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EE212 Passive AC Circuits
Lecture Notes 3
Transformers
EE 212 2010-2011
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Magnetic Circuit
• What is the relationship between magnetic
flux and magnetomotive force, Fm?
• What is the relationship between electric
current, i, and magnetomotive force?
Magnetic field strength, H = Fm / l
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Magnetic Circuit
• What is Faraday’s Law?
– The voltage induced in an electric circuit is
proportional to the rate of change of the
magnetic flux linking the circuit
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Magnetic Circuit
Consider a coil around a magnetic core. If a current i flows through the
coil, a magnetic flux  is generated in the core.

=
i
a
vab = L di
dt
L
b
Direction of flux by
1
N i in webers (Wb)
Rm
N = number of turns in the coil
Rm = constant known as reluctance
(depends on the magnetic path of the flux)
Rm =
l
in ampere-turns/Wb
.A
Right-Hand Rule
Fingers curled around coil
– direction of current
Thumb – direction of flux
l = length of magnetic path
A = cross-section area
 = permeability
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B-H Curve

Flux Density, B =
in teslas, T
A
Magnetic field strength, H = N.i in AT/m
l
B=H
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Coupled Circuits
Circuits that affect each other
by mutual magnetic fields
a
vab = L1
1
i1
vab
di 2
di1
vcd = L2
±M
dt
dt
L1
2
b
c
L2
di1
di
± M 2
dt
dt
i2
vcd
d
The flux 2 generated by
current i2 in Coil 2
induces a voltage in Coil 1,
and vice-versa.
± depending on whether the fluxes
add or oppose each other
L1, L2: self inductance
M: mutual inductance
ratio of induced voltage in one circuit
to the rate of change of current
in another circuit
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Coupled Circuits in Phasors
If input signals are sinusoidal waveforms,
coupled circuits can be in phasor representation
di1
di 2
vab = L1
± M
dt
dt
vcd = L2
di 2
di
±M 1
dt
dt
± depending on flux directions
I1
a
Vab
b
L1
I2
M
c
L2
Vcd
d
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Equivalent Circuit with Dependent
Sources
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Dot Convention
Dots are placed at one end of each coil, so that
currents entering the dots produce fluxes that
add each other.
The dots provide information on how the
coils are wound with respect to each other.
a
i1
L1
2
b
A current i entering a dotted terminal in one coil
di
induces a voltage M
with a positive polarity
dt
at the dotted terminal of the other coil.
(+) if both currents enter the dotted terminals (or
the undotted terminals).
(-) if one current enters a dotted terminal and the
other current enters an undotted terminal.
Vab = (jwL1) I1 + (jwM) I2
for the currents as shownEE 212 2010-2011
1
c
L2
i2
d
(currents entering the dots
produce upward fluxes)
I1
a
c
L1
L2
d
b
I2
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Coefficient of Coupling, k
k=
M
L1 L2
0≤k≤1
k depends on the magnetic properties of the flux path.
When k = 0,
no coupling
k = 0.01 to 0.1, loosely coupled
k > 0.5,
close coupled, e.g. air core
k ≈ 1.0,
e.g. power transformer
all the flux generated by one coil is linked to
the other coil (i.e. no leakage flux)
ideal transformer
k = 1.0
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Transformer

i
+
+
e1
v1
-
N1 N2
e2
secondary
winding
+
V1 = -E1
primary
winding
I
+
V1 E1
N1 N2
I
E2
-
When a voltage V1 is applied to
the primary winding, an emf e2
is induced in the secondary
winding. The induced emf lags
the inducing flux by 900.
E1
E2
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d
e2 = N2
dt

Faraday’s Law
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Transformer Application in Power System
-V/I step up/down
- Y/D conversion
- circuit (dc) isolation
- Z matching
(for max power transfer,
min. reflection from load)
Instrument Transformers:
CT (current transformer)
PT or VT (voltage transformer)12
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Ideal Transformer
 No leakage flux  Coupling Coefficient, k = 1 , i.e. the same flux 
goes through both windings
e1 = N 1
d
dt
and e2 = N2
d
dt
=>
e1 N 1
=
= a (turns ratio)
e2 N2
Turns ratio (a or n) is also known as the transformation ratio.
No Losses
No voltage drops in the windings: V1 = - e1
Instantaneous powers in primary and secondary are equal
(i.e. all the energy from the primary is transferred to the
secondary winding).
e1 i1 = e2 i2
i2
e1 N 1
=
Therefore,
=
i1
e2 N2
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= a
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Transformer Loading
• a secondary current I2 is drawn by the load
• I2 generates a flux that opposes the mutual flux 
(Lenz’s Law: effect opposes the cause)
• reduction in  would reduce induced emf e1
• since source voltage V1 is constant, and V1 = - e1,  must remain constant
• the primary winding must draw an additional current I’1 from the source
to neutralize the demagnetizing effect from the secondary
I1
I2
+
V1
e1
e2 V2
ZL
-
N1 N2
Primary current I1 = I’1 + I
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Ideal Transformer
• No leakage flux, i.e., k = 1
• Self-inductance, L1 = L2 = ∞ i.e., magnetizing current = 0
• Coil losses are negligible
• With polarities, dots and currents as shown:
V2 = V1 /a
I2 = a I1
Z1 = a2 Z2 where Z2 is the load impedance
I1
+
a
Vab
- b
I2
M
●
L1
●
L2
c +
Vcd
d EE 212 2010-2011
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Actual Transformer
i1
+
v
1
-

R1
R2
e
e
1
2
N1
+
v
2-
ZL
N2
- resistance in primary and secondary windings
- leakage reactance in pr. and sec. windings
- voltage drops in both windings (leakage impedance)
- losses
- copper loss
primary: I12·R1 secondary: I22·R2
- iron loss (core loss)
Core loss depends on voltage and frequency
eddy current
hysteresis loss
EEloss
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Iron (Core) Losses
Eddy Current Loss:
emf induced in core generates eddy currents which circulate in
the core material, generating heat.
laminations (silica sheets between core layers) –
to reduce eddy current, and minimize loss
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Iron (Core) Losses
Hysteresis Loss:
The direction of the magnetic flux in the core changes every cycle.
Power is consumed to move around the magnetic dipoles in the core
material, and energy is dissipated as heat.
Hyst. loss  (vol. of core) x (area of hyst. loop)
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Transformer Construction
•
•
•
•
•
Coil Winding
Core Assembly
Core-Coil Assembly
Tank-up
Accessories Mounting and Finishing
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Core Assembly
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Core-Coil Assembly
•
•
•
•
•
•
•
•
Core vertical sides – limbs, top horizontal side – yoke
Yoke is removed to insert the coils into the limbs
LV coil is first placed on the insulated core limbs
Insulating blocks are placed at the top and bottom of the
LV coil
Cylinder made out of corrugated paper is placed over
the LV Coil
HV coil is placed over the cylinder
The top yoke is fixed in position
LV and HV windings are connected as required
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Tank-up
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Accessories Mounting
• Connections of LV and HV coil ends to the terminal
bushings are made
• Transformer tap changer and protection accessories
(e.g. Buchholz relay, Conservator, Breather,
temperature indicator, etc.) are installed
• Tank is closed
Functions of Transformer Oil
• Cooling
• muffle noise
• displace moisture (avoid insulation degradation)
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Transformer Cooling
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Typical Power Transformers
Pole-mounted
Single-phase Transformer
Three-phase Transformer
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Transformer Rating
• Rated kVA
• Rated Voltage primary and secondary: (transformers are normally operated
close to their rated voltages)
• Rated Current (FL Current) is the maximum continuous current the
transformer can withstand
For single phase transformer:
Rated primary current = Rated VA / Rated Pr Voltage
When rated current flows through a transformer, it said to be fully loaded.
The actual current through a transformer varies depending on the load
connected at different times of the day.
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Equivalent Circuit
• Represent inductively coupled circuits by a conductively connected circuit
• Equivalence in terms of loop equations
• Assume a load impedance is connected to the secondary.
+
V1
-
R1
I1
L11
KVL at Loop 1:
R2
M
L22
+
V2
-
I2
N1 N2
-V1 + R1I1 + jwL11I1 - jwM I2 = 0
KVL at Loop 2:
R2I2 + V2+ jwL22I2 - jwM I1 = 0
inductively coupled circuits
turns ratio,
N1
= a
N2
The loop equations are:
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Equivalent Circuit (continued)
+
V
1
aM
I1
-
Consider following substitutions:
a2L2 a2R2
L1
R1
+
aV2
-
M  a·M
L2  a2·L2 R2  a2·R2
V2  a·V2
I2 
I2
a
conductively connected circuit
KVL at Loop 1:
-V1 + R1I1 + jwL1I1 + jwaM (I1-I2/a) = 0
KVL at Loop 2:
jwaM (I2/a - I1) + jw a2L2.I2/a + jw a2R2.I2/a + aV1 = 0
Let
and
The loop equations remain the same:
V1
- V2
R1 + jwL11
=
- jwM
- jwM
I1
R2 + jwL22
I2
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L11 = L1 + aM
L22 = L2 + M/a
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Transformer Equivalent Circuit
R1
+
I1
Rc
a2R2
I2
Ie
Ic
V1
a2X 2
X1
a
I
X
-
equivalent circuit referred to the primary side
a2ZL
ZL = load impedance
Ie = excitation current
I = magnetizing current
X = magnetizing
reactance
R1, R2= primary, secondary winding resistance
X1, X2= primary, secondary leakage reactance
I1, I2= primary, secondary current
Rc = core loss resistance (equivalent resistance contributing to core loss)
Ic = core loss equivalent current EE 212 2010-2011
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Example: Transformer Equivalent Circuit
A 50-kVA, 2400/240-V, 60-Hz distribution transformer has a leakage impedance of
(0.72 + j0.92) W in the high voltage (HV) side and (0.0070 + j0.0090) W in the low
voltage (LV) winding. At rated voltage and frequency, the admittance of the shunt
branch of the equivalent circuit is (0.324 – j2.24)x10-2 S [siemens] when viewed from
the LV side. Draw the circuit:
a) viewed from the LV side
b) viewed from the HV side
Turns ratio, a = 2400/240 = 10
a)
b)
0.72/a2 W
j0.92/a2 W
308.642 W
0.0070 W
j0.0090 W
0.72 W
j44.6429 W
j0.92 W
308.642a2 W
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0.0070a2 W j0.0090a2 W
j44.6429 a2 W
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Transformer: Approximate Equivalent Circuit
Re
Xe
R e = R1 + a 2R2
X e = X 1 + a2 X 2
Rc
X

where, R2 and X2 are referred
to the primary side
Re and Xe  obtained from Short Circuit Test
Rc and X  obtained from Open Circuit Test
Re
Xe
For transformers operating close to full load
Xe
For transformers in a large power network analysis
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Short Circuit Test
+
A
Connect meters on HV side as shown
Short circuit LV side
Energize HV side with a variable
voltage source and increase voltage
gradually to get rated current reading
on the Ammeter
Take the V, I and P readings from meters
W
Vsupply V
HV
LV
P = copper loss = I2 Re  Re =
+
P
I2
V
Ze =
V
I

Xe =
R
R 1 ≈ a2 R 2 ≈ e
2
Ze - R e
2
X1 ≈
-
2
I
Re
Rc
X
Xe

Equivalent circuit referred to HV side
a2 X
Xe
2≈ 2
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Open Circuit Test
A
W
V
H
V
Connect meters on LV side as shown
Open circuit HV side
Energize LV side with rated voltage
Take the V, I and P readings from meters
+
Vsupply
-
LV
2
V
P = core loss = R
c
I = I e - I c
2
V
 Rc =
P
2
V
-
V
Rc
Ie = I , Ic =
+
2
I
Ie
Ic
Rc
I
X
Equivalent circuit referred to LV side

V
X =
Iφ
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Transformer Efficiency
Efficiency, h = Power Output / Power Input
Power Output = |VL| |IL| cos  = |IL|2 RL
Power Input = Power Output + Cu losses + Core losses
Cu Loss – varies with load current
Core Loss – depends on voltage (usually a constant for practical
purposes)
Transformer efficiency is maximum when
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Cu loss = core loss
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Transformer Efficiency (continued)
Power transformers: - usually operate at rated capacity,
- designed to have max. h at full load
Distribution transformers: carry a widely varying load
- designed to have max. h at less than full load
- always energized despite load levels – designed to
have low core loss
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Example: Transformer
Find the equivalent circuit of a 50-kVA, 2400/240-V transformer. The
following readings were obtained from short circuit and open circuit
tests:
Test
V-meter (V) A-meter (A) W-meter
Short Circuit
Open Circuit
48
240
20.8
5.41
(W)
617
186
If rated voltage is available at the load terminals, calculate the transformer
efficiency at
(a) full load with 0.8 p.f. lagging
(b) 60% load with 0.8 p.f. lagging.
What is the voltage at the HV terminals of the transformer?
If rated voltage is applied at the primary terminals, calculate the
transformer efficiency at
(c) full load with 0.8 p.f. lagging
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Transformer Example 2
The transformer is used to step down the voltage at the load end
of a feeder whose impedance is 0.3 + j1.6 ohm. The voltage at
the sending end of the feeder is 2400 V. Find the voltage at the
load terminals when the connected load is
(a) Zero
(b)60% load at 0.8 p.f. lagging
(c) Full load at 0.8 p.f. laging
(d)Full load at 0.5 p.f. leading
What will be the current on the low voltage side if a short circuit
occurs at the load point?
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Voltage Regulation
When a constant rated voltage is applied to the primary,
At no load
- no current, and therefore no voltage drop in transformer
- secondary voltage, Vsec = rated voltage
As load (resistive or inductive) increases
- voltage drop in transformer (Vdrop) increases
- secondary voltage decreases (reverse is the case with capacitive
load increase)
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Voltage Regulation (continued)
To maintain rated voltage at the secondary,
At no load, primary voltage required, Vpr = rated voltage
At a certain load, required Vpr = rated voltage + Vdrop in transformer
Voltage regulation is the change in primary voltage required to keep
the secondary voltage constant from no load to full load, expressed as
a percentage of rated primary voltage.
Load p.f. has a big effect on voltage regulation.
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Example
A 50-kVA, 2400/240-V, 60-Hz transformer has a leakage impedance
of (1.42+j1.82) ohms on the HV side. The transformer is operating at
full load in all 3 cases below.
1. Calculate the voltage regulation at
a) 0.8 p.f. lagging
b) unity p.f.
c) 0.5 p.f. leading
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Voltage Control
The allowable voltage variation at the customer load point is usually ±
5% of the rated (nominal) value. Control measures are taken to
maintain the voltage within the limits.
- Use of transformer tap changers: on-load/off-load
- Injecting reactive power (Vars): series/shunt compensation
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Autotransformers
• Used where electrical isolation between primary
side and secondary side is not required.
• Usually relatively low power transformers
• Can be readily made to be variable.
• Can think of it as two separate windings
connected in series.
• Usually it is a single winding with a tap point.
• Can be used as either step-up or step-down
transformer
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Autotransformers (continued)
• Schematic diagrams:
From J.D. Irwin, “Basic Engineering Circuit
Analysis”, 3rd ed. Macmillan
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Autotransformers (continued)
From Jackson et al. “Introduction to
Electric Circuits” 8th ed., Oxford
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Autotransformers (continued)
Assume N1 = 200
and N2 = 100
Let Vsource = 120 V
Then what is V2?
Assume I2 is 15 A.
Then Sload = ?, and Ssource = ?
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Autotransformers (continued)
What is the node equation at the tap?
Node equation:
I1 + IZY = I2
therefore, IZY =
From transformer action:
N1I1 = N2IZY
therefore,
That is, only part of the load current is due to the
magnetomotive force that the current in the primary coil exerts
on the secondary coil. The remaining portion is direct
conduction of the source current to the load.
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