Elec467 Power Machines &
Transformers
Electric Machines by Hubert, Chapter 2
Topics: Exciting current, Ideal transformer,
Equivalent circuits, Per Unit calculation,
Open circuit/short circuit testing models
Typical physical designs
for a transformer
Basic transformer action
ex  N x d
dt
ex is the
counter emf
Ex  4.44 N x fmax
(Formula 2-3)
EP N P

ES N S
Voltage ratio = turns ratio
Calculating transformer flux
From Example 2.1
 Given that a transformer 50-kVA, 200
turns, 240V, 60 Hz primary, what is the
maximum flux, Φmax?
 Use formula 2.1 Ep = 4.44 Np f Φmax
 Insert knowns… 240 = 4.44·200·60· Φ
 Solve for the unknown
 Φmax = 240/(4.44·200·60)→.0045 Wb

Section 2-5—No load conditions
In this diagram is a simplified view of the primary side of a transformer. The input
current IP = I0 (the exciting current) during no load conditions which can be
measured directly with an ampere meter. We see the eddy loss component (Ife)
caused by the resistance Rfe (a fictitious resistance to account for core loss) and
the magnetizing current (IM) present during no load conditions whose reactance
component is jXM is in parallel. The value jXM is a fictitious reactance to account for
the magnetizing current. Their vector summation gives the current input when there
is no load: I0 = Ife + IM. Additional equations from formula 2-4:
Ife = VT/Rfe
IM = VT/jXM
Example 2.3—Calculating flux loss
for no-load conditions
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The exciting circuit, Io , is a no-load current. Let’s calculate the exciting circuit
in the primary of a transformer using data from Example 2.3:
Given a 25-kVA, 2400—240 V, 60 Hz, draws 138 W at no-load condition with a
.210 lagging power factor (current into a coil lags the voltage).
1st get the phase angle Ѳ…cos-1(.210) = 77.88°
2nd insert the power factor angle of 77.88° into the formula for the phase
difference between the voltage and current: Ѳ = Ѳv – Ѳi
Using 0° phase for Ѳv gives: 77.88° = – Ѳi
therefore Ѳi = – 77.88°
This value is the angle the current vector I0 makes with the vector Ife
We’ll need this information to calculate the IM current but first let’s do the
core-loss component that draws real power: Pcore loss = Vt ∙ Ife
Insert known values: 138 = 2400 ∙ Ife
Solve for the unknown: Ife = .0575 amps …giving us the amplitude for the
vector Ife
Utilize this value and a trig formula from the power triangle: cos Ѳi = Ife / Io
Insert known values: .210 = .0575/ Io yields up Io = .2738 amps
Using Io = Ife + IM we can solve for IM = Io - Ife → .2738 - .0575 = .268 amps
Summary: Io = .2738 amps, Ife = .0575 amps, IM = .268 amps
Formulas used for
no load calculations
NPIM
M 
R core

◦ ΦM =the mutual flux produced by the magnetizing component of the exciting current
◦ NpIM is aka magnomotive force, F

VT = IpRp + Ep
 Ip

= (VT – Ep)/Rp
Above formulas are 2-6, 2-7 & 2-8 from text, page 46.
Formulas used for calculation of
quadrature components
Vt = Ife ∙ Rfe multiply both sides by Ife and
we get real power loss in watts equal to
I2R
Pcore loss = Vt ∙ Ife
 Vt = IM ∙ j XM reaction power loss in watts
 Io = Ife + IM
 The vector diagram
from Fig. 2-4
→

(this is a vector formula)
A closer look at the phasors
Figure 2-4 phasor
drawing is seen at the
right. Ѳi is the power
factor angle. When
rearranged as seen on
the right
cosine = adj/hyp→Ife/Io
giving us one more
handy formula:
Io = Ife/cos Ѳi
or
Io = Ife/power factor
Io
IM
Ѳ
Ife
Mutual flux
When a load is placed on the secondary the primary side
changes from just the exciting current (Io = Ife + IM) to the
exciting current plus a load current, IP = I0 + IP,load (formula
2-11).
Section 2-7—Flux loss
Flux leakage results in a different induced voltage in the
secondary voltage than is no leakage occurred. It
diminishes the efficiency of the transformer.
Flux leakage is accounted for in the formulas:
Φp = ΦM + Φlp
formula 2-12
Φs = ΦM – Φls
formula 2-13
Section 2-8—Ideal transformer
The ideal transformer is represented above. The induced counteremf voltages and the input impedance are designated with primed
symbols. There is a long list of what is not included in the ideal
transformer. When the turns ratio is not available use the
nameplate voltage ratio. The following slide lists a number of
formulas used with an ideal transformer. The effects of flux
leakage and winding resistance are insignificant at no load.
Ideal transformer formulas
a
N HS E ' HS VHS


N LS E 'LS VLS
Assuming the primary is the High Side:
E 'P N P

a
E 's N S
E 'P  aE 'S
Input Impedance of an Ideal Transformer:
E 'S
E'
E'
Z

Z 'in  a 2 S
Z 'in  P
sinc load I '
I 'S
I 'P
S
e
Z 'in  a 2 Z load
The apparent power input must equal the apparent power output:
1 *
IS
a
E 'P I P*  E 'S I S*
I P* 
E 'S *
I 
IS
E 'P
therefore : I P 
*
P
1
IS
a
Equivalent circuit parameters
for a real transformer
Close up look at the ideal
transformer from Fig. 2-8
In the ideal: Ep = 4.44Np f Φp and Es = 4.44Ns f Φs
If the windings have the same polarity, the current will appear
to flow into the transformer and flow out the same end as if the
primary and secondary circuits were connected to together.
Equivalent circuit details
An equivalent circuit of a real transformer using an ideal
transformer and external components to represent what is
going on internal to the transformer. Losses are represented
by IPRP, Elp, IfeRfe and by the magnetizing current IMjXM.
Leakage Reactance
When we include the flux leakage (formulas 2-12 &13) in
the ideal’s equations we get:
Ep = 4.44Np f ΦM + 4.44Np f Φlp formula 2-20
Es = 4.44Ns f ΦM - 4.44Ns f Φ ls formula 2-21
Formulas we need to know to
calculate voltages around the loops

Expressing formulas 2-20 & 21 in simpler form
 EP = E’P + Elp
 VT = Ep + IpRp
 Ip = Ife + IM + Ip load
 Es = IsRs + Vs load

ES = E’S - Els
VT = E lp + E’p + IpRp
formula 2-26
E’s = Els + IsRs + Vs load
formula 2-28

The sinusoidal voltage generated:

E = Emaxcos(2πft) with Emax = 2πfNΦmax formula 2-30

L = N2/R
formula 2-32
Equivalent impedance using
referred parameters
Referring 2nd parameters to the primary side
Formulas used for Z’in

Ideal transformer: Z’in = a2E’s/Is

Where Is = E’s/(Rs + jXls + Zload)
(see Fig. 2-9)
Substituting Is into the first equation gives:
Z’in = a2(Rs + jXls + Zload) →
Z’in = a2Rs + a2jXls + a2Zload
 The impedance seen by the source is
Zin = Z’in + (RP +jXlP)

How to calculate the core losses
seen by the primary
Neglect exciting current at near rated
load conditions…the equivalent
impedance is…
 ZeqP = Rp + a2Rs + j(Xlp + a2Xls)

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and, Zin, HS = ZeqP + a2Zload
(note: this is not Z’in)
Equivalent impedance HS is defined
for a step-down transformer
Here they have taken the impedance in { Rp + a2Rs + j(Xlp + a2Xls) }
from the previous slide and changed it into Zeq,HS = Req,HS + jXeq,HS by
grouping the resistances and impedance together:
Req,HS = RP + a2RS
jXeq,HS = j(XlP + a2XlS)
(note: Req,HS & jXeq,HS are the parameters obtained in the short-circuit test)
Taking the core losses to the secondary
Notice that the loads when reflected thru the transformer use
a2 but current and voltage relations only use “a”. Current as
usual is inverted.
How to calculate the wiring and flux
losses in the secondary

ZeqS = Rs + Rp/a2 + j(Xls + Xlp/a2)

Here the primary values are adjusted by
1/a2 which is the result of going the other
direction across the turns ratio.
Transformer as a lens
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Think of the transformer as a telescope.
Telescopes make thing look larger or smaller
depending on which way you look thru it.
When the high voltage is stepped down, going
from primary to secondary make values smaller
Thus the primary losses when taken to the
secondary are divided by a2.
But bringing the secondary values to the
primary side makes them bigger thus Zload and
the secondary losses are multiplied by a2.
Equivalent circuit for a step-down
transformer—low side
The secondary impedance is changed from {Rs + Rp/a2 + j(Xls + Xlp/a2)}
seen in the last slide into Zeq,LS = Req,LS + jXeq,LS
by grouping the resistances & impedances together:
Req,LS = RLS + RHS/a2 & jXeqLS = j(XLS + XHS/a2)
Equivalent circuit for a step-up
transformer can go either way also
Changing from a Step-Down to a Step-Up
If we think of the inputs and output of a
transformer as primary (the driving circuit) and
secondary (the load) an adjustment to “a” is
necessary when changing from a step-down to a
step-up or visa-versa.
 The value of “a” is inverted because the turns
ratio is inverted.
 We can avoid this problem by using formulas
that refers to the opposite sides of the
transformer in terms of voltage high side or low
side.

Equivalent circuit for a step-up
transformer—low side
For the step-up mode the source is on the low side (smaller number
of windings). The equivalent impedance is
Zeq,LS = RLS+RHS/a2+ j(XlLS+ XlHS/a2) or
Zeq,LS = Req,LS + jXeq,LS
where Req,LS gather the real terms and jXeq,LS gathers the imaginary
terms.
Zin,LS = (1/a2 )*Zload,HS + Zeq,LS
Equivalent circuit for a step-up
transformer—high side
The equivalent impedance on the high side uses
Zeq,HS = RHS + a2RLS + j(XlHS + a2XlLS).
The impedance show above is Zeq,HS = Req,HS + jXeq,HS
where Req,HS gathers the real terms and jXeq,HS gathers the imaginary
terms.
Voltage Regulation

Regulation = (Enl – Vrated)/ Vrated

Enl is the voltage present at the output terminals when
no load is present. Vrated is the name plate voltage.
Enl = ILSZeq,LS + VLS
 ILS = Papparent/VLS
 Zeq,LS = Zeq,HS/a2

formula 2-45 (p. 62)
consider VLS as Vrated
formula 2-43a (p. 59)
Per Unit Parameters
Use by professionals in the power
industry, per unit parameters simplify
voltage calculation for transmission lines.
 ZPU = Irated Zeq)/Vrated
 RPU = Irated Req)/Vrated
 XPU = Irated Xeq)/Vrated
 Zbase = Vrated/Irated
also V2rated/Srated
 Phase angle α = tan-1(XPU/RPU)

Transformer losses and efficiency
η = Pout/Pin
 η = Pout/(Pout + Pcore + I2Req)

Pout is the apparent power (xxKVA)
 Pcore is the wattage reading showing the real power lost
under no load conditions. This happens to be the (HS)
value of the power reading taken during the open
circuit test!
 I2Req is the wattage reading taken during the short
circuit test!
 Therefore: η = Pout/(Pout + POC + PSC)

Open Circuit Test
Open circuit test is measured on the low side with the high side
open so that only the exciting current parameters are detected.
Formulas used in the
Open Circuit Test
Parameters measure during an open
circuit test are: POC,VOC, IOC
 POC = VOCIfe → Ife = POC/VOC
 IOC = √(Ife2 +IM2) → IM = √(IOC2 – Ife2)
 Rfe,LS = VOC/Ife
 XM,LS = VOC/IM
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Note: By conducting an open circuit test from both sides, Rfe and
XM can be determined for both primary and secondary sides.
These values are used in no-load calculations in step-down and
step-up transformers.
Short circuit test
Short-circuit test detects the high side equivalent parameters as
there is no-load seen and the exciting current is ignored.
Formulas for the short circuit test

Parameters measure during a short
circuit test are: PSC,VSC, ISC
 ISC
 PSC

= VSC/Zeq,HS → Zeq,HS = VSC/ISC
= ISC2 Req,HS → Req,HS = PSC/ISC2
Zeq,HS = √(Req,HS2 + Xeq,HS2) →
Xeq,HS = √(Zeq,HS2 - Req,HS2)
Homework Assignment
2-3
 2-5
 2-9
 2-11
 2-13
 2-17 a & b
 2-27
 2-34 a
 2-39 a
