---
title: "Copper saving in an autotransformer"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:autotransformer-copper-saving"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/autotransformer-copper-saving
site: BCT Engineering
---

# Copper saving in an autotransformer

Electrical Machine · Chapter 2: Transformer

- Asked in: 2074 Bhadra, 4 marks; 2070 Bhadra, 4 marks (asked 2 times)
- Video: [Saving of copper in an autotransformer](https://www.youtube.com/watch?v=hivePSSkhbU) — Dr.Jayaudhaya ,Simple and Easy Way
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/autotransformer-copper-saving
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

Derive an expression giving the amount of copper saving in an auto-transformer.

## Short answer

Copper weight ∝ turns × current. Autotransformer: Wₐ ∝ (N₁−N₂)I₁ + N₂(I₂−I₁) = 2N₁I₁(1−K). Two-winding: Wₜ ∝ 2N₁I₁. Saving = Wₜ − Wₐ = K·Wₜ, i.e., K×100 % of the copper is saved.

## Answer

An autotransformer has a single winding, part of which is common to the primary and the secondary. Because the common part carries only the difference current, less copper is needed than in a two-winding transformer.

**Circuit and notation** (step-down, N1 turns total, N2 turns across the output)
```text
 I1 -->  A o---------+---------o  (input)
                     |
             N1-N2   |
             turns   | I1
                 B o-+-----+---o  (output tap)
             N2      |     | I2
             turns   |     v
                 C o-+---------o  (common)
          V1 across AC, V2 across BC
```
- Transformation ratio K = N2/N1 = V2/V1 = I1/I2 (K < 1 for step-down).
- Section AB (N1 − N2 turns) carries current I1.
- Section BC (N2 turns) carries (I2 − I1).

**Weight of copper ∝ (number of turns × current)** for the same current density and mean turn length.

```text
Auto-transformer:
 W_a ∝ (N1 − N2) I1 + N2 (I2 − I1)

Two-winding transformer of the same output:
 W_t ∝ N1 I1 + N2 I2
```
Using N1 I1 = N2 I2 (ampere-turn balance, so I1 = K I2):
```text
W_a ∝ N1 I1 − N2 I1 + N2 I2 − N2 I1
    = N1 I1 + N2 I2 − 2 N2 I1
    = 2 N1 I1 − 2 N2 I1             (as N1 I1 = N2 I2)
    = 2 N1 I1 (1 − N2/N1) = 2 N1 I1 (1 − K)

W_t ∝ N1 I1 + N2 I2 = 2 N1 I1

W_a / W_t = 1 − K
```
**Saving of copper = W_t − W_a = K × W_t**

So the autotransformer needs (1 − K) times the copper of a two-winding unit, i.e. K × 100 % is saved. For K = 0.8 the saving is 80 %. The saving is large only when K is close to 1 (voltage ratios near 1).

## Check yourself

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

1. Why does an autotransformer need less copper?
   A. It uses thicker wire
   B. It has no iron core
   C. It has more turns
   D. Its common section carries only the difference current
   Answer: D. Section BC carries (I2 - I1), so less copper is needed. (Key point: Why less copper)

2. Copper weight is proportional to what?
   A. Turns only
   B. Turns x current (same current density)
   C. Current squared
   D. Voltage only
   Answer: B. Weight is proportional to turns x current for equal current density and mean turn length. (Key point: Copper weight)

3. What is Wa/Wt?
   A. 1 - K
   B. K
   C. 1 + K
   D. 1/K
   Answer: A. Wa = 2 N1 I1 (1 - K), Wt = 2 N1 I1. (Key point: Ratio)

4. For K = 0.8 what saving results?
   A. 20%
   B. 80%
   C. 64%
   D. 8%
   Answer: B. Saving = K x Wt = 80%. (Key point: Saving)

## Related in Electrical Machine

- [Copper saving in auto-transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aautotransformer-copper-saving) — Electrical Machines, chapter 2
- [Copper saving in auto transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aautotransformer-copper-saving) — Electrical Machines I, chapter 2
- [Auto transformer merits and Cu saving](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aautotransformer-merits-copper-saving) — Electrical Machines, chapter 2
- [Autotransformer working and Cu saving](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aautotransformer-principle-cu-saving) — Electrical Machine, chapter 2
- [CT secondary open; auto transformer economy](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Act-open-autotransformer-economical) — Electrical Machines, chapter 2
- [Transformer losses and efficiency expression](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-losses-efficiency-derivation) — Electrical Machine, chapter 2, asked 2×

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