Transformers (Especially For 12th Std)

WELCOME

TOPIC

TransformerA

• 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 Na

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 ti 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

aVV

1p

s a

II

• 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 = 4000w.k.t ( Es / Ep ) = ( ns / np )Therefore ( 230 / 2300 ) = ( ns / 4000 )

ns = 400Therefore 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 * VSo, Power available = (60/100) P * V = ( 3 / 5 ) ( ρ h g ) ( V ) where ρ = 103 kg/m3 for waterTherefore Power = 176.4 MW

THAT’S ALL FROM MY

PRESENTATIONTHANK YOU