1. TRANSFORMER

2. where S is the reluctance C vp ip P Np
A transformer consists of a core made of laminated iron separated by insulators and a coil of Np turns wound around the core. This coil is supplied with an a.c voltage supply vp which then produces a current ip. Due to this current , a flux F is produced which is given by an equation
= Npip/S …….(1)
where S is the reluctance

3. Substitute  = Npip/S into the above equation , then
Since current varied with time , F also varied with time. A back electromagnetic force (e.m.f) will be produced which is given by the equation.
……(2)
Substitute  = Npip/S into the above equation , then
……(3)

4. If ip is sinusoidal, the flux produced also sinusoidal, i.e
 = m sin 2ft
……(4)
therefore
vp = NP2fmcos 2ft = NP2fmsin (2ft + /2)
……(5)
The peak value = Vpm = NP2fm
and vp is leading the flux by p/2.
……(6)
The rms value
……(7)

5.  C vp ip vs is
Load P S Ns Np
primary
secondary
When another coil is wound on the other side of the core with no of turns Ns , then the fux will induce the e.m.f vS as given by
……(8)

6. From (2) and (8) we get Or in rms value
…….(9)
…….(10)
Or in rms value
With load , is will flow in the load, mmf at load will equal to the mmf at input, then (in rms value)
NpIp = NsIs
…….(11)
rearrange
…….(12)

7. Symbol for ideal transformer
For ideal transformer, the energy transferred will be the same as input. Thus power at primary is same power at secondary.
Pp = Ps
or IpVp = IsVs
NP : NS VP VS
Primary
Secondary
Symbol for ideal transformer

8. Example 1
A 250 kVA,11000V/400V, 50Hz single –phase transformer has 80 turns on the secondary. Calculate
(a) The appropriate values of the primary and secondary currents;
(b) The approximate number of primary turns;
(c) the maximum value of the flux.
(a) Full-load primary current
Full-load secondary current

9. (b) Number of primary turns
recall
(c) Maximum flux
recall

10. Ideal transformer with no loadEP VP VS NS NP IO
Ideal transformer with no load
Io is the no load current when the secondary is open circuit. This current consists of Iom that is required to produce the flux in the core (it is in phase) and Io1 is to compensate the hysteresis and eddy current losses.

11. VP= emf of supply to the primary coil and 90o leads the flux.
Phasor diagram for
no load transfomer 
IOm
IOI
EPVP IO VS fo
VP= emf of supply to the primary coil and 90o leads the flux.
EP=emf induced in the primary coil and same phase as VP.
VS=emf induced in the secondary coil and 90o lags the flux.
Iom=magnetizing current to produce flux and it is in phase with flux.
Io1=current to compensate the losses due to hysteresis and eddy current.
Io=the no load current and given by
Power factor

12. Application of transformer
Transformer converts the energy to high electrical voltage and transmits in the high voltage line. At the load, the high voltage energy is converted to low voltage. In this way, it will compensate the losses during transferring of the voltage energy.
Transformer 1
Transformer 2
Low voltage generator
Low voltage load
High voltage line

13. Example 2
A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in core is 1.8m and the joints are equivalent to the airgap of 0.1mm. The value of the magnetic field strength for 1.1 T in the core is 400A/m, the corresponding core loss is 1.7W/kg at 50Hz and the density of the core is 7800kg/m3.
If the maximum value of the flux is to be 1.1T when a p.d of 2200V at 50Hz is applied to the primary, calculate:
(a) the cross-sectional area of the core;
(b) the secondary voltage on no load;
(c) the primary current and power factor on no load

14. (a)
recall
recall
Practically 10% more allow for insulator
(b)
recall

15. (c)
magnetomotive force (mmf) for the core is
mmf for the airgap is
Total mmf is
recall
Maximum magnetizing current
Rms value

16. Volume of core
mass of core
Core loss= loss rate x mass
Core-loss component of current
No load current
Power factor

17. Loaded transformer E1 V1 V2 I1 I2 L(2) Loaded transformer
Phasor diagram ‘
L(q2)= load with power factor of cos q2
V1 = emf at supply
E1=induced voltage at primary
V2=emf at load
E2=induced voltage at secondary
I1= primary current
I2=secondary current

18. Example 3
A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3A at a power factor 0.2 lagging when secondary current is 280A at a power factor of 0.8 lagging. Calculate the primary current and the power factor. Assume the voltage drop in the windings to be negligible.
Recall Equation 12
therefore

19. Solve for horizontal and vertical components‘
Solve for horizontal and vertical components
Power factor

20. Equivalent circuit of transformer
Leakage fluxes
V2’ R1 L1 L2 R2 RC Lm V2 E1
I1’ I1
Equivalent circuit of transformer
Path of leakage
Flux leakage is due to secondary current that produce flux which encounter the primary flux. Some of the flux will link to its own windings and produce induction. This is represented by inductance L1. Similarly with the flux in secondary and represented by L2.

21. Equivalent circuit of transformer
There are four main losses
Dissipated power by wire resistance of the windings (I2R)
Power due to hysteresis
Power due to eddy current
Power via flux leakages. E2 R1 L1 L2 R2 RC Lm V2 V1
I1’ I1
Equivalent circuit of transformer E1
R1= wire resistance of primary windings
L1=inductance due to leakage flux in primary windings
RC=resistance represent power loss due to in hysteresis and eddy current
Lm= inductance due to magnetizing current Iom
L2=inductance due to leakage flux in secondary windings
R2=wire resistance of secondary windings

22. Phasor diagram for a transformer on loadV1 E1 I2
-I2’
I1X1 V2 E2 I1 I0
I1R1
I1Z1
I2X2
I2Z2
I2R2 f1 f2
Phasor diagram for a transformer on load

23. Simplified/approximate equivalent circuit
We can replace R2 by inserting R2’ in the primary thus the equivalent resistance is
Giving us
Similarly

24. Transformer simplified circuit thenZe V1
E1=V2’ I1 I2
To load
E2=V2
then
where
and

25. Magnified Ze portion Complete
Phasor diagram of simplified equivalent circuit of transformer
Complete
fo-f2
Magnified Ze portion

26. Voltage regulation of a transformer
recall
Secondary voltage on no-load
V2 is a secondary terminal voltage on full load
Substitute we have

27. Or
From phasor diagram can be proved that Or

28. Example 4
A 100kVA transformer has 400 turns on the primary and 800 turns on the secondary. The primary and secondary resistances are 0.3W and 0.01W respectively, and the corresponding leakage reactances are 1.1W and 0.035W respectively. The supply voltage is 2200V. Calculate:
The equivalent impedance referred to the primary circuit;
The voltage regulation and the secondary terminal voltage for full load having a power factor of (i) 0.8 lagging and (ii) 0.8 leading.
The percentage resistance and leakage reactance drops of the transformer

29. (a)
(b) (i)

30. Sec. terminal voltage on no-load
The decreasing of full-load voltage is
Therefore the secondary full-load voltage
(b) (ii) power factor 0.8 leading
The increasing of full-load voltage is
Therefore the secondary full-load voltage

31. Or
Per unit

32. Alternative
Recall Equation 12
Full load secondary current
Equivalent resistance referred to secondary
Secondary voltage on no-load