1. The Transformer - The Transformer Principle
When an alternating current is passed through a coil the magnetic
field produced will also alternate.
If a second coil is placed within this magnetic field the alternating flux will link
the two coils and induce an emf in the second coil. This process is called mutual
induction and is also known as the transformer effect.
AC Supply
Alternating
magnetic flux
AC Supply
Alternating
Induced
emf
Primary coil Secondary coil

2. The Double Wound Transformer
A transformer is commonly used to change (transform) an
alternating voltage from one value to another.
The basic transformer consists of two coils, called the ‘primary’ and
‘secondary’ each electrically ‘isolated’ from the other, there is no
electrical connection between the primary and secondary coils.
Primary
winding
Secondary
winding
Alternating
induced emf
Applied AC
supply
Laminated
iron core
Magnetic
flux
If both coils are mounted on an iron core the magnetic flux and coupling
between coils will increase, this assembly is called a transformer.

3. The Transformer – Turns Ratio
An alternating current flowing in one winding will produce an alternating flux in the core
and since the turns of both windings cut this flux the emf induced in each turn is the
same for each winding.
The volts per turn for primary and secondary windings are the same.
The ratio of secondary turns (NS) to primary turns (NP) is called the turns ratio.
The ratio of secondary voltage (VS) to primary voltage (VP) depends on the turns ratio.
VP
VSNP
NS
IP
IS
Primary
Winding Secondary
Winding
VP
VS
NP
NS
==
IS
IP
=

4. The Double Wound Transformer
Activity – Turns Ratio
1. A transformer is required to provide a 60V output from a 240V ac supply.
Calculate (a) the turns ratio required and (b) the number of primary turns, if the
secondary is wound with 500 turns.
2. A transformer operating from a 240V ac supply is required to provide two
secondary windings one of 12 volts and the other 18 volts. If the primary
winding has 2400 turns determine the turns on each secondary winding.
3. A transformer operates from a 240 volt supply and has a secondary voltage
of 24 volts, if this secondary winding has 96 turns determine the turns required
for the primary winding.

5. The Transformer – Power Conservation
If we neglect transformer losses since they are very small the efficiency can
be taken as 100%, making the secondary power (PS) equal to the primary
power (PP) .
VS
VP
NS
NP
==
PS = PP
VS IS = VP IP
but
IP
IS
=
VS
VP
IP
IS
=
NS
NP
This provides a more general expression linking voltage and turns to the
current in each winding.
VP
VS
NP
NS
==
IS
IP
=

6. Transformer Rating
When designing or specifying a transformer two main items of information are
needed – Secondary Voltage (V) and Secondary Current (I).
The rating of a transformer is defined in terms of the above information in the
form ‘VA’ (volt-amp).
The ‘VA’ rating is the maximum operating power level a transformer can
deliver to a load.
The above example assumes an ‘ideal’ transformer with no losses.
120VA
Transformer
IP 0.5A IS 6.6A
VP
240V
VS
18V

7. The Double Wound Transformer
Activity – Transformer Rating
1. A 600VA transformer operating from a 240 volt supply has a 6 volt
secondary winding. Determine the ‘full load’ current capability of the
secondary winding and the current drawn from the supply when operating
at full load.
2. A welding set transformer operating from a 240V mains supply has a
rating of 2000 VA (2kVA). If the full load secondary voltage is 3 volts
determine the a) the maximum welding current and b) the current drawn
from the mains supply under full load.
3. A 60VA transformer has two 12 volt secondary windings, determine the
secondary current capability.
NOTE: In all cases an ideal transformer is assumed.

8. The Transformer - Losses
Iron (core) loss
Iron losses are made up of ‘hysteresis’ loss and ‘eddy’ current loss
both cause heat to be generated. The iron loss is constant and not
dependant on load conditions meaning that the no-load iron loss is
the same as the full-load iron loss.
Copper (I2
R) loss
Copper losses are due to the effect of resistance on the currents of
primary and secondary windings causing heat to be generated. The
copper loss is a power loss and dependant on the square of the load
current. Therefore a transformer operating at half load will have only
a quarter of the copper loss it has on full-load.

9. The Transformer – Efficiency
The input power to a transformer provides both the output power and
the transformer losses.
input power = output power + power
losses
efficiency =
output power
input power
x 100%
=
output power
input power + power loses
x 100%efficiency
Primary
winding
Secondary
winding
Output power
PO
Input power
PI

10. The Transformer – Regulation
The resistance and inductance of a transformer provide an
impedance to the output current resulting in a volt drop due to the
load.
On no-load there will be no secondary current and no volt drop. On
full-load the output voltage will fall.
The difference between the no-load voltage and the full-load voltage,
expressed as a percentage of the no-load voltage is called the
‘voltage regulation’.
no load output voltage – full load output voltage
no load output voltage
x 100%Regulation =

11. Toroidal Equipment
Transformer
Common Types of Transformer for 50Hz Operation
Package styles vary but all have a laminated iron core to support the magnetic
flux linking the windings.
Chassis Mounted
Equipment
Transformer
Audio
Transformer
Circuit Board Mounted
Equipment Transformer