- Transformers allow the transfer of power between circuits at different voltage levels through electromagnetic induction. They work by changing the voltage and current in opposite proportions while keeping power constant. - An ideal transformer has no losses, but a real transformer is modeled with winding resistances and leakage reactances to account for energy losses. - Transformers can be connected in various configurations depending on the application, including wye-wye, wye-delta, delta-delta, etc. Proper transformer selection and connection is important for voltage regulation and power transmission efficiency.

1. TRANSFORMERS
FUNDAMENTALS
NYU EE 3824 ENERGY
CONVERSION SYSTEMS
INSTRUCTOR Prof. Z. Jan Bochynski
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2. Transformers in a Power Grid
An important safety feature is that the transformer insulates the primary and
secondary electric circuits and permits the separate grounding of the circuits, which
prevents high-voltage, high-power damage the secondary circuit.
Limit 20-26 kV
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3. Transformers Polarity
Each transformer coil can be wound either clockwise or
counter clockwise.
The winding polarity determine the primary and secondary
voltage polarity.
This is important when a pair of transformers work in in
parallel or series connection.
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4. THREE-PHASE TRANSFORMERS
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5. Large oil-cooled high-voltage
transformer
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6. Three-phase Transformer
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7. A three-phase transformer wound on a
single three-legged iron core.
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8. Transformers in Electronics
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Transformer allows matching impedance to obtain a maximum
power in electronic devices
Transformers allows electrical isolation between systems

9. TRANSFORMER PROPERTIES
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Transformer works only when a magnetic flux changes
Transformer does not convert energy
Transformer transmits power from one network to another
Transformer increasing voltage decreasing current and vice
versa
Transformer allows phase conversion in secondary circuit

10. Transformer Lamination
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11. Transformer Core Type
Construction
Iron core magnetic circuit
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12. Transformer Winding
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13. Transformer Construction
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14. Transformer Shell Type
Construction
Iron core magnetic circuit with an air gap
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15. H1
H2
X1
X2
Transformer Ratio
Transformer turns ratio
Ns
Np
Es
Ep

Ns
Np
Es
Ep

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Φ
Es = Ep (Ns /Np)
The secondary (output voltage )

16. Power, Currents, Voltage, Flux
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]
[
44
.
4
4
11
.
1 V
N
f
N
f
E m
m
rms 








17. IDEAL TRANSFORMER
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
m 
E1
4.44 fN1

TR  a 
e1
e2

N1
N2

i2
i1

18. Ideal Transformer
Es = Ns ΔФ / Δt
Ep = Np ΔФ /Δt
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19. Unloaded Transformer
Vp
Ic
Im
Io
θ
Io
Vp Ep
Es=Vs
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20. NO-LOAD CURRENTS
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Iex
Ic = Ih+e VP
Im
Φ
The magnetization current Im is the current required to produce the
flux Φ in the transformer core.
The core-loss current Ih+e is the current required to make up for
hysteresis and eddy current losses
The fundamental component of the magnetization current lags the
voltage applied to the core by 90°.
The total no-load current in the core is called the excitation current of the
transformer. This current is the sum of the magnetization current and the core-
loss current in the core:
Iex = Im + Ih+e

21. Real Transformer
The equivalent circuit contains an ideal transformer and both
network reactance X and resistances R.
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22. PARTS OF REAL TRANSFORMER
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Copper losses are the resistive heating losses in the pri-
mary and secondary windings of the transformer.
They are proportional to the square of the current in the
windings.(I2R).

23. REAL TRANSFORMER
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The real transformer can be represented by an ideal
transformer with winding resistance and leakage
reactance.

24. Loaded Transformer
Ns
Np
Is
Ip 

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25. LEAKAGE FLUXES
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The leakage fluxes ΦLP and ΦLS
are fluxes that escape the core and
pass through only one of the
transformer windings. They
produce a self-inductance L in the
primary and secondary coils, and
the effects of this inductance must
be accounted in calculation.
Φs = ΦM + ΦLS
ΦP = ΦM + ΦLS

26. EQUIVALENT CIRCUIT OF A REAL
TRANSFORMER
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Ideal
Transformer
Sec.
winding
Prim.
winding
Core

27. Real Transformer Equivalent Circuit
Ep =4.44 Фm f Np [V]
Es = 4.44 Фm f Ns [V]
Ф (Wb), f (Hz)
Ep = Vp + ΔVp = Vp + (IoRp + Io Xp )
Vp = Ep - ΔVp = Ep + (IoRp + Io Xp )
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28. TRANSFORMER MODEL REFERRED
TO ITS PRIMARY VOLTAGE LEVEL
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Approximate equivalent circuit of a transformer as viewed
from the primary side
R’2 X’2
I’2
V’2

29. TRANSFORMER MODEL REFLECTED
TO ITS SECONDARY VOLTAGE LEVEL
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An approximate equivalent circuit of a transformer as viewed from the secondary
side
R’1
X’1
I’1
V’1

30. APPROXIMATE EQUIVALENT
TRANSFORMER CIRCUIT
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Approximate transformer model referred to the primary side with
excitation branch and without excitation components.
Req1 = R1 + R’2
Xeq 1 = X1 +X’2

31. Clockwise Connection
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32. Counter Clockwise Connection
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33. Transformer with Center -Tapped
Secondary Windings
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34. Two Secondary Windings Transformer
Secondary winding connected in series – aiding voltage
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35. Two Secondary Winding Transformer
Secondary winding connected in series – opposing voltage
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36. Multi-output Transformers
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37. Transformers in Parallel Connection
In addition to the turn ratio, the primary and secondary
winding polarities are the same in both transformers.
The parallel connection requires that the respective
secondary and primary voltages must be equal that is
have the same turns ratio.
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38. CONDITIONS FOR
TRANSFORMERS IN PARALLEL
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The condition for connecting two transformers in parallel:
1. Two transformers must the same turns ratios
2. The nominal voltage and current ratings must be the same
3. The equivalent resistance, reactance, and impedance must be
the same
4. The polarities must be the same
5. Two transformers must have the same construction of core or
shell type.

39. THREE-PHASE TRANSFORMER
CONNECTIONS
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40. TRANSFORMERS CONNECTIONS
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wye-wye (Y-Y) Seldom used, unbalanced and 3rd harmonics problems
wye-wye-delta (Y-Y-Δ)
Frequently is used to interconnect high voltage networks
(240 kV/345 kV). The delta winding filters the 3rd
harmonics, equalizes the unbalanced current, and
provides a path for ground current
wye-delta (Y-Δ) Frequently is used as step down (345 kV/69 kV)
delta-delta (Δ-Δ)
Used for medium voltage (15 kV), one of the
transformers can be removed (open delta)
delta-wye (Δ-Y) Step-up transformer in a generation station

41. RATIO FOR Y-Y CONNECTION
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In the case of a wye (Y) connection, the rated winding current is the rated
line current and is calculated from the line-to-neutral equivalent of the
voltage rating
a
TR
I
I
N
N
V
V
V
V
V
V
p
s
s
p
ca
CA
bc
BC
ab
AB







42. Y-Y CONNECTION TYPE
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43. RATIO FOR Y-Δ CONNECTION
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This connection will cause the secondary voltage lagging if the
system phase sequence is abc.
If the system phase sequence is acb, then the secondary voltage
be leading the primary voltage by 30°.

44. PHASE SHIFTING IN Y-Δ
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l
l
l
l
l
l
l
l
V
V
V
V
V
/
V
V
V
V
V
a
2
2
2
1
2
1
3
3
3









VΦ = Vll /√3
3
a
V
V
ab
AB


45. RATIO FOR Δ-Y CONNECTION
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For a delta (Δ) connection,
the rated winding current is:
The line current for the delta connection is independent of the
connection type

46. RATIO FOR Δ-Δ CONNECTION
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a
V
V
Vbc
V
V
V
ca
CA
BC
ab
AB




47. PHASE SHIFTING IN Δ-Δ
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Ill =√3 IΦ

48. THREE-PHASE TO SIX-PHASE
CONNECTION
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49. THREE-PHASE TO SIX-PHASE
CONNECTION
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50. THREE-PHASE TO SIX-PHASE
CONNECTION
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51. Examples
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52. Example 1
The transformer has alternating input
of 100 V. The number of turns on
each winding are NP = 375, NS1 =
750, NS2 = 500, NS3 = 75.
1. Determine the output voltage from
each secondary winding.
2. Find the total output if all of three
secondary windings were connected in
series-adding.
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53. Solution
Example 1
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54. EXAMPLE 2
The three secondary in Problem 1 have following loads
attached: RS1 =1 kΩ, RS2 = 500 Ω, and RS3 = 100 Ω.
Assuming that the transformer is 100 % efficient find
the total primary current Ip.
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55. Solution Example 2
For the secondary winding in coil 1 current is
Is1 =Es1/Rs1 = 200V/1kV = 200 mA
Ip =Is x Ns / Np
Ip1 =Is1 x Ns1 / Np = 200 mA x (750/375 ) = 400 mA
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56. Example 3
The transformer input max voltage is 100 V, frequency 400 Hz
and the primary winding has 375 turns.
Determine the pick value of the flux Φ.
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57. Solution Example 3
Wb
Wb
Hz
V
N
f
E
p
p
m

150
10
150
375
400
44
.
4
100
44
.
4
6












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58. Example 4
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A transformer with two secondary windings has these turns
data: N1 = 4000, N2 = 1000, N3 = 500. The primary rms
voltage is 120 V. The secondary voltage is V2 = 30<00 and
V3 = 15<00. The secondary windings loads are: Z2 = 3 +j 4;
and Z3 = 5 +j 0.
Compute the followings:
1. Current in all three windings
2. Active power in the primary circuit
3. Apparent power in the primary winding
4. Reactive power in the primary winding
5. Power factor of the system

59. 1/12/2023 Prof. Z. Jan Bochynski 59
Solution Example 4
I2 = E2/Z2 = 30<00 / (3+j4) = 6<-53.10 A
I3 = E3/Z3 = 15<00 / (5+j00) = 3<00 A
Since the primary power is equal to the secondary power
I1 x N1 = I2 x N2 + I3 x N3
I1 x 4000 =(6<-53.10) x 1000 + 3 x 500
I1 = 1.276 – j 1.2 =1.75<-43.260 A
Apparent power of the source is
S1 = E1 x I1* = (120<00) (1.75<+43.260) = 153 + j 144 VA
Active power P = 153 W
Reactive power Q = 144 VAR
P.F. = cos 43.260 = 0.728 leading