1. transformer
Transformer is a device used for converting a low
alternating voltage to a high alternating voltage or a high
alternating voltage into a low alternating voltage. It is a static
electrical device that transfers energy by inductive coupling
between its winding circuits. Transformers range in size from a
thumbnail-sized coupling transformer hidden inside a stage
microphone to huge units weighing hundreds of tons used in
power plant substations or to interconnect portions of the
power grid. All operate on the same basic principles, although

2. therange of designs is wide.
While new technologies have eliminated the need for
transformers in some electronic circuits, transformers are still
found in many electronic devices. Transformers are essential
for high-voltage electric power transmission, which makes long-
distance transmission economically practical. A transformer is
most widely used devicein both low and high current circuit. In
a transformer, the electrical energy transfer from onecircuit to
another circuit takes place without the use of moving parts. A
transformer which increases the voltages is called a step-up
transformer. A transformer which decreases the A.C. voltages is
called a step-down transformer.
Transformer is, therefore, an essentialpiece of apparatus both
for high and low current circuits.
transformer.A transformer which decreases the
A.C. voltages is called a step-down transformer.

3. Transformer is, therefore, an essentialpiece of apparatus both
for high and low current circuits.
 PRINCIPLE
It is based on the principle of mutual induction thatis if a varying
current is set-up in a circuit then induced e.m.f. is produced in
the neighboring
circuit. The varying current in a circuit produce varying magnetic
flux which induces e.m.f. in the neighboring circuit.

4. CONTRUCTION
A transformer consists of a rectangular shaft iron core
made of laminated sheets, well insulated from one
another. Two coils p1 & p2 and s1 & s2 are wound on the
same core, but are well insulated
with each other. Note that the both the coils are insulated
from the core, the source of alternating
e.m.f is connected to p1 p2, the primary coil and a load
resistance R is connected to s1 s2, the secondary coil
through an open switch S. Thus, no current can be drawn
through the secondary coil aslong as the switch is open For
an ideal transformer, we assume that the resistance of the
primary & secondary windings is negligible.
Further, the energy loses due to magnetic flux andiron core
is also negligible. For operation at low frequency, we may
use a soft iron. The soft iron core is insulating by joining
thin iron strips coated with varnish to insulate them to
reduce energy losses by eddy currents. The input circuit is

5. calledprimary and the output circuit is called secondary.
An ideal voltage step-down transformer. The
secondary current arises from the action of the
secondary EMF on the (not shown) load impedance.

6. The ideal transformer as a circuit element

7.  THEORY AND WORKING
When an altering e.m.f. is supplied to the primary coil p1p2, an
alternating current starts falling in it. The altering current in the
primary produces a changing magnetic flux, which induces
altering voltage in the primary as well as in the secondary. In a
good-transformer, whole of the magnetic flux linked with
primary is also linked with thesecondary, and then the induced
e.m.f. induced in each turn of the secondary is equal to that
induced in each turn of the primary. Thus if Ep and Es be the
instantaneous values of the induced e.m.f in the primary and
the secondary coil and Np and Ns are the no. of turns of the
primary secondary coils of the transformer and, dфB/dt = rate of
change of flux in each turn of the coil at this instant, we have
Ep = -Np dфB/dt …. (1)
Es = -Ns dфB/dt …. (2)
Since the above relations are true at every instant, so by
dividing (2) by (1), we get
Es / Ep = - Ns / Np …. (3)
As, Ep is the instantaneous value of back e.m.f induced in the

8. primary coil p1, so the instantaneouscurrent in primary coil is
due to the difference (E – Ep) in the instantaneous values of the
applied and back e.m.f. further if Rp is the resistance of p1 p2
coil, then the instantaneous current Ip in the primary coil is
given by,
I =E – Ep / Rp
E – Ep = Ip Rp
When the resistance of the primary is small, Rp Ip can be
neglected so therefore,
E – Ep = 0 or Ep = E
Thus, back e.m.f = input e.m.f
Hence, equation (3) can be written as Es / Ep = Es /E = output
e.m.f / input e.m.f = Ns / Np = K
where K is constant, called turn or transformation ratio.

9. In a step down transformer
Es < E so K < 1, hence Ns < Np
If Ip=value of primary current at the same instant tand Is =value
of sec. current at this instant, then Input power at the
instant t = Ep Ip and Output power at the same instant t = Es Is
If there are no losses of power in the transformer, then Input
power = output power or
Ep Ip = Es Is Or Es / Ep = Ip / Is = K
In a step up transformer
As, k > 1, so Ip > Is or Is < Ip
i.e. current in sec. is weaker when secondary voltage is higher.
Hence, whatever we gain in voltage, we lose in current in the
same ratio. Similarly, it can be shown, that in a step down
transformer, whatever we lose in voltage, we gain in current in
the same ratio. Thus a step up

10. transformer in reality steps down the current & astep down
transformer steps up the current.
Efficiency
Efficiency of a transformer is defined as the ratio of output
power to the input power i.e. η = output power / input power =
Es Is / Ep Ip
Thus in an ideal transformer, where there are no power losses,
η = 1. But in actual practice, there are many power losses;
therefore, the efficiency oftransformer is less than one.

11. Energy losses
In practice, the output energy of a transformer is always
less than the input energy, because energylosses occur
due to a number of reasons as explained below,
Loss of Magnetic Flux: The coupling betweenthe coils
is seldom perfect. So, whole of the magnetic flux
produced by the primary coil is not linked up with the
secondary coil.
Iron Loss: In actual iron cores in spite of
lamination,
Eddy currents are produced. The magnitude of eddy
current may, however be small. And a part of energy is lost
as the heat produced in the iron core.

12. Copper Loss: In practice, the coils of the transformer
possess resistance. So a part of theenergy is lost due to
the heat produced in the resistance of the coil.
Hysteresis Loss: The alternating current in thecoil tapes
the iron core through complete cycle of magnetization. So
Energy is lost due to hysteresis.
Magneto restriction: The alternating current in the
Transformer may be set its parts in to vibrations and sound
may be produced. It is called humming. Thus, a part of
energy may be lost due to humming.
A Big Transformer
USES OF TRANSFORMER

13. A transformer is used in almost all a.c. operations
 In voltage regulator for T.V., refrigerator,
computer,air conditioner etc.
 In the induction furnaces.
 A step down transformer is used for welding
purposes.
 A step down transformer is used for obtaining
large current.
 A step up transformer is used for the
production of X-Rays and NEON advertisement.
 Transformers are used in voltage regulators
and stabilized power supplies.
 Transformers are used in the transmissions of
a.c over long distances.
Small transformers are used in Radio sets, telephones, loud
speakers and electric bells.