Transformers (Especially For 12th Std)

• The long distance transmission of electrical
energy is always done at higher voltages.
• But we receive only 220 V at home from a
high voltage wires of around 100 kV.
• How is that possible ? Who Converts it ?
• P = I2 * R
Recap

• The Loss of power transmission lines is I2R,
• To reduce the power loss AC is transmitted
over a long distances at extremely high
voltages.
• This reduces I in same ratio, therefore I2R
becomes comparatively negligible.
• Which reduce loss of energy during
transmission.

• High voltage currents were needed to be
generated and transmitted to deliver necessary
power over a great distance.
Why Transformer

• A device that used to change high voltage
current into low voltage current and
vice-versa.
Transformer

• It is based on principle of
MUTUAL INDUCTION.
• According to which an e.m.f. is induced in a
coil when current in the neighbouring coil
changes.
Principle Of Transformer

Let L1 & L2 be two current carrying coils,
where current passes through L1 & e.m.f is
induced in L2

• There are two types of core form
Core Form
Shell Form
Types Of Core

Core Form
• Windings are
wrapped
around two
sides of a
laminated
square core.

Shell Form
• Windings are
wrapped
around the
center leg of a
laminated core.

It consists of a laminated soft iron core.
On which two enameled copper wires are wound.
These coils are usually not connected electrically
together. However, they are connected through the
common magnetic flux confined to the core.
Constructions Of
Transformer

 One of the coil which is fed with input A.C. is
called primary coil
 Across the other coil where output supply is
taken out is called secondary coil.

• When current in the primary coil changes
being alternating in nature, a changing
magnetic field is produced in the primary coil.
• This changing magnetic field gets associated
with the secondary coil through the soft iron
core.
Working Of Transformer

• Hence magnetic flux linked with the secondary coil
changes.
• Which induces e.m.f. in the secondary coil.

• If Np is the number of turns of the primary coil
and Ns is the number of turns of the secondary
coil. Let the rate of change of magnetic flux is
• d ɸ
• dt
Mathematical Equation
Of Transformer

• The e.m.f of a primary coil is
• Ep = - Np d ɸ
• dt
• The e.m.f in secondary coil is
• Es = - Ns d ɸ
• dt

• Then ratio of e.m.f’s of primary and secondary
coils is
• ( Ep / Es ) = ( Np / Ns )
• Hence e.m.f are directly proportional to their
respective no. of turns.

• These ratios are taken or possible only after
accepting three assumption.
The Primary Resistance and current are
small.
There Is no leakage of magnetic flux.
The secondary current is small.

• We consider a transformer without any loss of
energy with the above assumptions applicable.
Ideal Transformer

• The relationship between the voltage applied
to the primary winding vp(t) and the voltage
produced on the secondary winding vs(t) is
• Where ‘a’ represents Turn Ratio or
Transformation Ratio
( )
( )
p p
s s
v t N
a
v t N

• The relationship between the primary ip(t) and
secondary is(t) currents is
• Where ‘a’ represents Turn Ratio or
Transformation Ratio
( ) 1
( )
p
s
i t
i t a

• In the phasor notation:
• The phase difference between Current &
Voltage is always same. The ideal transformer
changes magnitudes of voltages and currents
but not their angles.
p
s
a
V
V
1p
s a
I
I

• As we assume that there is no loss of energy,
• Input Power = Output Power
• Ep Ip = Es Is
• Therefore; Is = ( Ep Ip ) / Es
• Is = ( Ip /a )

• On Solving Ip = Is * a
• We get Ip = ( Ep / Req )
• Which implies Req = (1/a)2 * R
• Therefore Req = ( Np / Ns )2 * R

• There are two types of transformer
a.Step Up Transformer.
b.Step Down Transformer.
Types Of Transformer

• If a Transformer converts a low voltage A.C
into high voltages A.C is called STEP UP
TRANSFORMER.
For A Step Up Transformer Es > Ep ,
• Therefore ‘a’ > 1.
• Which implies Is < Ip
Step Up Transformer

• If a Transformer converts a high voltage A.C
into low voltages A.C is called STEP DOWN
TRANSFORMER.
For A Step Up Transformer Es < Ep ,
• Therefore ‘a’ < 1.
• Which implies Is > Ip
Step Down Transformer

• Efficiency of a transformer is defined as the
ratio of output power to that of the input
power.
• Therefore ɳ = ( Es Is ) / (Ep Ip )
Efficiency Of Transformer

 Note:
• When A.C Voltage is Raised to ‘n’ times, the
corresponding current reduces to ‘1/n’ time.

Loss Of Magnetic Flux :
• The coupling
between the coils is seldom perfect. So whole
of magnetic flux produced by primary coil doe
not get linked with the secondary. However in
a shell type transformer these losses are less.
In shell type transformer the primary and
secondary are wound over each other.
Energy Losses In
Transformer

Iron losses:
• In actual iron cores, inspite of
lamination, some heat is still produced by the
eddy currents.
Copper Losses:
• In actual practice, coils of the
transformer possess some resistance. So a part
of energy is lost due to heat produced by the
resistance of the coils.

Hysteresis loss:
• The alternating current in the
coils repeatedly takes the iron core through
complete cycle of magnetization. So energy is
lost due to hysteresis.
Humming Loss:
• The alternating current in the
transformer may set its parts into vibrations
and sound may be produced. This sound
produced is called humming. Thus a part of
energy is lost in the form of sound energy.

• Transformer is used for transmission off A..C..
over long distances by stepping it up.
• It reduces current for a given power
requirement, hence reduces losses due to
• Joulle’’s heating along the resistance off the
transmission line.
• At the city A..C.. is again stepped down to
220V for the consumption...
Uses Of Transformer

 Example 1. :
• A Power Transmission line feeds input
power of 220 V to a step down transformer,
with its primary windings having 4000
turns. What should be the number of turns
in the secondary winding in order to get
output power as 230 V.
Examples

Answer :
Given
Ep = 2300 V, Es = 230 V, np = 4000
w.k.t ( Es / Ep ) = ( ns / np )
Therefore
( 230 / 2300 ) = ( ns / 4000 )
 ns = 400
Therefore secondary winding should have 400
turns.

 Example 2. :
• At a hydroelectric power plant, the water
pressure head is at a height of 300m and the
water flow available is 100m3s-1 . If the
turbine generator efficiency is 60%,
estimate the electric power available from
the plant ( g = 9.8ms-2 ).

Answer :
Given
h = 300 m, ɳ = 60 %, V ( Vol / time ) = 100 m3s-1,
g = 9.8ms-2 .
w.k.t Power = Work / time
Therefore
Power = ( force * dist ) / time
Power = force * velocity
Power = pressure * area * velocity
But ( area * velocity ) = Volume / time

Therefore
Hydroelectric power =
( pressure * volume ) / time
Hydroelectric power = P * V
So,
Power available = (60/100) P * V
= ( 3 / 5 ) ( ρ h g ) ( V )
where ρ = 103 kg/m3 for water
Therefore
Power = 176.4 MW

Transformers (Especially For 12th Std)

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