Chapter Topics to be Covered
1. Introduction
2. Types and Construction of Transformers.
3. The Ideal Transformer
4. Theory of Operation of Real Single Phase Transformer
5. The Equivalent Circuit of a Transformer
6. The Per Unit System of Measurements
7. Transformer Voltage Regulation and Efficiency
8. Transformer Taps and Voltage Regulation
9. The Autotransformer
10. Three Phase Transformers
TRANSFORMERS
Dr. Abdul Khaliq
2
2.1: Why Transformers are Important
•
•
•
•
•
Introduction to Transformers
The first power distribution system in U.S. was 120V dc
system invented by Thomas A. Edison to supply power for
incandescent light bulbs.
Edison’s first central power station went into operation in New
York city in 1882.
The power was transmitted at very low voltage level, thus
requiring a very large current to supply significant amount of
power resulting in high power losses.
To avoid this problem central power stations were located
every few blocks of the city.
The invention of transformer and concurrent development of
ac power sources eliminated these restrictions forever.
3
•
A transformer is a device that changes ac electric power from
one voltage level to another voltage level through the action
of magnetic field. The transformer has two windings:
i. Primary winding: The winding connected to the power
source is called the primary winding.
ii. Secondary winding: The winding connected to the load is
called secondary winding.
4
1
2.2: Types of Transformers w. r. t. Core Design
2.2: Types of Transformers w.r.t. Core Design
•
Core Type Transformer
Shell Type Transformer
•
•
With respect to the core of the transformer there are two
types of the transformers:
i. Core type transformer: Winding wrapped around two
sides of the core.
ii. Shell type transformer: This is three legged laminated
core, with the winding wrapped around the center leg.
Core is made up of laminations with the High voltage
winding outside & Low voltage winding inside.
Advantages:
i. Easy to insulate H.V. winding from core
ii. Less leakage than if they are at distance from each other.
5
Ideal Transformer
Transformer Types w.r.t. its Applications
1.
2.
3.
4.
5.
6
Unit transformer: To step up the voltage at generation.
Substation transformer: To step down voltage from
transmission to distribution level.
Distribution transformer: Takes distribution voltage and
steps it down to final voltage at which the power is used.
Potential transformer: Is used to change the high voltage to
low voltage, but can handle very small current.
Current transformer: Is used to change the primary current
to much smaller secondary current but the current is directly
proportional to the primary current.
Types 4 and 5 are used for instrumentation.
Ideal transformer is a lossless
device
Np: no of turns on primary
side
Ns: no of turns on the
secondary side
Then Voltage eq. is:
V P (t ) N P
=
=a
V S (t ) N S
Where a is the turn ratio, and
the Current eq. is:
N p i p (t ) = N s i s (t )
i p (t ) N s 1
=
=
i s (t ) N p a
7
8
2
Ideal Transformer
Power in Ideal Transformer
Phasor
Primary side:
Vp
N
I
= P = s =a
NS I p
Vs
•
•
•
Where θp is the angle between Vp and Ip
Vp. & Vs have same phase angle
Ip & Is have same phase angle
Turn ratio just effects only the magnitude of voltage &
current.
Secondary side:
Since voltage and current angles are unaffected
for an ideal transformer
⇒ θ P − θS = 0
Power factor on primary and secondary side is
the same
9
Power in Ideal Transformer
10
Impedance Transformation Through a Transformer
•
Pout = Vs I s cos θ
VP
& I S = aI P
a
V
⇒ Pout = P (aI P ) cos θ
a
Pout = VP I P cos θ = Pin
VS =
Pout = Vs I s cos θ
Where θs is angle between Vs and Is
1. If Vp is +ve @ dotted end w.r.t un-dotted end then Vs
will be also +ve @ dotted end.
2. If Ip flows into the dotted end of primary then Is it will
flow out of dotted end of secondary.
Pin ⇒ Pout
Pin = VP I P cos θ
For an Ideal Transformer
Output power =Input power
Likewise
•
Qin = VP I P sin θ = VS I S sin θ = Qout
and
S in=VP I P = VS I S = S out
11
With a transformer it is possible to match the magnitude of
the load impedance to a source simply by picking the
proper turn ratio.
To analyze the circuit replace the portion of circuit on one
side of the transformer by an equivalent circuit with the
same terminal characteristics. This process is known as
referring the first to the second side.
12
3
Class Activity 1: Effect of Transformer on the losses
Class Activity 1: Effect of Transformer on the losses
Example 2-1: A single phase power system consists of a 480V 60 Hz generator supplying a load Zload=4+j3Ω through a
transmission line of impedance Zline=0.18+j0.24Ω. Answer
the following questions about this system.
a) If the power system is exactly as described above what
will the line losses be.
b) Suppose a 1:10 step up transformer is placed at the
generator end of transmission line and 10:1 step down
transformer is placed at the load end of the line, What will
the load voltage be now? what will the transmission line
losses be now?
13
14
Real Single Phase Transformer
Class Activity 1: Effect of Transformer on the losses
•
•
•
•
Primary connected to an AC source and
secondary is open circuited
λ is the flux linkage in the coil across which
voltage is being induced.
Total flux linkage is the sum of the flux
passing through each turn in the coil added
over all turns of the coil.
However, average flux can be given by:
eind =
dλ
dt
N
λ = ∑ φi
i =1
__
φ=
λ
N
thus
__
ein = N
15
dφ
dt
16
4
The Voltage Ratio Across a Real Transformer
•
When the voltage is applied on the primary
side of the transformer then the flux produced
by this voltage is:
__
φ=
The Voltage Ratio Across a Real Transformer
1
v P (t )dt
NP ∫
•
Some of this flux produced
in the primary links with the
secondary winding and
some goes into the leakage.
There is similar division of
flux in the secondary side:
•
•
Thus the Faraday’s law for
the primary winding can be
written as:
φP = φM + φLP
φS = φM + φLS
v p (t ) = N P
dφP
Dt
dφM
dφ
+ N P LP
dt
dt
= e p (t ) + eLP (t )
= NP
17
The Voltage Ratio Across a Real Transformer
The Voltage Ratio Across a Real Transformer
Like wise for secondary side:
Vs (t ) = N s
18
•
dφs
dt
dφ
dφ
= N s M + N S LS
dt
dt
Vs (t ) = es (t ) + eLS (t )
• From eqs. above the primary
and secondary voltages due to
mutual flux are given by:
• The ratio of the primary voltage
producing the mutual flux to the
secondary voltage induced by
the mutual flux is equal to the
turn ratio of the transformer.
dφ M
dt
dφ M
eS (t ) = N s
dt
e P ( t ) dφ M e S ( t )
⇒
=
=
NP
dt
NS
eP (t ) = N P
⇒
e P (t ) N P
=
=a
eS (t ) N S
•
19
In a good design ØM >> ØLP
and ØM >> ØLS. Therefore,
Ratio of the total voltage on
the primary of transformer to
the ratio of the total voltage
on the secondary of a
transformer is approximately.
Which was the case for the
ideal transformer.
The smaller the leakage flux,
the closer the transformer will
be to the ideal one.
V p (t )
Vs (t )
=
Np
Ns
=a
20
5
The Magnetization Current in Real Transformer
•
•
When AC power is
connected to the
transformer a current
flows even when
secondary is open
circuited.
This is the current
required to produce the
flux in real ferromagnetic
material and it consists of
two components:
1. Magnetization current
iM, Current required to
produce the flux in the
core of transformer.
2. Core losses current ih+e,
current required for
hystersis and eddy
current losses.
φ =
The Magnetization Current in Real Transformer
1
Ndφ
V p (t )d (t ) Q (e =
)
∫
Np
dt
V p (t ) = VM cos ωt
φ =
1
VM cosωtdt
Np ∫
φ =
VM
sin ωt
ωN p
21
22
The Magnetization Current in Real Transformer
The Hysteresis and Eddy Current in Real
Transformer
•
•
If the value of the current required to produce a given flux
is known then it is possible to construct the sketch of
magnetizing current in the winding of the core.
Observations:
1) im is not sinusoidal
2) Øm reaches max; a small increase in Ø requires very
large increase in im.
3) im lags voltage applied by 90˚.
23
•
This is the current required to supply power to make up the
hysteresis and eddy current losses in the core, known as core
loss current.
since the eddy current are proportional to the rate of change
of flux. Therefore, the eddy current are largest when the flux
is passing through the 0 wb.
24
6
The Total No Load (Excitation) Current in Real Transformer
The Current Ratio & Dot Convention
•
•
•
Total no load current is called excitation current which is
sum of magnetization and core loss current.
Connect a load on the secondary side of the transformer.
A current flowing into the dotted end of the transformer
produces a positive mmf.
iex = im + ih+ e
25
26
The Current Ratio & Dot Convention
The Current Ratio & Dot Convention
•
Assumptions to convert a
real transformer into an
ideal transformer.
1. The core must have no
hysteresis or eddy
current.
2. The Magnetization
curve must be an ideal
one.
3. The leakage flux in the
core must be zero,
implying that all the
flux in the core
couples both windings.
4. The resistance of the
transformer winding
must be zero.
•
•
•
•
A current flowing into the dotted
end produces +ve mmf, while a
current flowing into the undotted
end produce a –ve mmf.
If both currents are entering the
dotted end then the mmf will add
to each other.
If one current enters and the
other one leaves then the mmf
will subtract from each other.
For a good designed transform, R
should be very small nearly zero,
as long as the core is operating in
unsaturated region.
In order for mmf to be zero,
current must flow in to the one
dotted end and out of the other
dotted end.
Fp = N p i p & Fs = − N s i s
Fnet = Fp − Fs = N p i p − N s is
Fnet = N p i p − N s is = φR
Fnet = N p i p − N s is ≅ 0
N p i p ≈ N s is
ip
is
≈
Ns 1
=
Np a
27
28
7
Driving The Equivalent Circuit Model
2.5: The Equivalent Circuit of a Transformer
An accurate model of the transformer, have to account for
all the losses of the transformer.
2
1. Copper Losses( I R ): The resistive heating losses in the
primary and secondary winding of the transformer. They
are proportional to the square of the current in the
winding.
2. Leakage Flux( φ LP & φ LS ): The flux which escapes the
core and pass only through one of the transformer
winding. These escaped fluxes produce a self inductance
in the primary and secondary coil.
3. Eddy Current Losses: These are resistive heating losses
in the core and are proportional to the square of the
voltage applied to the transformer.
4. Hysteresis loses: These are associated with the
rearrangement of the magnetic domains in the core
during each half cycle and are nonlinear function of
29
applied voltage.
Driving The Equivalent Circuit Model
3.
4.
1. Copper Losses: The
resistive copper losses are
modeled by placing the a
resistance RP in the primary
and RS in the secondary
winding of the transformer.
2. Leakage Flux: which
escapes the core and pass
only through one of the
transformer winding.
• LP: self inductance of
primary coil
• LS: self inductance of
secondary coil
Therefore, Leakage flux will be
modeled by primary and
secondary inductances.
eLP (t ) = N P
dφLP
dt
dφ LS
dt
= ( ρN P )iP & φ LS = ( ρN S )iS
eLS (t ) = N s
φLP
di
d
2
( ρN P )iP = N p ρ p
dt
dt
di p
2
eLP (t ) = L p
Q LP = N P ρ
dt
d
di
2
eLS (t ) = N S ( ρN S )iS = N S ρ S
dt
dt
diS
2
eLS (t ) = LS
Q LS = N S ρ
dt
eLP (t ) = N p
30
The Final Equivalent Circuit Model
(Eddy Current +Hysteresis Losses): The core loss current,
ih+e, is proportional to voltage applied to the core and is in
phase with voltage. Therefore, can be modeled by Rc
across primary.
The Magnetization Current: Is proportional to the voltage
applied to the core (in unsaturated region) and is lagging
the applied voltage by 90o. So it can be modeled by a
reactance connected across the winding, represented by
XM.
The XM & RC represent the excitation effect which includes
the core loss current (eddy+hysteresis) and the
magnetization current.
The XM & RC are placed inside, after LP and RP, because the
voltage applied to the core is input voltage.
31
•
•
To analyze practical circuits containing transformers,
it is important to convert the entire circuit to a single
voltage level.
Therefore the circuit must be referred either to
primary or to its secondary side.
32
8
The Final Equivalent Circuit Model Referred to
One Side of Transformer
Approximate Equivalent Circuit of a Transformer
•
I.
Referred to Primary
Side
II.
Referred to Secondary
Side
Excitation branch adds complexity. However, current through
this branch is very small, as compared to the load current.
It causes negligible voltage drop in the Rp and Xp and hence
can be neglected.
•
ReqP = R p + a 2 Rs
X eqP = X p + a 2 X s
ReqS =
Rp
a
2
+ Rs
X eqS =
Xp
a2
+ Xs
33
34
The Per-Unit Analysis
The Per-Unit Analysis
• The advantage of per unit analysis is that by
properly specifying the base quantities, the
transformer equivalent circuit can be
simplified.
• The ideal transformer winding can be
eliminated, such that the voltages, currents,
and external impedances and admittances
expressed in per-unit do not change when
they are referred from one side of
transformer to the other.
• This can be a significant advantage even in a
power system of moderate size, where
hundreds of transformers may be
encountered.
•
The following two rules are adopted for the base
quantities:
1. The value of Sbase is the same for entire power system
of concern.
2. The ratio of voltage bases on either side of
transformer is selected to be the same as the ratio of
transformer voltage ratings.
per − unit quantiy =
actual value
base value of the quantity
35
36
9
Per-unit Impedance
Per-unit Impedance
• Three zones of a single phase circuit are identified in the figure
below. These zones are connected by transformers T1 and T2
whose ratings are also shown. Using base values of 30 KVA and
240 volts in zone 1, draw the per unit circuit and determine the
per unit impedances and per unit source voltage. The calculate
the load current both in per unit and in amperes. Transformer
winding resistances and shunt admittances are neglected.
• A single phase two-winding transformer is rated 20
kVA, 480/120 volts, 60 Hz. The equivalent leakage
impedance of transformer referred to the 120-volt
winding, denoted winding 2, isZ = 0.0525∠78.13 Ω. Using
the transformer ratings as base values, determine
the per-unit leakage impedance referred to the
winding 2 and the winding 1.
o
eq 2
37
38
The Open Circuit Test
The Open Circuit Test
•
•
•
Rp & Xp are too small as compared to Rc and Xm.
Therefore, significant voltage drops in Rc and Xm,
approximately all the voltage drop across the excitation
impedance
Easy to look at the admittance and conductance to
calculate Rc and Xm. [Parallel equivalent of Gc and BM].
•
•
1
GC =
RC
YE = GC − jBM
YE =
BM
1
=
XM
1
1
−j
…….(i)
Rc
XM
Magnitude of YE referred to
primary side.
The P.F. for real
transformer is always
lagging. Therefore: I lags V
by θ
Finally , compare (i) and
(ii) and you can obtain the
values of RC and XM
directly.
YE =
I OC
VOC
POC
VOC I OC
P
θ = cos −1 ( OC )
VOC I OC
PF = cos θ =
YE =
I OC
∠ −θ
VOC
YE =
I OC
∠ − cos −1 PF
VOC
Y E = A - jB...... (ii)
39
40
10
Short Circuit Test
Short Circuit Test
•
•
•
•
In this test the secondary terminals of the transformer are
short circuited.
Apply a fairly low voltage to primary side, till current in
short circuited winding is equal to rated current.
Caution: Make sure to keep the primary voltage low,
otherwise you could burn the transformer’s winding.
•
In this test the input
voltage is so low.
Therefore, no current
flows though the
excitation branch.=>
All the voltage drop is
in the series elements
in the circuit.
The magnitude and
angle of the series
impedance referred to
primary side are given
by:
Z SE =
VSC
I SC
PSC
VSC I SC
Psc
I SCVSC
PF = cos θ =
θ = cos −1
Z SE =
VSC ∠0
V
= SC ∠ + θ
I SC ∠ − θ I SC
Z SE = Req + jX eq
Z SE = ( RP + a 2 Rs ) + j ( X P + a 2 X S )
41
Class Activity 2:
Determining Transformer Model’s Parameters
2.7: Transformer Voltage Regulation and Efficiency
Example 2-2: The equivalent circuit impedance of a 2-kVA,
8000/240 V. 60 HZ transformer are to be determined. The
open circuit test and short circuit test were performed on
the primary side of the transformer, and the following data
were taken. Find the impedance of the approximate
equivalent circuit referred to primary side, and sketch the
circuit.
Open-circuit test (on primary)
•
•
Due to series impedance in the transformer, the output
voltage of the transformer varies with the load even if
input voltage remains constant.
The voltage regulation can be calculated at any load.
However, we define full load voltage regulation as a
quantity which compares the output voltage of transformer
at no load with the output voltage at full load to
conveniently compare various transformers.
Short-circuit test (on primary)
Voc=8000 V
Vsc=489 V
Ioc=0.214 A
Isc=2.5 A
Poc= 400 W
42
VR =
Since at no load:
VS ,nl − VS , fl
VS , fl
× 100%
VP
VS = VP / a ⇒ VR =
Psc= 240 W
43
− VS , fl
a
× 100%
VS , fl
44
11
Transformer Voltage Regulation
VR with Lagging P.F.
• Low voltage regulation means less impedance of
the winding. So is mostly desirable.
• However, some times high impedances are used
to reduce the short circuit current.
•
Vp
a
Vp
= V s + Req I s + jX eq I s
a
> Vs
Figure 2-18(b) and 2-18(d)
The voltage regulation of a transformer with lagging loads
must be greater than zero.
45
VR with Unity P.F.
Vp
a
= V s + Req I s + jX eq I s
46
VR with Leading P.F.
Vp
a
Vp
> Vs
a
The voltage regulation of a transformer with unity loads must be
greater than zero. However, it is a smaller number compared to
lagging P.F. load.
47
= V s + Req I s + jX eq I s
 Vp

< Vs ⇒  − Vs  < 0
a
 a

Vp
If the secondary current is leading , the secondary voltage can
be actually higher than the referred primary voltage. Which
means a -ve voltage regulation.
48
12
Simplified Voltage Regulation
•
Transformer Efficiency
Considering a lagging p.f. because this is the practical case
mostly, we will drive an approximate equation for
calculating the primary voltage.
In general:
η=
Pout
Pout
× 100% =
× 100%
Pin
Pout + Plosses
From equivalent circuit we can calculate η easily.
1. Copper losses
-in series resistance, Rp and Rs
2. Hystersis losses
-accounted for by Rc
3. Eddy current
-accounted for by Rc
Therefore, at any given load efficiency can be calculated as:
Vp
a
Pout = Vs I s Cosθ s & η =
= Vs + Req I s cos θ + jX eq I s sin θ
Vs I s Cosθ s
× 100
Pcu + Pcore + Vs I s Cosθ
49
Class Activity 4: Transformer Taps
2.8: Transformer Taps
•
•
•
•
50
So far we considered turn ratio to be fixed. Practically, almost
all the distribution transformers have taps in the winding
which allows small changes in transformer’s turns ratio in the
field.
Typically 4 taps are designed, in addition to nominal setting
with spacing of ± 2.5 percent of full load voltage between
them.
Taps accommodate the variations in local voltages. These taps
normally can not be changed while the power system is
operating and power is being supplied to the load.
Tap Changing Transformer Under Load (TCUL): It is
transformer which has the ability to change taps while power
is connected to it.
51
Example 2-6: A 500 KVA, 13,200/480 V distribution
transformer has four 2.5% taps on the primary winding.
What are the voltage ratios of transformer at each tap
setting.
+ 5.0% taps
13,860/480
+2.5% taps
13,530/480
Nominal rating
13,200/480
-2.5% taps
12,870/480
-5.0% taps
12,540/480
52
13
Class Activity 3:
2.9: The Auto Transformer
Example 2-5: A 15 kVA, 2300/230-V transformer is to be
tested to determine its excitation branch components, its
series impedance and its voltage regulation. The following
test data have been taken from the primary side of the
transformer.
a) Eq ckt. Referred to HV.
b) Eq ckt. Referred to LV.
c) Full load VR at 0.8 pf
Lagging, 1.0, and 0.8
leading.
d) Perform the above using
approximate Eq.
e) Plot VR versus load from no
load to full load at 0.8 pf
Lagging, 1.0, and 0.8
leading.
f) Efficiency at full load and
0.8 pf.
Open-circuit
test (on
primary)
Short-circuit
test (on
primary)
Voc=2300 V
Vsc=47 V
Ioc=0.21 A
Isc=6.0 A
Poc= 50 W
Psc= 160 W
•
•
•
•
Some times it is desirable to change voltage levels by only
small amount. For example from 110 V to 120 V or 13.2 kv to
13.8 kv.
In this case it is wasteful and excessively expensive to wind a
transformer with two full windings each rated at about the
same voltage and a special purpose transformer called
Autotransformer is used.
The voltage at the output of the whole transformer is the sum
of the voltage on the first winding and voltage on the second
winding.
The winding across which both the primary and secondary
voltages appear is called Common Winding and the smaller
winding which is connected in series with the common winding
and across which only one voltage (primary/secondary)
appears is called Series Winding.
53
The Auto Transformer
54
A Step-down Auto Transformer
55
56
14
The Auto Transformer
Voltage Relationship
between two sides of
the transformer
VH = VC + VSE
Application of Auto Transformers
Current Relationship
between two sides of
the transformer
I L = I C + I SE
VH = VC +
N SE
VC
NC
IL =
N SE
I SE + I SE
NC
V H = VL +
N SE
VL
NC
IL =
N SE
IH + IH
NC
VL
N + N SE
= C
VH
N SE
•
Autotransformer is used when the turn ratio is nearly equal to
1 and where there is no objection to electrical connection
between the primary and secondary winding. Hence such
transformers are used:
1) To give small boost to a distribution cable to correct the
voltage drop.
2) As autotransformer to give 50 to 60 % of full voltage to
an induction motor during starting.
3) As furnace transformers for giving convenient supply to
suit the furnace winding from a 230 V supply.
IL
N + N SE
= C
IH
N SE
57
58
Issues to Consider while using the Auto Transformer
Three Phase Transformers
Since there is only one winding both for the primary and the
secondary side. Therefore, the winding has to be designed
based on the higher voltage and current ratings.
The autotransformer can’t be used for the applications where
the electrical isolation is desired because there is a physical
connection between the primary and the secondary circuits,
so the electrical isolation of the two windings is lost.
The common practice is to use autotransformer whenever the
two voltages are fairly close to each other.
The autotransformer is also used as variable transformer,
where the low voltage taps move up and down the winding.
The major power generation, transmission, and distribution
systems are three phase systems. Therefore, a it is necessary
to understand the design and operation of a three phase
transformer. A three phase transformer can be constructed in
one of the two ways:
i. Simply take three single phase transformers and connect
them in three phase bank.
ii. Make a three phase transformer consisting of three sets of
windings wrapped around a common core.
A single three phase transformer is the preferred practice
today because it is lighter, smaller, cheaper, and slightly
more efficient.
59
60
15
Three Phase Transformer Bank
Three phase transformers
bank with independent
transformers.
Three Phase Transformer Connections
Three phase transformers
bank on a single three
legged core.
The Primary and secondary winding of a three phase
transformer can be independently connected in either wye (Y)
or delta (∆). This gives a total of four possible connections for
a three phase transformer.
1. Y-Y
2. Y-∆
3. ∆-Y
4. ∆-∆
Any single transformer in the bank behaves exactly like a
stand alone single phase transformer. Thus the impedance,
voltage regulation, efficiency, and similar calculations for
three phase transformer can be done on per phase basis
using the technique established for single phase transformer.
61
Three Phase Transformer Y Connections
62
Three Phase Transformer ∆ Connections
63
64
16
Three Phase Transformer Connections
1) Y-Y
VLP = 3VφP
VφP
=a
VφS
2) Y-∆
∆
VLP = 3VφP
VφP
=a
VφS
3) ∆-Y
VLP = VφP
VφP
=a
VφS
4) ∆-∆
∆
VLP = VφP
VφP
=a
VφS
Instrument Transformers: CT & VT/PT
Two special purpose transformers used in the power
system are Potential Transformer (PT) and Current
Transformer (CT).
I. A potential transformer is specially wound transformer with
the high voltage primary and low secondary winding. It has
very low power rating because its sole purpose is to provide
a sample of power system voltage to the instruments
monitoring it. It must be very accurate to get the true
value of primary voltage because based on that protection
and control decisions are taken.
II. A Current Transformer samples the current in the line and
reduces it to a safe and measurable level. The current
transformer consists of a secondary winding wrapped
around a ferromagnetic ring, with the single primary line
running through the center of the coil.
VLS = 3VφS
VLP
=a
VLS
VLS = VφS
VLP
= 3a
VLS
VLS = 3VφS
VLP
a
=
VLS
3
VLS = VφS
VLP
=a
VLS
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Instrument Transformers: CT & VT/PT
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Characteristics of a CT
•
I.
Current Transformer (CT)
II. Voltage /Potential
Transformer (VT/PT)
•
•
•
•
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Thus the winding in a CT is loosely coupled and mutual
flux is smaller than the leakage flux.
The conventional voltage and current equation does not
apply for a CT. However, the secondary current is
proportional to the much larger primary current and the
device can provide the sample of the line’s current for
measurement, monitoring, protection, and control
purposes.
Typical CT ratio might be 600:5, 800:5, 1000:5. A 5
amps rating is standard on the secondary of the
transformer.
It is important to keep the CT short-circuited all the
time. Because
Extremely high voltage can appear across the open
terminals, mostly there is an interlock on the secondary
side.
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Questions to Ponder
Why the discovery of the transformer accelerated the
development of the use of electricity ?
Why the iron or magnetization losses are more important
than the losses caused by the winding resistance ?
Why the transformer has to be cooled ? How it is done ?
What is the connection of the transformer that supplies your
house and where is it ?
What is an Isolation transformer?
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