are per phase. Copper loss is found from winding currents; the voltage drop is neglected when finding currents.

#### Secondary currents (delta, 11 kV)

$$
\begin{aligned}
I_{L2} &= \frac{P}{\sqrt3 V_L\cos\phi} = \frac{5\times10^6}{\sqrt3\times11000\times0.8} = 328.04\ \text{A} \\
I_{ph2} &= \frac{I_{L2}}{\sqrt3} = \frac{328.04}{\sqrt3} = 189.39\ \text{A}
\end{aligned}
$$

#### Primary current (star, 132 kV)

Phase voltages: primary $132/\sqrt3 = 76.21$ kV, secondary 11 kV, so the turns ratio per phase is $76.21/11 = 6.928$.

$$
I_{L1} = I_{ph1} = \frac{189.39}{6.928} = 27.34\ \text{A}
$$

Check: $5\times10^6/(\sqrt3\times132000\times0.8) = 27.34$ A.

#### Losses

$$
\begin{aligned}
P_{cu1} &= 3 I_{ph1}^2 R_1 = 3\times27.34^2\times10 = 22419\ \text{W} \\
P_{cu2} &= 3 I_{ph2}^2 R_2 = 3\times189.39^2\times0.02 = 2152\ \text{W} \\
P_{cu} &= 22419 + 2152 = 24571\ \text{W} \\
P_{loss} &= 24571 + 20000 = 44571\ \text{W}
\end{aligned}
$$

(The leakage reactances cause voltage drop but no loss, so they do not enter the efficiency.)

#### Efficiency

$$
\eta = \frac{5000}{5000 + 44.571}\times100 = 99.12\%
$$

**Answer:** secondary phase current = **189.4 A**, secondary line current = **328.0 A**, primary line current = **27.34 A**, efficiency = **99.12%**.

## Check yourself

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

1. How is the secondary line current found from the 5 MW load?
   A. P/(sqrt3 x VL x cos phi) = 328 A
   B. P/(3 VL) = 152 A
   C. P/(VL cos phi) = 568 A
   D. P/(sqrt3 x VL) = 262 A
   Answer: A. For three-phase power P = sqrt3 VL IL cos phi, so IL = 5e6/(sqrt3 x 11000 x 0.8) = 328.04 A. (Key point: Secondary line current)

2. The secondary is delta. How does phase current relate to line current?
   A. Phase current = line current / sqrt3
   B. Phase current = line current
   C. Phase current = line current / 3
   D. Phase current = line current x sqrt3
   Answer: A. In delta, line current = sqrt3 x phase current, so Iph2 = 328.04/sqrt3 = 189.39 A. (Key point: Delta phase current)

3. How is the primary (star) current obtained?
   A. Divide the line current by sqrt3 only
   B. Multiply Iph2 by 12
   C. Divide by 12 giving 15.8 A
   D. Divide Iph2 by phase turns ratio 6.928, giving 27.34 A
   Answer: D. Phase voltage ratio is (132/sqrt3)/11 = 6.928; star line = phase current = 189.39/6.928 = 27.34 A. (Key point: Primary line current)

4. Which losses enter the efficiency calculation?
   A. Only iron loss
   B. Copper loss in both windings (I^2R) plus iron loss, not leakage reactance
   C. Only primary copper loss
   D. Reactance drops included as losses
   Answer: B. Reactance causes voltage drop but no real power loss; total loss = 24.57 kW copper + 20 kW iron, giving 99.12%. (Key point: Efficiency losses)

## Close variants set in other exams

- [Currents and efficiency of 132/11 kV transformer at 4 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-4mw-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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