# Single Phase Transformer

Topics: Transformer, Alternating current, Switched-mode power supply Pages: 28 (3677 words) Published: March 27, 2013
Transformer
BEE2123 ELECTRICAL MACHINES
Mohd Rusllim Bin Mohamed Ext: 2080 A1-E10-C09 rusllim@ump.edu.my © MRM 05

Learning Outcomes
 At the end of the lecture, student should to:

Understand the principle and the nature of static machines of transformer. Perform an analysis on transformers which their principles are basic to the understanding of electrical machines.

Introduction
  

A transformer is a static machines. The word „transformer‟ comes form the word „transform‟. Transformer is not an energy conversion device, but is a device that changes AC electrical power at one voltage level into AC electrical power at another voltage level through the action of magnetic field, without a change in frequency. It can be either to step-up or step down. Transmission System

TX1 TX1

Generation Station
33/13.5kV 13.5/6.6kV

Distributions
TX1

TX1

6.6kV/415V Consumer

Transformer Construction

Two types of iron-core construction:
a) b)

Core - type construction Shell - type construction

Core - type construction

Transformer Construction

Shell - type construction

Ideal Transformer
 An ideal transformer is a transformer which has no loses,

i.e. it‟s winding has no ohmic resistance, no magnetic leakage, and therefore no I2 R and core loses.  However, it is impossible to realize such a transformer in practice.  Yet, the approximate characteristic of ideal transformer will be used in characterized the practical transformer. N1 : N2 I1 V1 E1 E2 I2 V2

V1 – Primary Voltage V2 – Secondary Voltage E1 – Primary induced Voltage E2 – secondary induced Voltage N1:N2 – Transformer ratio

Transformer Equation
 Faraday‟s Law states that,  If the flux passes through a coil of wire, a voltage will be induced in the turns of wire. This voltage is directly proportional to the rate of change in the flux with respect of time.

Vind  Emf ind

d(t )  dt

Lenz‟s Law

If we have N turns of wire,

Vind  Emf ind

d(t )  N dt

Transformer Equation
 For an ac sources,  Let V(t) = Vm sint i(t) = im sint Since the flux is a sinusoidal function;

(t )   m sin t Then: Therefore: d m sin t
Vind  Emf ind   N dt   N m cos t

Thus:

Vind  Emfind (max)  N m  2fN m
N m 2fN m    4.44 fN m 2 2

Emf ind ( rms)

Transformer Equation
 For an ideal transformer

E1  4.44 fN1 m

………………… (i)  In the equilibrium condition, both the input power will be equaled to the output power, and this condition is said to ideal condition of a transformer.

E2  4.44 fN 2 m

Input power  output power V1 I1 cos   V2 I 2 cos   V1 I 2  V2 I1

 From the ideal transformer circuit, note that,

E1  V1 and E2  V2
 Hence, substitute in (i)

Transformer Equation
Therefore, E1 N1 I 2   a E2 N 2 I1

Where, „a‟ is the Voltage Transformation Ratio; which will determine whether the transformer is going to be step-up or step-down For a >1 For a E2 E1 < E2 © MRM 05

Step-down Step-up

Transformer Rating
 Transformer rating is normally written in terms of Apparent

Power.  Apparent power is actually the product of its rated current and rated voltage.

VA  V1I1  V2 I 2
 Where,  I1 and I2 = rated current on primary and secondary winding.  V1 and V2 = rated voltage on primary and secondary winding. 

Rated currents are actually the full load currents in transformer © MRM 05

Example
1.

1.5kVA single phase transformer has rated voltage of 144/240 V. Finds its full load current. Solution

1500 I1FL   10.45 A 144 1500 I 2 FL   6A 240

Example
2. A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60m2. If the primary winding is connected to a 50Hz supply at 520V, calculate: a) The induced voltage in the secondary winding b)...