---
title: "Primary current and efficiency of 11 kV/380 V Δ/Y transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:delta-star-11kv-380v-40kw-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/delta-star-11kv-380v-40kw-efficiency
site: BCT Engineering
---

# Primary current and efficiency of 11 kV/380 V Δ/Y transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2070 Magh, 8 marks
- Video: [Tutorial 1 - Three Phase Transformer line, phase currents and power calculations](https://www.youtube.com/watch?v=zHBlovJyXiQ) — CMTEQ
- 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-11kv-380v-40kw-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

An 11 kV/380 V delta/star 3-phase transformer has balanced star connected 3-phase load of 40 kW at p.f. of 0.8 lagging per phase. Calculate the primary line current. If the transformer has iron loss of 1.0 kW, calculate the approximate efficiency of the transformer. Given that primary winding resistance and leakage reactance are 25 Ω per phase and 40 Ω per phase respectively. Secondary winding resistance and leakage reactance are 0.01 Ω per phase and 0.02 Ω per phase respectively.

## Short answer

Total load is 40 kW at 0.8 pf, so apparent power = 50 kVA. Primary line current = 50000/(√3×11000) = 2.624 A. Copper losses total 345.31 W, iron loss 1 kW, efficiency = 40000/(40000+1345.31) = 96.75%.

## Answer

Primary is **delta** (phase voltage = 11 kV, phase current = line current/$\sqrt3$); secondary is **star** (phase current = line current). Winding resistances are per phase.

Load apparent power:
$$
S = \frac{40}{0.8} = 50\ \text{kVA}
$$

#### Primary line current
Neglecting no-load current, input VA = output VA:
$$
\begin{aligned}
I_{1L} &= \frac{S}{\sqrt3\,V_{1L}} = \frac{50000}{\sqrt3 \times 11000} = 2.624\ \text{A} \\
I_{1ph} &= \frac{I_{1L}}{\sqrt3} = 1.515\ \text{A}
\end{aligned}
$$

#### Secondary current
$$
I_{2L} = I_{2ph} = \frac{50000}{\sqrt3 \times 380} = 75.97\ \text{A}
$$
Check by turns ratio: $K = \frac{380/\sqrt3}{11000} = 0.01995$, $I_{1ph} = 0.01995 \times 75.97 = 1.515$ A.

#### Copper losses
$$
\begin{aligned}
P_{cu1} &= 3I_{1ph}^2R_1 = 3 \times 1.515^2 \times 25 = 172.18\ \text{W} \\
P_{cu2} &= 3I_{2ph}^2R_2 = 3 \times 75.97^2 \times 0.01 = 173.13\ \text{W} \\
P_{cu} &= 172.18 + 173.13 = 345.31\ \text{W}
\end{aligned}
$$
(Leakage reactances cause no power loss.)

#### Efficiency
$$
\begin{aligned}
\text{Losses} &= P_i + P_{cu} = 1000 + 345.31 = 1345.31\ \text{W} \\
\eta &= \frac{40000}{40000 + 1345.31}\times 100 = 96.75\%
\end{aligned}
$$

| Quantity | Value |
|---|---|
| Primary line current | 2.624 A |
| Primary phase current | 1.515 A |
| Secondary line/phase current | 75.97 A |
| Copper loss | 345.31 W |
| Efficiency | 96.75% |

**Answer:** Primary line current = **2.624 A**; approximate efficiency = **96.75%**.

## Check yourself

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

1. For 40 kW at 0.8 pf the apparent power is:
   A. 32 kVA
   B. 40 kVA
   C. 50 kVA
   D. 64 kVA
   Answer: C. S = 40/0.8 = 50 kVA. (Key point: Load VA)

2. The primary line current is:
   A. S/(sqrt3 x 11000) = 2.624 A
   B. S/11000
   C. S/(sqrt3 x 380)
   D. 1.515 A
   Answer: A. Input VA equals output VA; the primary line voltage is 11 kV. (Key point: Primary line current)

3. In the delta primary the phase current is:
   A. Equal to line current
   B. sqrt3 times line current
   C. Line current divided by sqrt3 = 1.515 A
   D. Line current divided by 3
   Answer: C. In delta, line current = sqrt3 x phase current. (Key point: Delta phase current)

4. Total copper loss uses:
   A. Line currents and reactances
   B. 3 Iph^2 R for each winding, with phase currents
   C. Only the primary winding
   D. The iron loss
   Answer: B. Pcu = 3 x 1.515^2 x 25 + 3 x 75.97^2 x 0.01 = 345.31 W, giving efficiency 96.75%. (Key point: Copper loss)

## 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×
- [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×

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