GATE EE

Auto-Transformer GATE Electrical Engineering - Quick Notes

Lecture Notes

SEC 01

Auto-Transformer Fundamentals

1Auto-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)

1GATE Important Point

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

1GATE Exam Focus

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

1Working Principle
1Energy Distribution
\[\begin{aligned} P_{total} &= P_{electromagnetic} + P_{conduction}\\ P_{electromagnetic} &= V_2 \times I_c \times \cos\phi\\ P_{conduction} &= V_2 \times (I_2 - I_c) \times \cos\phi \end{aligned}\]
SEC 02

Voltage and Current Relations

1Transformation Ratio and Voltage Relations
1GATE Key Formulas
\[\begin{aligned} \text{Transformation ratio: } K &= \dfrac{V_1}{V_2} = \dfrac{N_1}{N_2}\\ \text{For step-down: } V_1 &> V_2 \text{ and } K > 1\\ \text{For step-up: } V_2 &> V_1 \text{ and } K < 1 \end{aligned}\]
1GATE Important
1Current Relations - Step Down Auto-Transformer
1Current Equations (Step-Down)
\[\begin{aligned} I_1 &= \dfrac{I_2}{K} \text{ (from load)}\\ I_c &= I_2 - I_1 = I_2\left(1 - \dfrac{1}{K}\right)\\ I_c &= I_2\left(\dfrac{K-1}{K}\right) \end{aligned}\]
1Current Equations (Step-Up)
\[\begin{aligned} I_2 &= \dfrac{I_1}{K}\\ I_c &= I_1 - I_2 = I_1\left(1 - \dfrac{1}{K}\right)\\ I_c &= I_1\left(\dfrac{K-1}{K}\right) \end{aligned}\]
1GATE Tip

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

SEC 03

Power and Efficiency Analysis

1Power Relations
1Power Equations
\[\begin{aligned} P_{input} &= V_1 I_1 \cos\phi_1\\ P_{output} &= V_2 I_2 \cos\phi_2\\ P_{electromagnetic} &= V_2 I_c \cos\phi\\ P_{conduction} &= V_2 (I_2 - I_c) \cos\phi \end{aligned}\]
1GATE Important Formula

Copper Saving Ratio:

\[\text{Copper Saving} = \left(1 - \dfrac{1}{K}\right) \times 100\%\]
1Efficiency Analysis
1Efficiency Formula
\[\eta = \dfrac{P_{output}}{P_{input}} = \dfrac{V_2 I_2 \cos\phi_2}{V_1 I_1 \cos\phi_1}\]
1GATE Key Point

Auto-transformer efficiency is higher because:

SEC 04

Advantages and Disadvantages

1Advantages - GATE Perspective
1Economic Benefits

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

  • Weight reduction due to less copper

  • Smaller size for same power rating

1Disadvantages - GATE Important
1GATE Critical Point

Auto-transformers are NOT used when:

SEC 05

GATE Problem Solving

1GATE Problem Type 1
1Problem 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

1Solution
\[\begin{aligned} K &= \dfrac{400}{200} = 2\\ I_1 &= \dfrac{I_2}{K} = \dfrac{50}{2} = 25 \text{ A}\\ I_c &= I_2 - I_1 = 50 - 25 = 25 \text{ A}\\ P_{em} &= 200 \times 25 \times 0.8 = 4000 \text{ W}\\ P_{cond} &= 200 \times 25 \times 0.8 = 4000 \text{ W} \end{aligned}\]
1GATE Problem Type 2
1Problem Statement

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

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

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

1GATE Important Formulas - Quick Reference
1Must Remember for GATE
\[\begin{aligned} K &= \dfrac{V_1}{V_2} = \dfrac{N_1}{N_2} \text{ (Transformation ratio)}\\ I_c &= I_2\left(1 - \dfrac{1}{K}\right) \text{ (Common winding current)}\\ \text{Copper Saving} &= \left(1 - \dfrac{1}{K}\right) \times 100\%\\ P_{em} &= V_2 I_c \cos\phi \text{ (Electromagnetic power)}\\ P_{cond} &= V_2(I_2 - I_c)\cos\phi \text{ (Conduction power)}\\ \eta &= \dfrac{P_{out}}{P_{in}} \text{ (Efficiency)} \end{aligned}\]
1GATE Memory Tips
SEC 06

Applications

1GATE 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

1GATE Focus Areas
SEC 07

Summary

1GATE Exam Summary
1GATE 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?