---
title: "Delta/star 120 kVA transformer values"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:delta-star-120kva-line-phase-values"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/delta-star-120kva-line-phase-values
site: BCT Engineering
---

# Delta/star 120 kVA transformer values

Electrical Machine · Chapter 2: Transformer

- Asked in: 2074 Bhadra, 6 marks
- Video: [Delta and Wye - Volts, Amps, & VA](https://www.youtube.com/watch?v=dLqz_zPpgHI) — Dave Gordon
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/delta-star-120kva-line-phase-values
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

An ideal 3-phase delta/star step-down transformer delivers power to a balanced 3-phase load of 120 kVA at 0.8 power factor. The input line voltage is 11 kV and the turns ratio of the transformer, phase to phase is 10. Determine the line voltage, line currents, phase voltages and phase currents on both the primary and secondary sides.

## Short answer

Primary (delta): VL=11 kV, Vph=11 kV, IL=6.30 A, Iph=3.64 A. Secondary (star): VL=1905 V, Vph=1100 V, IL=36.4 A, Iph=36.4 A.

## Answer

A balanced three-phase load of 120 kVA is supplied. For a delta/star step-down transformer, the primary is delta-connected and the secondary is star-connected.

**Given:** S = 120 kVA, cos φ = 0.8 lag; primary line voltage VL1 = 11 kV; phase turns ratio N1/N2 = 10.

**Assumption:** the transformer is ideal, so input kVA = output kVA = 120 kVA.

#### Primary side (delta)
```text
Line voltage       VL1 = 11 000 V
Phase voltage      Vph1 = VL1 = 11 000 V   (delta)
Line current       IL1 = S/(√3 VL1)
                       = 120 000/(√3 × 11 000) = 6.30 A
Phase current      Iph1 = IL1/√3 = 6.30/1.732 = 3.64 A
```

#### Secondary side (star)
```text
Phase voltage      Vph2 = Vph1/10 = 1100 V
Line voltage       VL2 = √3 Vph2 = 1.732 × 1100 = 1905 V
Line current       IL2 = S/(√3 VL2)
                       = 120 000/(√3 × 1905) = 36.4 A
Phase current      Iph2 = IL2 = 36.4 A   (star)
```
Check: Iph2/Iph1 = 36.4/3.64 = 10, as required by the turns ratio.

| Quantity | Primary (delta) | Secondary (star) |
|---|---|---|
| Line voltage | 11 000 V | 1905 V |
| Phase voltage | 11 000 V | 1100 V |
| Line current | 6.30 A | 36.4 A |
| Phase current | 3.64 A | 36.4 A |

The power factor of 0.8 gives the load power of 96 kW, which does not change the above currents, because they are determined from the kVA.

## Check yourself

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

1. For the delta-connected primary, what is the relation between line and phase quantities?
   A. Vph = VL, Iph = IL/√3
   B. Vph = VL/√3, Iph = IL
   C. Vph = VL, Iph = IL
   D. Vph = √3 VL, Iph = IL/√3
   Answer: A. In delta, phase voltage equals line voltage and line current is √3 times phase current. (Key point: Delta line-phase relation)

2. What is the primary line current for 120 kVA at 11 kV?
   A. 10.9 A
   B. 6.30 A
   C. 3.64 A
   D. 36.4 A
   Answer: B. IL1 = 120000/(√3 × 11000) = 6.30 A. (Key point: Line current from kVA)

3. With phase turns ratio 10 and 11 kV on delta, what is the secondary line voltage (star)?
   A. 1100 V
   B. 110 V
   C. 1905 V
   D. 3300 V
   Answer: C. Vph2 = 1100 V and VL2 = √3 × 1100 = 1905 V. (Key point: Secondary line voltage)

## Related in Electrical Machine

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