---
title: "Star-delta transformer cannot parallel with star-star"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:parallel-star-delta-justify"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/parallel-star-delta-justify
site: BCT Engineering
---

# Star-delta transformer cannot parallel with star-star

Electrical Machines · Chapter 2: Transformer

- Asked in: 2067 Mangsir (old course), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/parallel-star-delta-justify
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

Justify the statement: It is not possible to operate star delta transformer in parallel with star-star or delta-delta transformer.

## Short answer

A star–delta transformer introduces a 30° phase shift between primary and secondary line voltages, while star–star and delta–delta have 0° shift. Paralleling them creates a voltage difference of about 52% of rated voltage across very low impedance, causing a large circulating current that can damage windings.

## Answer

Parallel operation requires that the secondary line voltages of the two banks be exactly equal in magnitude and in phase. A star–delta transformer does not meet this condition with a star–star or a delta–delta bank.

**Reason: phase displacement**

- In a star–star or delta–delta transformer, the secondary line voltage is in phase with the primary line voltage (0° shift).
- In a star–delta (or delta–star) transformer, the secondary line voltage is displaced by 30° from the primary (lagging for Yd1, leading for Dy11).

When the primaries are on the same supply, the secondary line voltages of the two banks differ by 30°.

```text
  V (Y-Y secondary)
   ^
   |    30 deg
   |  /  V (Y-D secondary)
   | /
   +----------------->
```

The voltage difference across the paralleled terminals is

$$
\Delta V = 2V\sin\frac{30^\circ}{2} = 0.518\,V
$$

This is about 52% of the rated secondary voltage. It appears across the very low internal impedances of the two transformers, so a huge circulating current flows, even at no load. The windings overheat and may be damaged, and the load cannot be shared properly.

**Also:** the voltage ratios differ in form. Star–delta has $V_{L2}/V_{L1} = 1/(\sqrt3\,a)$ compared with $1/a$ for star–star, so the turns would have to be redesigned. Even if the magnitudes matched, the 30° displacement cannot be removed by any change of turns ratio.

Hence star–delta can be paralleled only with another star–delta, or with a transformer that has the same phase displacement (the same vector group), such as delta–star of the right group.

## Check yourself

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

1. A star-delta bank differs from a star-star or delta-delta bank by:
   A. A 30 degree phase shift
   B. A different frequency
   C. Opposite polarity only
   D. A 90 degree shift
   Answer: A. YY and DD have 0 degrees; YD has 30 degrees. (Key point: Phase displacement)

2. The voltage difference between the two secondaries is:
   A. V sin 30 = 0.5 V
   B. 2V sin(15 degrees) = 0.518 V
   C. 0
   D. 2V
   Answer: B. Two equal phasors displaced by 30 degrees. (Key point: Voltage difference)

3. Because the internal impedances are small, this difference causes:
   A. No current
   B. Small load sharing error
   C. Huge circulating current, even at no load
   D. Voltage rise
   Answer: C. About 52% of rated voltage across a very small impedance. (Key point: Consequence)

4. Can a change of turns ratio remove the problem?
   A. Yes, by adjusting turns
   B. Yes, by lowering the frequency
   C. Yes, by a resistor
   D. No, a 30 degree shift cannot be removed by any turns ratio
   Answer: D. Only the same vector group can be paralleled. (Key point: Fix)

## Related in Electrical Machines

- [Star-delta starting torque ratio and full-load slip](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-star-delta-starting-torque-ratio) — Electrical Machines, chapter 5
- [Parallel transformers and per-unit impedance load sharing](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aparallel-operation-per-unit-impedance) — Electrical Machines I, chapter 2
- [CT secondary should not be kept open](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Act-secondary-open-true-false) — Electrical Machines, chapter 2
- [Load sharing of 50 kVA and 25 kVA transformers in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aparallel-50kva-25kva-60kva-load) — Electrical Machines I, chapter 2
- [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×
- [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×

## Related in other subjects

- [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
- [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
- [CT ratio for 33/6.6 kV star/delta (300:5)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Adifferential-ct-33-6-6kv-300) — Switchgear and Protection, chapter 7
- [CT ratio for 33/6.6 kV star/delta (400/5)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Adifferential-ct-33-6-6kv-400) — Switchgear and Protection, chapter 7