---
title: "Primary current and efficiency of 100 kVA Δ/Y transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:delta-star-100kva-11kv-100a-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/delta-star-100kva-11kv-100a-efficiency
site: BCT Engineering
---

# Primary current and efficiency of 100 kVA Δ/Y transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2068 Bhadra, 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/delta-star-100kva-11kv-100a-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 100 kVA, 11 kV/400 V Delta/star 3-phase transformer has following parameters: R1 = 25 Ω, X1 = 50 Ω, R2 = 0.04 Ω, X2 = 0.4 Ω, Iron loss = 1000 Watts. A 3-phase balanced load draws per phase current of 100 Amp at 0.85 power factor lagging. Calculate primary line current and efficiency of the transformer.

## Short answer

Per-phase V2ph = 400/√3 = 230.94 V, turns ratio K = 230.94/11000 = 0.020995. Primary line current = √3 * K * 100 A = 3.636 A. Output = 3*230.94*100*0.85 = 58.89 kW. Total copper loss = 3*(2.0995^2*25 + 100^2*0.04) = 1530.58 W. Iron loss = 1000 W. Efficiency = 58890/(58890+2530.58)*100 = 95.88%.

## Answer

Primary **delta** (phase voltage 11 kV), secondary **star** (phase voltage $400/\sqrt3$). $R_1, X_1, R_2, X_2$ are per-phase values. Load current 100 A per phase (= line current, star) at 0.85 pf lagging.

#### Turns ratio per phase
$$
\begin{aligned}
V_{2ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\
K &= \frac{V_{2ph}}{V_{1ph}} = \frac{230.94}{11000} = 0.020995
\end{aligned}
$$

#### Primary line current
Neglecting no-load current:
$$
\begin{aligned}
I_{1ph} &= KI_{2ph} = 0.020995 \times 100 = 2.0995\ \text{A} \\
I_{1L} &= \sqrt3 \times 2.0995 = 3.636\ \text{A}
\end{aligned}
$$
Check: $I_{1L} = \frac{\sqrt3 \times 400 \times 100}{\sqrt3 \times 11000} = 3.636$ A.

#### Output power
$$
P_{out} = 3V_{2ph}I_{2ph}\cos\phi = 3 \times 230.94 \times 100 \times 0.85 = 58890\ \text{W} = 58.89\ \text{kW}
$$

#### Copper losses
$$
\begin{aligned}
P_{cu1} &= 3I_{1ph}^2R_1 = 3 \times 2.0995^2 \times 25 = 330.58\ \text{W} \\
P_{cu2} &= 3I_{2ph}^2R_2 = 3 \times 100^2 \times 0.04 = 1200\ \text{W} \\
P_{cu} &= 330.58 + 1200 = 1530.58\ \text{W}
\end{aligned}
$$

#### Efficiency
$$
\begin{aligned}
\text{Losses} &= 1000 + 1530.58 = 2530.58\ \text{W} \\
\eta &= \frac{58890}{58890 + 2530.58}\times 100 = 95.88\%
\end{aligned}
$$

| Quantity | Value |
|---|---|
| Primary phase current | 2.0995 A |
| Primary line current | 3.636 A |
| Output | 58.89 kW |
| Copper loss | 1530.58 W |
| Efficiency | 95.88% |

**Answer:** Primary line current = **3.636 A**; efficiency = **95.88%**.

## Check yourself

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

1. For a 11 kV/400 V delta/star transformer, the per-phase voltages are:
   A. 11000 V and 400 V
   B. 11000 V and 230.94 V
   C. 6351 V and 230.94 V
   D. 11000 V and 692.8 V
   Answer: B. Delta primary phase voltage = line voltage 11 kV; star secondary phase voltage = 400/sqrt3 = 230.94 V. (Key point: Phase voltages)

2. The primary line current for 100 A secondary line current is:
   A. 3.636 A
   B. 2.0995 A
   C. 9.09 A
   D. 5.25 A
   Answer: A. I1ph = K x 100 = 2.0995 A in delta; line current = sqrt3 x 2.0995 = 3.636 A. (Key point: Primary line current)

3. Total copper loss is computed as:
   A. 3 I1ph^2 R1 + 3 I2ph^2 R2 = 1530.58 W
   B. I1^2 R1 only
   C. 3 I2^2 R2 only = 1200 W
   D. 1000 W + 1200 W
   Answer: A. Both windings' per-phase I^2R are multiplied by 3: 330.58 + 1200 = 1530.58 W. (Key point: Copper loss calculation)

4. Output power and efficiency are:
   A. 58.89 kW and 95.88%
   B. 100 kW and 97%
   C. 58.89 kW and 98%
   D. 85 kW and 95.88%
   Answer: A. Output = 3 x 230.94 x 100 x 0.85 = 58.89 kW; losses 2530.58 W give 95.88%. (Key point: Efficiency)

## Related in Electrical Machines I

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