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Electrical Machines
G.C University, Lahore
Transformer
• A transformer is a device that changes ac electric power
at one voltage level to ac electric power at another
volt age level through the action of a magnetic field.
• It’s a static device
• It consists of two or more coils
• Wrapped around a common ferromagnetic core
• Coils not directly connected
Why Transformer
• First power distribution system in the United States was a
120-V dc system invented by Thomas A. Edison
• At such low voltage, large currents needed to be
transmitted with high transmission losses
• Generation units needed after every few block
• But invention of transformer and AC power source
eliminated the drawbacks of DC power system
• AC power can be generated at one central location
Why transformer
• The voltage stepped up for transmission over long distances at
very low losses
• Voltage stepped down again for final use
• Transmission losses are proportional to the square of the
current in the lines
• Raising voltage and reducing currents by a factor of 10 with
transformers reduces power transmission losses by a factor of
100
• Today power is generated at 12-25 kV. Transformers step up the
voltage to 110 kV or even 1000 kV for over long distances, then
step down 12- to 34.5-kV range for local distribution
Transformer Usage
• The principal purpose of a transformer is to convert
ac power at one voltage level to ac power of the
same frequency at another voltage level.
Transformers are also used for
• voltage sampling
• current sampling
• impedance transformation
• Circuit isolation
Construction of Transformer
• Two coils, called primary & secondary
• Primary attached with the source
• Secondary to the load
• No electrical connection b/w two coils
• So, energy travels from electrical to magnetic and
back to electrical domains
Construction
• Two core types
• One with simple
rectangular laminated
piece of steel with the
transformer windings
wrapped around two
sides
• Called core form
Construction
• other type consists of a
three-legged laminated
core with the windings
wrapped around the
center leg
• Called Shell type
Construction
• The primary and secondary windings in a physical
transformer are wrapped one on top of the other with the
low-voltage winding innermost.
• Such an arrangement serves two purposes:
• It simplifies the problem of insulating the high-voltage
winding from the core.
• It results in much less leakage flux than would be the case
if the two windings were separated by a distance on the
core.
THE IDEAL TRANSFORMER
A Lossless device
i.e. no leakage flux,
Cu or eddy current loss
THE IDEAL TRANSFORMER
𝑑∅
• On primary side Vp = 𝑁𝑝
𝑑𝑡
(Faradays Law)
𝑑∅
• 𝑑𝑡
(change of flux) travels through the core towards
secondary coil. (No leakage flux, ideal transformer)
𝑑∅
• 𝑑𝑡
is due to the AC source as it changing frequency
𝑑∅
• So on secondary side, Vs = 𝑁𝑠
𝑑𝑡
THE IDEAL TRANSFORMER
• Ratio of primary and secondary voltages gives us
𝑉𝑝 𝑁𝑝 𝑁𝑝
• 𝑉𝑠
=
𝑁𝑠
= 𝑎, ℎ𝑒𝑟𝑒 𝑎 𝑖𝑠 𝑡𝑢𝑟𝑚 𝑟𝑎𝑡𝑖𝑜 𝑎 =
𝑁𝑠
• Since power at both ends is same for ideal transformer
𝑉𝑝 𝐼𝑆 𝐼𝑝 1
• 𝑉𝑝 𝐼𝑝 = 𝑉𝑠 𝐼𝑠 or 𝑉𝑠
=
𝐼𝑃
𝑔𝑖𝑣𝑒𝑠 𝑢𝑠
𝐼𝑠
=
𝑎
𝑉𝑝 𝑁𝑝
• 𝑉𝑠
=
𝑁𝑠
= Effort or Potential Variables
𝑁𝑝 𝐼𝑆
• 𝑁𝑠
=
𝐼𝑃
= Kinetic or flow variables
THE IDEAL TRANSFORMER
• In terms of phasor quantities
𝑽𝑷 𝑰𝑷 1
• 𝑽𝒔
=a and
𝑰𝒔
=
𝑎
• Phase angle of Vp is the same as of Vs and the
phase angle of Ip is the same as of Is
• In ideal transformer, turn ratio affect the magnitude
not the phase angles
THE IDEAL TRANSFORMER
THE IDEAL TRANSFORMER
• If the primary voltage is positive at the dotted end of
the winding with respect to the undotted end, then
the secondary voltage will be positive at the dotted
end also.
• If the primary current of the transformer flows into
the dotted end of the primary winding, the
secondary current will flow out of the dotted end of
the secondary winding
Power in an Ideal Transformer
• The power supplied to the transformer by the
primary circuit is given by the equation
• 𝑃𝐼𝑁 = 𝑉𝑃 𝐼𝑃 cosθp
• θ p is angle between the primary voltage and the
primary current.
• At secondary 𝑃𝑜𝑢𝑡 = 𝑉𝑠 𝐼𝑠 cosθs
• Since voltage and current angles are unaffected by an
ideal transformer, θ p - θ s = θ
Power in an Ideal Transformer
• The power out of a transformer is 𝑃𝑜𝑢𝑡 = 𝑉𝑠 𝐼𝑠 cosθs
• Applying the turns-ratio Vs=Vp/a or Is=a Ip
𝑉𝑝
• 𝑃𝑜𝑢𝑡 = 𝑎 × 𝑎𝐼𝑝 cosθ
• 𝑃𝑜𝑢𝑡 = 𝑉𝑃 𝐼𝑃 cosθ = Pin
• The same relationship applies to reactive power Q and
apparent power S
• Qin= 𝑉𝑃 𝐼𝑃 sin θ = 𝑉𝑠 𝐼𝑠 sin θ = Qout
• Sin = 𝑉𝑃 𝐼𝑃 = 𝑉𝑠 𝐼𝑠 = Sout
Impedance Transformation
• The impedance of a device or an element is defined as
the ratio of the phasor voltage across it to the phasor
current flowing through it
𝑽𝑳
• 𝑍𝐿 = 𝑰𝑳
• Since transformer changes voltage and current levels,
it changes the ratio between voltage and current and
hence the apparent impedance of an element
Impedance Transformation
Impedance Transformation
• If the secondary current is Is and the secondary voltage
Vs, and the impedance of the load is given by
𝑽𝑳
• 𝑍𝐿 = 𝑰𝑳
• The apparent impedance of the primary circuit is
𝑽𝑷 𝐼𝑠
• 𝑍𝐿 ′ = 𝑰𝑷
Also we have 𝑉𝑝 = 𝑎𝑉𝑠 𝑎𝑛𝑑 𝐼𝑝 =
𝑎
Impedance Transformation
• Apparent impedance of the primary is
𝑽𝑷 𝒂 𝑽𝒔 𝑽𝒔
• 𝑍𝐿′ =
𝑰𝑷
=
𝑰𝒔/𝒂
= 𝒂𝟐
𝑰𝒔
, therefore 𝑍𝐿′ = 𝑎2 𝑍𝐿
• With a transformer, it is possible to match the
magnitude of a load impedance to a source
impedance simply by picking the proper turns ratio
Example
• A single-phase power system consists of a 480-V 60-
Hz generator supplying a load 𝑍𝑙𝑜𝑎𝑑 = 4 + j3 ohms
through a transmission line of impedance 𝑍𝐿𝑖𝑛𝑒 =
0.18 + j0.24 Ohm. Answer the following questions
about this system
• If the power system is exactly as described above
(Figure 1), what will the voltage at the load be? What
will the transmission line losses be?
• Figure a shows the power system without transformers.
Here 𝐼𝐺 = 𝐼𝑙𝑖𝑛𝑒 .The line current in this system is given
by
𝑉 480∠0°
• 𝐼𝑙𝑖𝑛𝑒 = 𝑍 = =90.8∠ − 37.8°𝐴
𝑙𝑖𝑛𝑒 +𝑍𝑙𝑜𝑎𝑑 (0.18+𝑗0.24)+(4+𝑗3)
• Load Voltage is
• 𝑉𝑙𝑜𝑎𝑑 = 𝐼𝑙𝑖𝑛𝑒 𝑍𝑙𝑜𝑎𝑑 = 90.8∠ − 37.8° × (4+j3) = 454∠ −
0.9°V
• 𝑃𝑙𝑜𝑠𝑠 = 𝐼2 𝑅𝑙𝑖𝑛𝑒 = 90.82 × 0.18 = 1484𝑤𝑎𝑡𝑡𝑠
• Suppose a 1: 10 step-up transformer is placed at the
generator end of the transmission line and a 10: 1
step-down transformer is placed at the load end of
the line (Figure b). What will the load voltage be
now? What will the transmission line losses be now?
• Here we convert system to a common voltage level
• Eliminate transformer T2 by referring the load over
to the transmission line's voltage level.
• Eliminate transformer T1 by referring the
transmission line's elements and the equivalent load
at the transmission line's voltage over to the source
side
• The value of the load's impedance when reflected to
the transmission system's voltage is
10
• 𝑍′𝑙𝑜𝑎𝑑 = 𝑎2 𝑍𝑙𝑜𝑎𝑑 =
1
4 + 𝑗3 = 400 + 𝑗300
• The total impedance at the transmission line level is
now
• 𝑍𝑒𝑞 = 𝑍𝑙𝑖𝑛𝑒 +𝑍′𝑙𝑜𝑎𝑑 = 400.18 + 𝑗300.24 = 500.3∠36.88°
• The total impedance at the transmission line level
𝑍𝑙𝑖𝑛𝑒 + 𝑍′𝑙𝑜𝑎𝑑 is now reflected across T1 to the source's
voltage level
1
• 𝑍′𝑒𝑞 = 𝑎2 𝑍𝑒𝑞 = 10
0.18 + 𝑗0.24 + 400 + 𝑗300
= 0.0018 + 𝑗0.0024 + 4 + 𝑗3 = 5.003∠36.88°
• The generator current is
480∠0°
• Ig =5.003∠36.88° = 95.94∠−36.88°
• Now 𝑁𝑃1 𝐼𝑔 = 𝑁𝑠1 𝑙𝑙𝑖𝑛𝑒
• 𝑙𝑙𝑖𝑛𝑒 = (𝑁𝑃1 /𝑁𝑆1 )𝐼𝑔 =
1
95.94∠−36.88°=9.594∠−36.88°
10
• Also 𝑁𝑃2 𝐼𝑙𝑖𝑛𝑒 = 𝑁𝑠2 𝑙𝑙𝑜𝑎𝑑
• 𝑙𝑙𝑜𝑎𝑑 = (𝑁𝑃2 /𝑁𝑆2 )𝐼𝑙𝑖𝑛𝑒 =
10 × 9.594∠−36.88°=95.94∠−36.88°
• 𝑉𝑙𝑜𝑎𝑑 = 𝐼𝑙𝑜𝑎𝑑 𝑍𝑙𝑜𝑎𝑑 = (95.94∠−36.88°)
(5∠36.88°) = 479.7∠−0.01° V
• Line loss
• 𝑃𝑙𝑜𝑠𝑠 = (𝑙𝑙𝑖𝑛𝑒 )2 𝑅𝑙𝑖𝑛𝑒 = 9.594 2 0.184 = 16.7𝑤
Real Transformer
• Two or more coils of wire physically wrapped around
a ferromagnetic core.
• The characteristics of a real transformer approximate
the characteristics of an ideal transformer, but only
to a degree.
• The fig. on next slide shows a real transformer
• Here
𝑑∅
• 𝑒𝑖𝑛𝑑 = 𝑁 𝑑𝑡 where ∅ 𝑖𝑠 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑓𝑙𝑢𝑥 𝑜𝑣𝑒𝑟 𝑁 𝑡𝑢𝑟𝑛𝑠
• This is because the flux passing through each turn of a
coil is slightly different from the flux in the other turns
• For voltage across primary Vp, we will have
1
∅= 𝑉𝑝 𝑡 𝑑𝑡
𝑁𝑝
• This flux is present in the primary coil of the transformer.
What effect does it have on the secondary coil of the
transformer?
• The effect depends on how much of the flux reaches the
secondary coil
• Some of the flux lines leave the iron core and pass
through the air instead, it is called leakage flux
• Thus we have mutual and leakage flux
• ∅𝑝 = ∅𝑚 + ∅𝐿𝑝
• So, on secondary side we have also: ∅𝑠 = ∅𝑚 + ∅𝐿𝑠
• We can also write for primary
𝑑∅𝑝 𝑑∅𝐿𝑝
𝑉𝑝 = 𝑁𝑝 + 𝑁𝑝 = 𝑒𝑝 𝑡 + 𝑒𝐿𝑝 (𝑡)
𝑑𝑡 𝑑𝑡
• Similarly for secondary 𝑉𝑠 = 𝑒𝑠 𝑡 + 𝑒𝐿𝑠 𝑡
• Voltage due to mutual flux
𝑑∅𝑚
• At primary 𝑒𝑝 (𝑡) = 𝑁𝑝
𝑑𝑡
𝑑∅
• At secondary 𝑒𝑠 (𝑡) = 𝑁𝑠 𝑚
𝑑𝑡
𝑒𝑝 (𝑡) 𝑑∅𝑚 𝑒𝑠 (𝑡)
• We can also write 𝑁𝑝
=
𝑑𝑡
=
𝑁𝑠
𝑒𝑝 (𝑡) 𝑁𝑝
• Hence 𝑒 (𝑡) = 𝑁𝑠
=a
𝑠
• Since ∅𝑚 ≫
∅𝐿 𝑓𝑜𝑟 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑎𝑛𝑑 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 therefore
𝑉𝑝 𝑡 𝑁𝑝
• 𝑉𝑠 𝑡
=
𝑁𝑠
=𝑎
Magnetization Current
• When AC source is connected on primary side and
secondary left open, current is required to produce
flux. Two components of current are
• Magnetization current 𝑖𝑀 , which is the current required
to produce the flux in the transformer core
• Core-loss current 𝑖𝑐𝑢+𝐻 which is the current required
to make up for hysteresis and eddy current losses
1
• Average flux in the core ∅ = 𝑁𝑝
𝑉𝑝 𝑡 𝑑𝑡
Magnetization Current
• If vp(t) = 𝑉𝑀 cos ωt , so we can write
1 𝑉𝑀
• ∅= 𝑁𝑝
𝑉𝑀 cos ωt𝑑𝑡 =
ω𝑁𝑝
𝑠𝑖𝑛ω𝑡
• If the values of current required to produce a given
flux are compared to the flux in the core at different
times, it is possible to construct a sketch of the
magnetization current in the winding on the core.
Magnetization Current
• The magnetization current is not sinusoidal.
• Once the peak flux reaches the saturation point in the core, a
small increase in peak flux requires a very large increase in the
peak magnetization current.
• The fundamental component of the magnetization current lags
the voltage applied to the core by 90°.
• The higher-frequency components in the magnetization current
can be quite large compared to the fundamental component.
• In general, the further a transformer core is driven into
saturation, the larger the harmonic components will become.
Core loss component
• The other component of the no-load current in the
transformer is the current required to supply power
to make up the hysteresis and eddy current losses in
the core.
Core loss component
• Notice the following points about the core- loss current
• The core- loss current is nonlinear because of the
nonlinear effects of hysteresis.
• The fundamental component of the core- loss current is
in phase with the voltage applied to the core.
• The total excitation current in a typical transformer core is
𝑖𝑒𝑥 = 𝑖𝑚 𝑖ℎ+𝑒
EQUIVALENT CIRCUIT OF
A TRANSFORMER
• The major items to be considered in the construction
of such a model are
• Copper losses
• Eddy current
• Hysteresis losses
• Leakage flux
EQUIVALENT CIRCUIT OF
A TRANSFORMER
• Copper losses are resistive losses in the primary and
secondary windings and modeled by placing a
resistor Rp in the primary circuit of the transformer
and a resistor Rs in the secondary circuit.
• The leakage flux in the primary and secondary
windings produces a voltage given by
𝑑∅𝐿𝑃 𝑑∅𝐿𝑆
• 𝑒𝑝 𝑡 = 𝑁𝑝
𝑑𝑡
and 𝑒𝑠 𝑡 = 𝑁𝑠
𝑑𝑡
EQUIVALENT CIRCUIT OF
A TRANSFORMER
• Flux is directly proportional to current flow,
therefore we can assume that leakage flux is also
proportional to current flow in the primary and
secondary windings
• ∅𝐿𝑃 = 𝑷𝑁𝑃 𝑖𝑝 and ∅𝐿𝑆 = 𝑷𝑁𝑠 𝑖𝑠
𝑑 𝑑𝑖𝑝
• 𝑒𝐿𝑃 𝑡 = 𝑁𝑝 𝑑𝑡 𝑷𝑁𝑃 𝑖𝑝 = 𝑁𝑝2 𝑃 𝑑𝑡
=
𝑑𝑖𝑝
𝐿𝑝 (𝑎𝑠𝐿𝑃 = 𝑁𝑝2 𝑷)
𝑑𝑡
EQUIVALENT CIRCUIT OF
A TRANSFORMER
𝑑𝑖𝑝
• Similarly 𝑒𝐿𝑆 = 𝐿𝑝
𝑑𝑡
• Therefore the leakage element may be modeled as an
inductance connected together in series with the
primary and secondary circuit respectively
EQUIVALENT CIRCUIT OF
A TRANSFORMER
• Core excitation effects
• The magnetization current Im is a current proportional
(in the unsaturated region) to the voltage applied to the
core and lagging the applied voltage by 90° - modeled as
reactance Xm across the primary voltage source.
• The core loss current 𝐼ℎ+𝑒 is a current proportional to the
voltage applied to the core that is in phase with the
applied voltage – modeled as a resistance Rc across the
primary voltage source
Transformer Tests
• Open Circuit Test: The transformer’s secondary
winding is open-circuited, and its primary winding is
connected to a full rated line voltage.
Open Circuit Test
• Full line voltage is applied to the primary – input voltage,
input current, input power measured.
• Then, power factor of the input current and magnitude
and angle of the excitation impedance can be calculated
• All the input current will be flowing through the
excitation branch of the transformer.
• The series element 𝑅𝑃 𝑎𝑛𝑑 𝑋𝑃 are too small in
comparison to 𝑅𝐶 𝑎𝑛𝑑 𝑋𝑀 to cause a significant voltage
drop. Essentially all input voltage is dropped across the
excitation branch
Open Circuit Test
• To obtain the values of RC and XM , the easiest way is to
find the admittance of the branch
• Total excitation admittance,
1 1
• 𝑌𝐸 = 𝐺𝑐 − 𝑗𝐵𝑀 = 𝑅𝑐 − 𝑗 𝑋𝑚
𝐼𝑜𝑐
• 𝑌𝐸 = 𝑉𝑜𝑐
𝑃𝑜𝑐 𝑃𝑜𝑐
• 𝑃. 𝐹 = 𝑐𝑜𝑠θ = 𝑉𝑜𝑐×𝐼𝑜𝑐 so, θ = 𝑐𝑜𝑠 −1 𝑉𝑜𝑐×𝐼𝑜𝑐
• The power factor is always lagging for a real transformer
Short-circuit Test
• Short-circuit Test: The secondary terminals are
short circuited, and the primary terminals are
connected to a fairly low voltage source.
Short-circuit Test
• The input voltage is adjusted until the current in the
short circuited windings is equal to its rated value.
• The input voltage, current and power are measured.
• The excitation branch is ignored, because negligible
current flows through it due to low input voltage
during this test.
• Thus, the magnitude of the series impedances
𝑉𝑠𝑐
referred to the primary is 𝑍𝑆𝐸 =
𝐼𝑠𝑐
Short-circuit Test
𝑃𝑠𝑐
• 𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑃𝐹 = 𝑐𝑜𝑠 𝜃 = 𝑉𝑠𝑐×𝐼𝑠𝑐
• The series impedance 𝑍𝑆𝐸 = 𝑅𝑒𝑞 + 𝑗𝑋𝑒𝑞
Example
• The equivalent circuit impedances of a 20-kVA,
800CV240- V, 6O-Hz transformer are to be
determined. The open-circuit test and the short-
circuit test were performed on the primary side of
the transformer, and the following data were taken
• Find the impedances of the approximate equivalent circuit
referred to the primary side, and sketch that circuit
• The power factor during the open circuit test is
𝑃𝑜𝑐 400
• 𝑃. 𝐹 = 𝑐𝑜𝑠θ = 𝑉𝑜𝑐×𝐼𝑜𝑐 = 8000×0.24 = 0.234 𝑙𝑎𝑔𝑔𝑖𝑛𝑔
• The excitation admittance is given by
𝐼𝑜𝑐 1 1
• 𝑌𝐸 = 𝑉𝑜𝑐∠-𝑐𝑜𝑠 −1 P.F =0.000063-j0.0000261 =𝑅𝑐 − 𝑗 𝑋𝑚
• Rc=159k and Xm=38.4k
• Now find the series elements
• Req=38.4 and Xeq=192
Per Unit System
• The process of solving circuits containing transformers
using the referring method where all the different voltage
levels on different sides of the transformers are referred
to a common level, can be quite tedious.
• The Per-unit System of measurements eliminates this
problem. The required conversions are handled
automatically by the method.
• In per-unit system, each electrical quantity is measured as
a decimal fraction of some base level. Any quantity can be
expressed on a per-unit basis by the equation
𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑎𝑙𝑢𝑒
• 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑝𝑒𝑟 𝑈𝑛𝑖𝑡 = 𝐵𝑎𝑠𝑒 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
• Two base quantities are selected to define a given
per-unit system. The ones usually selected are voltage
and power.
• In a single phase system, the relationship are
𝑃𝑏𝑎𝑠𝑒 𝑄𝑏𝑎𝑠𝑒 𝑜𝑟 𝑆𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 𝐼𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒
• 𝑍𝑏𝑎𝑠𝑒 = 𝐼𝑏𝑎𝑠𝑒
Example
• A simple power system is shown in Figure . This system contains a
480-V generator connected to an ideal 1 : 10 step-up transformer, a
transmission line, an ideal 20: 1 step-down transformer, and a load.
The impedance of the transmission line is 20 + j60 Ω, and the
impedance of the load is 10∠30°Ω. The base values for this system
are chosen to be 480 V and 10 kVA at the generator.
• (a) Find the base voltage, current, impedance, and apparent power
at every point in the power system.
• (b) Convert this system to its per-unit equivalent circuit.
• (c) Find the power supplied to the load in this system.
• (d) Find the power lost in the transmission line.
• In the generator region. 𝑉𝑏𝑎𝑠𝑒 = 480𝑉 and 𝑆𝑏𝑎𝑠𝑒 = 10𝑘𝑉𝐴, 𝑠𝑜
𝑆 10000
• 𝐼𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 = 480
= 20.83𝐴
𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒 480
• 𝑍𝑏𝑎𝑠𝑒 = 𝐼𝑏𝑎𝑠𝑒
= 20.83 = 23.04Ω
• The turns ratio of transformer T1 is a = 1/10 = 0.1, so the base voltage in
the transmission line region is
𝑉𝑏𝑎𝑠𝑒 1 480
• 𝑉𝑏𝑎𝑠𝑒 2 = 𝑎
=
0.1
= 4800𝑉
• Other quantities are
• 𝑆𝑏𝑎𝑠𝑒 2 = 10𝑘𝑉𝐴
10000
• 𝐼𝑏𝑎𝑠𝑒 2 = 4800
= 2.083𝐴
4800
• 𝑍𝑏𝑎𝑠𝑒 2 = 2.083 = 2304Ω
• The turns ratio of transformer T1 is a = 20/1 = 20, so the base
voltage in the load region is
𝑉𝑏𝑎𝑠𝑒 1 480
• 𝑉𝑏𝑎𝑠𝑒 2 = 𝑎
=
0.1
= 4800𝑉
• The other base quantities are
• 𝑆𝑏𝑎𝑠𝑒 3 = 10𝑘𝑉𝐴
10000
• 𝐼𝑏𝑎𝑠𝑒 3 = 240
= 41.67𝐴
240
• 𝑍𝑏𝑎𝑠𝑒 3 = 41.67 = 5.76Ω
• B) To convert a power system to a per-unit system, each component must
be divided by its base value
• The generator s per-unit voltage is its actual value divided by its base value
480∠0°
• 𝑉𝐺 , 𝑝𝑢 = 480
=1∠0° pu
• The transmission line s per-unit impedance is its actual value
divided by its base value:
20+𝑗60
• 𝑍𝑙𝑖𝑛𝑒 , 𝑝𝑢 = 2304
= 0.0087 + 𝑗0.0260 𝑝𝑢
• The loads per-unit impedance is also given by actual value
divided by base value
10∠30°
• 𝑍𝑙𝑜𝑎𝑑 , 𝑝𝑢 = 5.76
= 1.736∠30° pu
• The per-unit equivalent circuit of the power system is
shown in Figure
TRANSFORMER VOLTAGE
REGULATION AND
EFFICIENCY
• The output voltage of a transformer varies with the load
even if the input voltage remains constant.
• This is because a real transformer has series impedance
within it.
• Full load Voltage Regulation is a quantity that compares
the output voltage at no load with the output voltage at
full load, defined by this equation
𝑉𝑠 ,𝑛𝑙−𝑉𝑠,𝑓𝑙
• 𝑉𝑅 = 𝑉𝑠 ,𝑓𝑙
× 100
The Transformer Phasor
Diagram
• Consider the simplified
equivalent circuit referred
to the secondary side
• Voltage Regulation
depends on magnitude of
the series impedance and
the phase angle of the
current flowing through
the transformer
• Assume that the reference phasor is the secondary
voltage, VS. Therefore the reference phasor will have
0 degrees in terms of angle
• Based upon the equivalent circuit, apply Kirchoff
Voltage Law
𝑉𝑃
• 𝑎
= 𝑉𝑆 + 𝑅𝑒𝑞 𝐼𝑆 + 𝑗𝑋𝑒𝑞 𝐼𝑠
• Figure below shows a phasor diagram of a
transformer operating at a lagging power factor.
• For lagging loads, VP / a > VS so the voltage
regulation with lagging loads is > 0
• When the power factor is unity, VS is lower than VP
so VR > 0. But, VR is smaller than before (during
lagging PF).
• With a leading power factor, VS is higher than the
referred VP so VR < 0
Transformer Efficiency
• Transformer efficiency is defined as
𝑃𝑜𝑢𝑡 𝑃𝑜𝑢𝑡
• η= 𝑃𝑖𝑛
× 100 =
𝑃𝑜𝑢𝑡 +𝑃𝑙𝑜𝑠𝑠
× 100
• Types of losses incurred in a transformer:
• Copper 𝐼2 𝑅 losses
• Hysteresis losses
• Eddy current losses
𝑉𝑠 𝐼𝑠 𝑐𝑜𝑠∅
• η=𝑃 × 100
𝑐𝑢 +𝑃𝑐𝑜𝑟𝑒 +𝑉𝑠 𝐼𝑠 𝑐𝑜𝑠∅
TRANSFORMER TAPS
• Until now, transformers were described by their constant
turns ratios
• However, its not quite true for practical transformer
• Distribution transformers have a series of taps in the
windings to permit small changes in their turns ratio
• The taps on a transformer permit to accommodate
variations in local voltages
• TCUL is a transformer with the ability to change taps
while power is connected to it
AUTOTRANSFORMER
• Used to change voltage levels by only a small amount
• The first winding is shown connected in an additive
manner to the second winding.
• The voltage at the output of the whole transformer is the sum of the
voltage on the first winding and the voltage on the second winding
• The first winding is called the common winding and smaller
winding is called the series winding
𝑉𝑐 𝑁𝑐
• 𝑉𝑆𝐸
=
𝑁𝑆𝐸
• Similarly 𝑁𝑐 𝐼𝑐 = 𝑁𝑆𝐸 𝐼𝑆𝐸
• The voltages in the coils are related to the voltages at the
terminals
• 𝑉𝐿 = 𝑉𝑐
𝑁𝑆𝐸 𝑁𝑆𝐸
• 𝑉𝐻 = 𝑉𝑐 + 𝑉𝑆𝐸 = 𝑉𝑐 + 𝑁𝑐 𝑐
𝑉 = 𝑉𝐿 +
𝑁𝐶 𝐿
𝑉
𝑉 𝑁𝐶
• Hence 𝑉 𝐿 = 𝑁
𝑆𝐸 𝑆𝐸 +𝑁𝐶
𝐼 𝑁𝑆𝐸 +𝑁𝐶
• Similarly we can show 𝐼 𝐿 = 𝑁𝐶
𝑆𝐸
Apparent Power Rating Advantage
• Notice that the input and apparent power to the
autotransformer is given by
• 𝑆𝐼𝑁 = 𝑉𝐶 𝐼𝐶 𝑎𝑛𝑑 𝑆𝑂𝑈𝑇 = 𝑉𝐻 𝐼𝐻
• Input apparent power is equal to the output
• 𝑆𝐼𝑁 = 𝑆𝑂𝑈𝑇 = 𝑆𝐼𝑂
• However, the apparent power in the transformer windings is
• 𝑆𝑊 = 𝑉𝐶 𝐼𝐶 = 𝑉𝑆𝐸 𝐼𝑆𝐸
• 𝑆𝑊 = 𝑉𝑐 𝐼𝑐 = 𝑉𝐿 (𝐼𝐿 − 𝐼𝐻 )
• = 𝑉𝐿 𝐼𝐿 − 𝑉𝐻 𝐼𝐻
• Now,
𝑁𝐶 𝑁𝑆𝐸 +𝑁𝐶 −𝑁𝐶
• 𝑆𝑊 = 𝑉𝐿 𝐼𝐿 − 𝑉𝐿 𝐼𝐿 𝑁 = 𝑉𝐿 𝐼𝐿
𝑆𝐸 +𝑁𝐶 𝑁𝑆𝐸 +𝑁𝐶
𝑁𝑆𝐸
• 𝑆𝑊 = 𝑆𝐼𝑂 𝑁
𝑆𝐸 +𝑁𝐶
𝑆𝐼𝑂 𝑁𝑆𝐸 +𝑁𝐶
• 𝑆𝑊
=
𝑁𝑆𝐸
Internal Impedance
• Compared to a given transformer connected in the
conventional manner, the effective per-unit
impedance of an autotransformer is smaller by a
factor equal to the reciprocal of the power advantage
of the autotransformer connection
• Can be a serious problem in some applications where
the series impedance is needed to limit current flows
during power system faults
• The impedance of a transformer can be found by
shorting the secondary winding and determining the
ratio of the voltage to the current of its primary
winding. For the transformer connected as an
ordinary transformer, the impedance referred to the
primary ( 𝑁𝐶 ) is:
• 𝑍𝑒𝑞 = 𝑍1 + 𝑎2 𝑍2
• Equivalent circuit is shown
• When this transformer is
connected as an autotransformer,
the circuit is as shown below.
• If the output windings of the autotransformer are
shorted out, the voltages 𝑽𝑯 will be zero, and the voltage
𝑽𝑳 will be
• 𝑉𝐿 = 𝐼𝐶 𝑍𝐸𝑄
𝑁
• 𝐼𝐿 = 𝐼𝐶 + 𝐼𝑆𝐸 = 𝐼𝐶 + 𝑁 𝐶 𝐼𝐶
𝑆𝐸
𝑁𝑆𝐸
• 𝐼𝐶 = 𝑁 𝐼𝐿
𝑆𝐸 +𝑁𝐶
• so the input voltage can be
expressed in terms of the input current as
𝑁𝑆𝐸
• 𝑉𝐿 = 𝐼𝐶 𝑍𝑒𝑞 = 𝑁 +𝑁
𝐼𝐿 𝑍𝑒𝑞
𝑆𝐸 𝐶
𝑉𝐿 𝑁𝑆𝐸
• 𝑍𝑒𝑞
′
=
𝐼𝐿
= 𝑍
𝑁𝑆𝐸 +𝑁𝐶 𝑒𝑞
Three phase Transformers
• Transformers for 3-phase circuits can be constructed
in two ways
• connect 3 single phase transformers
• Three sets of windings wrapped around a common
core.
• The primaries and secondaries of any three-phase
transformer can be independently connected in
either a wye (Y) or a delta (Δ)
• Possible connection arrangements
• Wye-wye
• Wye-delta
• Delta-wye
• Delta-delta
• The impedance, voltage regulation, efficiency, and
similar calculations for three phase transformers are
done on a per-phase basis, using same techniques
as single-phase transformers
• A simple concept that all students must remember is
that, for a Delta configuration
𝐼𝐿 𝑆
• 𝑉∅𝑝 = 𝑉𝐿 𝐼∅𝐿 =
3
𝑆∅𝑃 =
3
• For Wye configuration
𝑉𝐿 𝑆
• 𝑉∅𝑝 = 3
𝐼∅𝐿 = 𝐼𝐿 𝑆∅𝑃 = 3
• The Per-unit System for 3-Phase Transformer
𝑆𝑏𝑎𝑠𝑒
• 𝑆1∅,𝑏𝑎𝑠𝑒 = 3
𝑆1∅,𝑏𝑎𝑠𝑒 (𝑉1∅,𝑏𝑎𝑠𝑒 )2
• 𝐼1∅,𝑏𝑎𝑠𝑒 = 𝑉 𝑎𝑛𝑑 𝑍𝑏𝑎𝑠𝑒 =
𝑆1∅,𝑏𝑎𝑠𝑒
1∅,𝑏𝑎𝑠𝑒
2
𝑆𝑏𝑎𝑠𝑒 3 𝑉∅,𝑏𝑎𝑠𝑒
• 𝐼1∅,𝑏𝑎𝑠𝑒 = 3𝑉𝑏𝑎𝑠𝑒
𝑎𝑛𝑑 𝑍𝑏𝑎𝑠𝑒 =
𝑆𝑏𝑎𝑠𝑒
WYE-DELTA CONNECTION
• WYE-WYE CONNECTION
𝑉L𝑝 3𝑉∅𝑃
• 𝑉𝐿𝑆
=
3𝑉∅𝑆
= 𝑎
• If loads on the transformer circuit are unbalanced, then the voltages on the
phases of the transformer can become severely unbalanced
• Third-harmonic voltages can be large.
• WYE-DELTA CONNECTION
𝑉L𝑝 3𝑉∅𝑃
• 𝑉𝐿𝑆
= 𝑉∅𝑆
= 3𝑎
• the secondary voltage is shifted 30" relative to the primary voltage of the
transformer
