Lecture 5

THE TRANSFORMER

INTRODUCTION
The transformer is a valuable apparatus in electrical power systems, for it enables
(allows) us to utilize different voltage levels across the system for the most economical value.
Generation of power at the synchronous machine level is normally at a relatively low voltage,
typically 20 kV, which is most desirable economically. Stepping up of this generated voltage
to high voltage, extra-high voltage, or even to ultra-high voltage is done through power
transformers to suit the power transmission requirement to minimize losses and increase the
transmission capacity of the lines. This transmission voltage level, typically ranging from 115
to 750 kV, is then stepped down in many stages for distribution and utilization purposes in the
range of 115/230 v for home use to 13,200 v in large industrial equipment.

GENERAL THEORY OF TRANSFORMER OPERATION

A transformer contains two or more windings linked by a mutual field. The primary
winding is connected to an alternating voltage source, which results in the flow of an
alternating flux whose magnitude depends on the voltage and number of turns of the primary
winding. The alternating flux links the secondary winding and induces a voltage in it with a
value that depends on the number of turns of the secondary winding. The operation of the
transformer depends on Faraday’s voltage law and Ampere’s law. If the primary voltage is v1,
the core flux ϕ is established such that the counter EMF e equals the impressed voltage
(neglecting winding resistance). Thus,

Here N1 is the number of turns of the primary winding. The EMF e2 is induced in the
secondary by the alternating core flux ϕ

Taking the ratio of Eqs. (4.1) to (4.2), we obtain

Neglecting losses, the instantaneous power is equal on both sides of the transformer, as shown
below:
P1=P 2

i1 v2
=
i2 v1
Combining Eqs. (4.3) and (4.4), we get

Thus the current ratio is the inverse of the voltage ratio. We can conclude that almost any
desired voltage ratio, or ratio of transformation, can be obtained by adjusting the number of
turns of the transformer winding. Transformer action requires a flux to link the two windings.
This will be obtained more effectively if an iron core is used because an iron core confines the
flux to a definite path linking both windings. A magnetic material such as iron undergoes a
loss of energy due to the application of alternating voltage to its B-H loop. The losses are
composed of two parts. The first is called the eddy-current loss, and the second is the
hysteresis loss. Eddy-current loss is basically an I2R loss due to the induced currents in the
magnetic material. To reduce these losses, the magnetic circuit is usually made of a stack of
thin iron-alloy laminations. Hysteresis loss is caused by the energy used in orienting the
magnetic domains of the material along the field. The loss depends on the material used. Two
types of construction are used, as shown in Figure 4.1. The first is called the core type, which
is a single ring encircled by one or more groups of windings. The mean length of the magnetic
circuit for this type is long, whereas the mean length of windings is short. The reverse is true
for the shell type, where the magnetic circuit encloses the windings.
 no load

Due to the nonlinearity of the B-H curve of the magnetic material, the primary current
on no-load (for illustration purposes) will not be a sinusoid but rather a certain distorted
version, which is still periodic. For analysis purposes, a Fourier analysis shows that the
fundamental component is out of phase with the applied voltage. This fundamental primary
current is basically made of two components. The first is in phase with the voltage and is
attributed to the active power taken by eddy-current and hysteresis losses and is called the
core-loss component Ic of the exciting current Iϕ.The current component that lags e by 90o is
called the magnetizing current Im. Higher harmonics are neglected. Figure 4.2 shows the no-
load phasor diagram for a single-phase transformer. Consider an ideal transformer (with
negligible winding resistances and reactances and no exciting losses) connected to a load as
shown in Figure 4.3. Clearly Eqs. (4.1)-(4.5) apply. The dot markings indicate terminals of
corresponding polarity in the sense that both windings encircle the core in the same direction
if we begin at the dots. Thus comparing the voltage of the two windings shows that the
voltages from a dot-marked terminal to an unmarked terminal will be of the same polarity for
the primary and secondary windings (i.e., v1 and v2 are in phase). From Eqs. (4.3) and (4.5)
we can write for sinusoidal steady state operation
v1 N1 N 1 v2 v1 N 1 v2
= v 1= =
Eq (4.3)
v2 N2 N2 (divided by i1)
i1 N 2 i1
i1 N2 N 2 i2
= i 1=
Eq (4.5) i2 N1 N1

But the load impedance Z2 is

Thus,
The result is that Z2 can be replaced by an equivalent impedance Z2' in the primary circuit.
Thus,

The equivalence is shown in Figure 4.3.

Representations of the transformer that are more realistic must account for winding
parameters as well as the exciting current. The equivalent circuit of the transformer can be
visualized by following the chain of events as we proceed from the primary winding to

the secondary winding in Figure 4.4. First the impressed voltage V1 will be reduced by a drop
I1 R1 due to the primary winding resistance as well as a drop jI1X1 due to the primary leakage
represented by the inductive reactance X1. The resulting voltage is denoted E1 . The current I ,
will supply the exciting current Iϕ as well as the current I2', which will be transformed through
to the secondary winding. Thus

Since Iϕ has two components (Ic in phase with El and Im lagging E1 by 90°), we can model its
effect by the parallel combination of the elements Gc and Bm as shown in the circuit. Next E1
and I1 are transformed by an ideal transformer with turns ratio N1/N2. As a result, E2 and I2
emerge on the secondary side. E2 undergoes drops I2'R2 and jI2'X2 in the secondary winding to
result in the terminal voltage V2.
 Figure 4.4(B) shows the transformer’s equivalent circuit in terms of variables referred
to the primary. This circuit is called “circuit referred to the primary side.” Note that

Although the equivalent circuit illustrated above is simply a T-network, it is customary to use
approximate circuits such as shown in Figure 4.5. In the first two circuits we move the shunt
branch either to the secondary or primary sides to form inverted L-circuits. Further
simplifications are shown where the shunt branch is neglected in Figure 4.5(C) and finally
with the resistances neglected in Figure 4.5(D). These last two circuits are of sufficient
accuracy in most power system applications. In Figure 4.5 note that

Ic Im
Example 4.1
A 100-kVA, 400/2000 V, single-phase transformer has the following parameters

Note that Gc. and Bm are given in terms of primary reference. The transformer supplies a load
of 90 kVA at 2000 V and 0.8 PF lagging. Calculate the primary voltage and current using the
equivalent circuits shown in Figure 4.5.
Solution:
Let us refer all the data to the primary (400 V) side:

Thus, the equivalent series resistance Req and series reactance Xeq are

The voltage V2 = 2000 V; thus

The current I2' is obtained from the VA rating and voltage as:
 P2
|I '2|=
V '2

The power factor (PF) of 0.8 lagging implies that

−1 o
cos 0. 8=36 . 87 , lagging = negative

For ease of computation, we start with the simplest circuit of Figure 4.5(D) . Let us denote the
primary voltage calculated through this circuit by V1d. It is clear then that

V 1d =400⟨0o ⟩+13.5⟨−36.87 o⟩ j
=( r 1 cos φ1 +r 1 sin φ1 j )

=( 13 . 5 cos−36 .87 o +13 . 5 sin−36 . 87 o j ) j

=( 10.8−8 .1 j ) j
=8.1+10.8 j
V 1d =400+0 j+8 .1+10 . 8 j=408. 1+10 . 8 j =x + yj

r= √ x 2 + y 2 =√ 408.12 +10.82 =√166545.61+116.64=408.243

y 10. 8 V 1d =408 .243⟨1.516 o⟩
φ=tan −1
()
x
=tan−1 (
408 .1 )
=tan −1 ( 0 . 02646 ) =1. 516o

Thus,
Comparing circuits (C) and (D) in Figure 4.5, we deduce (assume) that

' ' '

V 1c =V 2 +I 2 Req + jI 2 X eq
'
V 1c =V 1 d+I 2 R eq

V 1c =408.243 ⟨1.516 o ⟩+ ( 225 ⟨−36.87o ⟩) ( 0.02 )

 V 1c =408.243⟨1.516 o ⟩+4.5⟨−36.87o ⟩
=( r 1 cos φ1 +r 1 sin φ1 j ) + ( r 2 cos φ2 + r 2 sin φ2 j )

=( 408 .243 cos1 .516 o +408 .243 sin 1 .516 o j ) + ( 4 . 5 cos−36 .87 o + 4 . 5 sin−36 . 87 o j )
=( 408. 098+10. 8 j ) + ( 3 .6−2 .7 j )
=( 411.698+8. 1 j )

r= √ x 2+ y 2=√ 411.6982 +8.12 =√ 169495.24+65.61=411.78

V 1c =411.78⟨1.127o ⟩
φ=tan −1 ( xy )=tan ( 411.8 .698
−1 1
)=tan −1
( 0 .01967 )=1 .127 o

Let us consider circuit (A) in Figure 4.5. We can see that

I1a I2'

Iϕ

But

Circuit (B) is a bit different since we start with V2' impressed on the shunt branch. Thus,

Now
core-loss component Ic

ANY QUESTIONS?