Auto-Transformer GATE Electrical Engineering - Quick Notes

Auto-Transformer Fundamentals

Auto-Transformer Definition

  • Definition: Single winding transformer with electrical connection between primary and secondary

  • Key Feature: No electrical isolation between input and output

  • Construction: Continuous winding with intermediate tap

  • Terminals: Three terminals - High (H), Low (L), and Common (X)

GATE Important Point

Auto-transformer combines both electromagnetic induction and electrical conduction for energy transfer.

GATE Exam Focus

Remember: Auto-transformer is NOT suitable when electrical isolation is required.

Working Principle

  • Electromagnetic Induction: Flux linkage between turns

  • Electrical Conduction: Direct current flow through common winding

  • Energy Transfer: Approximately 85% electromagnetic + 15% conduction

Energy Distribution

\[\begin{aligned} P_{total} &= P_{electromagnetic} + P_{conduction}\\ P_{electromagnetic} &= V_2 \times I_c \times \cos\phi\\ P_{conduction} &= V_2 \times (I_2 - I_c) \times \cos\phi \end{aligned}\]

Voltage and Current Relations

Transformation Ratio and Voltage Relations

GATE Key Formulas

\[\begin{aligned} \text{Transformation ratio: } K &= \dfrac{V_1}{V_2} = \dfrac{N_1}{N_2}\\ \text{For step-down: } V_1 &> V_2 \text{ and } K > 1\\ \text{For step-up: } V_2 &> V_1 \text{ and } K < 1 \end{aligned}\]

GATE Important

  • Voltage transformation follows same law as conventional transformer

  • Current distribution is different due to common winding

  • Typical voltage ratios: 2:1 to 3:1 for economic benefits

Current Relations - Step Down Auto-Transformer

Current Equations (Step-Down)

\[\begin{aligned} I_1 &= \dfrac{I_2}{K} \text{ (from load)}\\ I_c &= I_2 - I_1 = I_2\left(1 - \dfrac{1}{K}\right)\\ I_c &= I_2\left(\dfrac{K-1}{K}\right) \end{aligned}\]

Current Equations (Step-Up)

\[\begin{aligned} I_2 &= \dfrac{I_1}{K}\\ I_c &= I_1 - I_2 = I_1\left(1 - \dfrac{1}{K}\right)\\ I_c &= I_1\left(\dfrac{K-1}{K}\right) \end{aligned}\]

GATE Tip

Common winding current \(I_c\) carries the difference between primary and secondary currents.

Power and Efficiency Analysis

Power Relations

Power Equations

\[\begin{aligned} P_{input} &= V_1 I_1 \cos\phi_1\\ P_{output} &= V_2 I_2 \cos\phi_2\\ P_{electromagnetic} &= V_2 I_c \cos\phi\\ P_{conduction} &= V_2 (I_2 - I_c) \cos\phi \end{aligned}\]

GATE Important Formula

Copper Saving Ratio:

\[\text{Copper Saving} = \left(1 - \dfrac{1}{K}\right) \times 100\%\]

Efficiency Analysis

  • Higher efficiency than conventional transformers (95-98%)

  • Reduced copper losses in common winding section

  • Same iron losses as conventional transformers

Efficiency Formula

\[\eta = \dfrac{P_{output}}{P_{input}} = \dfrac{V_2 I_2 \cos\phi_2}{V_1 I_1 \cos\phi_1}\]

GATE Key Point

Auto-transformer efficiency is higher because:

  • Reduced \(I^2R\) losses in common winding

  • Less copper required for same power rating

Advantages and Disadvantages

Advantages - GATE Perspective

  • Higher efficiency \((2-3\%\) better than conventional)

  • Reduced size and cost due to copper saving

  • Better voltage regulation (lower impedance)

  • Lower losses in common winding

  • Improved power factor at light loads

Economic Benefits

  • Cost reduction proportional to \((1 - 1/K)\)

  • Weight reduction due to less copper

  • Smaller size for same power rating

Disadvantages - GATE Important

  • No electrical isolation - major limitation

  • Higher short-circuit current due to direct connection

  • Limited voltage ratio applications (economical up to \(3:1\))

  • Safety concerns - common neutral point

  • Harmonic transfer between primary and secondary

GATE Critical Point

Auto-transformers are NOT used when:

  • Electrical isolation is required

  • High voltage ratios are needed (\(>3:1\))

  • Safety is a primary concern

GATE Problem Solving

GATE Problem Type 1

Problem Statement

A 10 kVA, 400V/200V auto-transformer is connected as step-down. Load current is 50 A at 0.8 pf lagging. Calculate:

  1. Transformation ratio

  2. Primary current

  3. Common winding current

  4. Power transferred electromagnetically and conductively

Solution

\[\begin{aligned} K &= \dfrac{400}{200} = 2\\ I_1 &= \dfrac{I_2}{K} = \dfrac{50}{2} = 25 \text{ A}\\ I_c &= I_2 - I_1 = 50 - 25 = 25 \text{ A}\\ P_{em} &= 200 \times 25 \times 0.8 = 4000 \text{ W}\\ P_{cond} &= 200 \times 25 \times 0.8 = 4000 \text{ W} \end{aligned}\]

GATE Problem Type 2

Problem Statement

Calculate the copper saving in an auto-transformer with transformation ratio \(2.5:1\) compared to a conventional transformer of same rating.

Solution

\[\begin{aligned} K &= 2.5\\ \text{Copper Saving} &= \left(1 - \dfrac{1}{K}\right) \times 100\%\\ &= \left(1 - \dfrac{1}{2.5}\right) \times 100\%\\ &= \left(1 - 0.4\right) \times 100\%\\ &= 0.6 \times 100\% = 60\% \end{aligned}\]

GATE Tip

Higher the transformation ratio, greater the copper saving (up to practical limits).

GATE Important Formulas - Quick Reference

Must Remember for GATE

\[\begin{aligned} K &= \dfrac{V_1}{V_2} = \dfrac{N_1}{N_2} \text{ (Transformation ratio)}\\ I_c &= I_2\left(1 - \dfrac{1}{K}\right) \text{ (Common winding current)}\\ \text{Copper Saving} &= \left(1 - \dfrac{1}{K}\right) \times 100\%\\ P_{em} &= V_2 I_c \cos\phi \text{ (Electromagnetic power)}\\ P_{cond} &= V_2(I_2 - I_c)\cos\phi \text{ (Conduction power)}\\ \eta &= \dfrac{P_{out}}{P_{in}} \text{ (Efficiency)} \end{aligned}\]

GATE Memory Tips

  • Auto-transformer = Higher efficiency + No isolation

  • Copper saving increases with transformation ratio

  • Suitable for voltage ratios up to \(3:1\)

Applications

GATE Relevant Applications

Power Systems:

  • Voltage regulation

  • Transmission interconnection

  • Distribution voltage control

  • Reactive power compensation

Industrial Applications:

  • Motor starting (reduced current)

  • Voltage stabilizers

  • Laboratory equipment

  • Lighting control

GATE Focus Areas

  • Motor starting applications (current reduction)

  • Voltage regulation in power systems

  • Efficiency calculations and comparisons

  • Economic analysis (copper saving)

Summary

GATE Exam Summary

  • Definition: Single winding transformer with electrical connection

  • Key Feature: No electrical isolation between primary and secondary

  • Efficiency: Higher than conventional transformers (95-98%)

  • Copper Saving: \((1 - 1/K) \times 100\%\)

  • Applications: Limited by lack of isolation

  • Voltage Ratio: Economical up to 3:1

GATE Quick Check

  • Can you calculate transformation ratio, currents, and power distribution?

  • Do you know when NOT to use auto-transformers?

  • Can you compare efficiency with conventional transformers?