RGPV BE 104 Unit II

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Prof M D Dutt HOD EX Department SRCT Bhopal
READING MATERIAL FOR B.E STUDENTS
OF RGPV AFFILIATED ENGINEERING COLLEGES
SUBJECT BASIC ELECTRICAL AND ELECTRONICS
Professor MD Dutt
Addl General Manager (Retd)
BHARAT HEAVY ELECTRICALS LIMITED
Professor(Ex) of EX Department
Bansal Institute of Science and Technology
KOKTA ANANAD NAGAR BHOPAL
Presently Head of The Department ( EX)
Shri Ram College Of Technology
Thuakheda BHOPAL
Sub Code BE 104 Subject Basic Electrical & Electronics
UNIT II Magnetic circuits and Transformers

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RGPV Syllabus
BE 104 BASIC ELECTRICAL & ELECTRONICS ENGINEERING
UNIT II
MAGNETIC CIRCUITS AND TRANSFORMER
Review of laws of magnetism flux and their relation. Analysis of magnetic circuit and single phase transformer, Basic concepts and construction feature of transformer. Voltage , current and impedance transformation,EMF equation, equivalent circuits and phasor diagrams, Voltage regulation, Losses and efficiency, Open circuit test, Short circuit test.
INDEX
S No
Topic
Page
1
Review of laws of magnetism flux and their relation
3,4,5
2
Analysis of magnetic circuit and single phase transformer
6,7,8
3
Basic concepts and construction feature of transformer
9,10,11
4
Voltage , current and impedance transformation EMF equation
11,12,13
5
Equivalent circuits and phasor diagrams
13, to 17
6
Voltage regulation
18,
7
Losses and efficiency
19,20,21
8
Open circuit test
22,23
9
Short circuit test.
23,24

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REVIEW OF LAWS OF ELECTROMAGNETISM M.M.F FLUX AND THEIR RELATION
The space around the poles of a magnet is called magnetic field. The force in the space around a magnet can be pictured by examining the pattern made by iron fillings. These chain of iron fillings to the assumption that the region (field) contains invisible lines of force. The total number of lines of force surrounding a magnet , is called the total flux.
The lines of flux of N and S pole attract each other
The lines of same pole that is N N gives the lines which repel each other
LINE OF INDUCTION
Lines of flux travels from N pole to S pole and continues to travel through the magnet and finally they reach the N pole again forming closed curves. The portion of the curves within the magnetic material are called LINES OF INDUCTION

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MAGNETIC FLUX
Magnetic flux is the total number of lines of force comprising the magnetic field . It is represented by ᶲ and is measured in Weber.
MAGNETIC FLUX DENSITY
Magnetic flux density is defined as the magnetic flux passing through per unit area of material through a plane right angles to the direction of flux. This is also known as magnetic induction It is represented by B.
B= ᶲ/a Magnetic flux density is scalar quantity.
RELATION BETWEEN MAGNETIC FIELD INTENSITY H AND INDUCTION DENSITY
The field intensity H is the because of the flux density B ‘s effect . Thus flux density can be assumed to be proportional to the field intensity in a magnetic field i.e free space
Β =μ0 H
B = Webers per square meter
H= in newton per weber ( amper turn/meter
μ0 = is the magnetic space constant
for free space the value of μ0 = 4π 10¯7 H/M

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MAGNETO MOTIVE FORCE
M.MF of the magnetic circuit is defined as the magnetic potential that derives or tends to derive flux around the magnetic circuit and is analogous to the e.m.f in an electric circuit. It is measured is ampere turns AT.
MAGNETIC FIELD INTENSITY
M.M F per unit length ( along the path of magnetic flux) is called the magnetic field intensity H is given by
H = M.M.F /Length AT/mtr
RELUCTANCE
It is the name given to that property of material which opposes the creation of flux in it, It is analogous to resistance of an electric circuit. It is measured in ampere turn/Wb
PERMEABILITY
It is the measure of receptiveness of material of having magnetic flux developed in it.
Every substance posses a certain power of conducting magnetic flux. For example iron is better conductor for magnetic flux than air. It is the ratio of flux density B and magnetic field strength H
μ = B/H
ANALYSIS OF MAGNETIC CIRCUITS SINGLE PHASE TRANSFORMER
Magnetic Circuits with Air Gap
Energy conversion devices which incorporates a moving element have air gaps in heir magnetic circuits. Airgaps are also provided in the magnetic circuits to avoid saturation. The length of air gap is Lg is equal to the distance between two magnetic surfaces. When the air gap length is very smaller than the adjacent core faces, the magnetic flux Φ is constrained essentially to reside in the core and air gap is continuous through magnetic circuits. Then configuration A can be analyzed as the magnetic circuit li and air gap permeability μo and length lg. Since the permeability of the air is constant, the air gap is linear part of the magnetic circuit and flux density in the air gap is proportional to the m.m.f is calculated separately for the air gap and iron portion and than added together to determine the total m.m.f

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COMPOSITE CIRCUIT Consider a circular ring made from different material of l1, l2 and l3 having cross sectional are a1, a2 and a3 and relative permeability μr1,μr2 and μr3 respectively with a cut of length lg known as air gap. The total reluctance as they are joined in series.
There fore
Total reluctance = l1 + l2 + l3
μ0μr1 a1 μ0μr2 a2 μ0μr3 a3
Total M M F =Φ Xs = Φ [ l1 + l2 + l3 ]
μ0μr1 a1 μ0μr2 a2 μ0μr3 a3
Total ampere turn required = H1l1 + H2 l2 + H3 l3 + Hglg
Sum of ampere turns required for individual parts of magnetic circuit

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PARALLEL MAGNETIC CIRCUITS
In series circuit all parts of the magnetic circuit carry same flux and total ampere turns required to create a given flux is the arithmetic sum of the ampere turns required for individual parts of circuit .
But if the various paths of the magnetic circuit are parallel as shown in figure, the ampere turns required for the combination is equal to the ampere turns required to create the given flux in one path.
In circuit ABCD and AFED are in parallel, so ampere turns required to create flux Φ, in path ABCD is equal to ampere turns required create flux Φ2 in path AFED and also equal to the ampere turns required for both the parts.
Hence total ampere turns required for the magnetic circuit= AT for path DA +AT for path ABCD = AT for path DA +AT for path AFED.
SELF INDUCED E.M.F
When current flowing through the coil is changed, the flux linking with its own winding changes and due to the change in linking flux with the coil, an E.M.F is induced. This known as self induced EMF.

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MUTUALLY INDUCED E.M.F
When ever current in coil A changes, the flux linkage with coil B changes and an EMF in induced. This induced EMF in coil B is known as mutually induced EMF.
SINGLE PHASE TRANSFORMER
It is a static machine.
i) Transfers electric energy from one electric circuit to another electric circuit.
ii) It does not change the frequency
iii) It works on the principle of electro magnetic induction
iv) It has electric circuits which are linked by a common magnetic circuits.
STEP UP TRANSFORMER
When the transformer raises the out put voltage compare to input voltage it is called the step up transformer.
STEP DOWN TRANSFORMER
When the transformer reduces the out put voltage compare to input voltage it is called the step down transformer.
BASIC CONCEPTS AND CONSTRUCTION FEATURES OF TRANSFORMER
It essentially consists of two separate windings placed over the laminated silicon steel core, The windings to which a.c supply is connected is called primary winding and the winding to which load is connected is called secondary winding.

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When a.c supply of voltage V1 is connected to the primary winding, an alternating flux is set up in the core. This alternating flux when links with secondary winding, an EMF is induced in it called mutually induced EMF. The direction of this EMF is opposite to the applied voltage.
The same alternating flux links with primary winding and produces self induced EMF E1, The EMF E1 also acts opposite to applied voltage V1 as per LENZ’s Law.
Although there is no connection between primary and secondary winding but the electrical power is transferred from one circuit to another circuit through mutual flux .
The induced EMF in the primary and secondary winding depends upon the rate of change of flux linkage
That is N dΦ/dt
dΦ/dt is Same for primary and secondary windings. The induced EMF in primary winding E1 is proportional to N1 ( Number of turns in primary windings) and secondary winding E2 α N2.
TURN RATIO
The ratio of number of turns in primary winding N1 and secondary winding N2 is called turn ratio. = N1/N2
TRANSFORMATION RATIO
The ratio of secondary voltage E2 to the primary voltage E1 is called transformation ratio. It is represented by K.

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K= E2/E1 = N2/N1
CONSTRUCTION FEATURES OF TRANSFORMER
The transformer consists of following parts
i) Magnetic Circuit core of transformer
ii) Electric circuit Primary and secondary windings
iii) Insulation of windings
iv) Tanks, cooling methods, conservator, bushings and protective relay
TYPES OF TRANSFORMER AS PER CONSTRUCTION
The type of transformer according to the core construction and the manner in which primary and secondary windings are placed around it, there are two type of transformers.
a) CORE TYPE TRANSFORMER
b) SHELL TYPE TRANSFORMER
CORE TYPE TRANSFORMER
In a core type transformer the magnetic core is built up of lamination to form a rectangular frame. The laminations are cut in L shape. In order to avoid high reluctance at the joints where laminations are butted against each other, the alternate layers are stacked differently to eliminate continuous joints. While placing primary winding an insulation layer ( Bakelite former) is provided between core an winding.

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SHELL TYPE TRANSFORMER
The laminations are cut and formed E’s and I’s . In order to avoid high reluctance at the joints where laminations are butted against each other, the alternate layers are stacked differently to eliminate continuous joints.
In shell type transformer there are three limbs, The central limb carries the whole flux while the side limbs carries half flux. The width of central limb is double than the outer limb. Windings both primary and secondary are placed on the central limb.
VOLTAGE, CURRENT AND IMPEDANCE TRANSFORMATION
IDEAL TRANSFORMER
An Ideal transformer is one which there is no ohmic resistance and no magnetic leakage flux, i.e all flux produced in the core links with primary as well as secondary . Hence transformer has no copper losses and core losses. It means an ideal transformer consists of two purely inductive coils wound on loss free core. In actual practice it is not possible In an ideal transformer there is no power loss.
E2I2cosΦ = E1I1 cosΦ
E2I2c = E1I1
E2 /E1 =I1/ I2
E2 α N2, E1 αN1 , E2 ≡ V2 , E1≡ V1
E2 /E1 =I1/ I2 = N2/ N1 = V2/ V1
The currents are inversely proportional to the transformation ratio.
E.M.F EQUATION

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When a sinusoidal Voltage is applied to the primary winding of a transformer a sinusoidal flux is set up in the iron core which links with primary and secondary windings.
Let
Φm = Maximum value of flux in weber
f = supply frequency in Hz
N1 = No of turns in primary winding
N2 = No of turns in secondary winding
Flux change from Φm to – Φmin half cycle i.e 1/2f seconds
Average rate of change of flux = Φm – (– Φmin)
1
2f
= (Φm + Φm) 2f webers
= 4 Φmf
Now the rate of change of flux per turn is the average induced emf per turn in volts
Therefore Average e.m.f induced / turn = 4 Φmf volts
Form factor for sinusoidal wave =1.11 = R.M.S value/Average value
Considering form factor = 1.11
R.M.S value of induced emf /turn = 1.11X4 Φmf = 4.44 Φmf
E.M.F induced in primary winding E1 depends on number of turn of primary winding N1
So
E1 = 4.44 ΦmfN1

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Similarly for secondary winding
E2 = 4.44 Φmf N2
In the form of flux density Bm
Φm = BmAi Ai is the iron core area
E1 = 4.44 Bm Aif N1 Volts
E2 = 4.44Bm A2 f N2 volts
TRANSFORMER ON NOLOAD
On no load a small current I0 is drawn by the primary winding when the secondary winding is open. This current is called exciting current, magnetizing current. This current has to supply the iron losses in the core and a very small amount of copper loss.
The no load current I0 has two components
1) Iw is in the phase with voltage V1 called active voltage or working component it supplies iron loss and small component of copper loss.
2) Im is in the quadrature with the applied voltage V1 , called reactive or magnetizing component
Working component = Iw = I0cosΦᵒ
Im = I0sinΦᵒ
No load current I0 =√Iw² + Im²
No load power P0= V1 I0cosΦᵒ

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TRANSFORMER ON LOAD
When the secondary circuit of a transformer is completed through an impedance or load, the secondary current I2 starts flowing through secondary winding. The magnitude of I2 and voltage V2 will depend upon the load characteristics. I2 will be in phase with V2 or it can lag or lead the V2 . The Secondary current I2 sets up its own m.m.f and hence create secondary flux Φ2 , which opposes main flux Φ momentarily gap between V1 and back emf E1 increases. The I1’ primary current increase to maintain the value of Φ and the gap between V1 and E1 reduces , until the original value of flux Φ is achieved . The current The I1’ is in phase opposition to I2 and is called COUNTER BALANCING CURRENT. The phasor representation is given below.
EQUIVALENT CIRCUITS AND PHASOR DIAGRAMS
Equivalent resistance and reactance:- The two independent circuits of a transformer can be resolved into an equivalent circuit to make the calculation simple.
Let the resistance and reactance of primary and secondary windings of transformer be R1 ,R2 , X1 and X 2 ohms and transformation ratio K

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Resistance drop in primary winding I1 R1
Reactive drop in primary winding I1 X1
Resistance drop in secondary winding I2 R2
Reactive drop in secondary winding I2 X2
REFERRED TO SENDARY SIDE Since K is the transformation ratio, resistive and reactive drop referred to secondary side shall be K times , i.e KI1 R1 and K I1 X1
If I1 Is substituted as equal to KI2 than we get K² I2 R1 K²I2 X1
Total resistive drop = K² I2 R1 + I2 R1 = I2 (K² R1 + R1) = I2 R02
Total reactive drop = K²I2 X1 + I2 X2 = I2(K² X1 + X2) = I2 X02
From Phasor diagram
KV1 =√( V2 + I2 R02 cosΦ + I2 X02 sinΦ)² + (I2 X02 cosΦ -- I2 R02 sinΦ)²
I2 secondary current which lags V2 by angle Φ
Since the term (I2 X02 cosΦ -- I2 R02 sinΦ)² is very small to compare to (I2 R02 cosΦ + I2 X02 sinΦ)
Neglecting small value we get
KV1= V2 + I2 R02 cosΦ + I2 X02 sinΦ
V2 = KV1- I2 R02 cosΦ - I2 X02 sinΦ
If the load is pure resistive Φ=0

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V2 = KV1- I2 R02
If the load is capacitive than Φ is negative
V2 = KV1- I2 R02 cosΦ + I2 X02 sinΦ
REFERRED TO PRIMARY SIDE
Resistance drop in secondary winding I2 R2 EVIDED /K
Reactive drop in secondary winding I2 X2/K
REFERRED TO PRIMARY SIDE Since K is the transformation ratio, resistive and reactive drop referred to PRIMARY side shall be divided by K , i.e I2 R2 /K and I1 X1 /K
If I2 Is substituted as equal to I1/K than we get I1 R2 / K² K²I1 X2/ K²
Total resistive drop = I1 R2/K² + I1 R1 = I1 (R1 + R1 / K²) = I1 R01
Total reactive drop = I1 X1 + I1 X2 / K² = I2(X1 + X2//K²) = I1 X01
From Phasor diagram
V1 =√( V2 /K+ I1 R01 cosΦ + I1 X01 sinΦ)² + (I1 X01 cosΦ -- I1 R01 sinΦ)²
I2 secondary current which lags V2 by angle Φ
Since the term (I1 X01 cosΦ -- I1 R01 sinΦ)² is very small, neglecting small value
we get
V1= V2/K+ I1 R01 cosΦ + I1 X01 sinΦ

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EQUIVALENT CIRCUITS OF TRANSFORMER
The equivalent circuit of any device can be quite helpful in predetermination of the behavior of the device under various condition of operation.
Equivalent circuit of a transformer having transformation ratio K
E2/E1=K
E1 is the induced emf due to V1 less primary voltage drop. This voltage causes iron loss current Ie and magnetizing current Im, these two components are represented by R0 and X0 as pure resistance and pure reactance X0
Secondary current I2 =I1’/K
Equivalent Diagram of a transformer with all secondary impedance transferred to primary side.

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Approximate equivalent diagram of a transformer.
PHASOR DIAGRAM OF TRANSFORMER WITH RESISTIVE,REACTIVE AND CAPACITIVE LOAD
VOLTAGE REGULATION
The way in which the secondary terminal voltage with load depends upon load current, the internal impedance and the load power factor. The change in secondary voltage from no load to load with primary voltage and frequency held constant, this termed as the inherent regulation.

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The voltage characteristics depends on the voltage regulation of transformer. It is expressed in percentage of rated terminal voltage (secondary). It is defined as the change in magnitude of terminal voltage (secondary) when the load is thrown off
( reduced to no load) with primary voltage and frequency constant.
If V2 is the secondary voltage at full load power factor and E2 is the secondary voltage at no load
Voltage regulation = (E2-V2)/E2
As per IS the secondary rated voltage of transformer is equal to secondary voltage at no load.
% Voltage regulation = {(E2-V2)/E2}100
= Voltage drop / No load secondary rated voltage
The voltage drop = I2 R02 cosΦ + I2 X02 sinΦ
So % regulation = (I2 R02 cosΦ + I2 X02 sinΦ)100
E2
When the power factor is leading the percentage regulation becomes
%Regulation = (I2 R02 cosΦ - I2 X02 sinΦ)100
E2
So we can write
%Regulation = (I2 R02 cosΦ ± I2 X02 sinΦ)100
E2
+ve sign for lagging P.F
-ve sign for leading P.F
Voltage regulation on an average in transformer is 4%. From consumer point of view, voltage regulation due to variation in load is not desirable.

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DETERMINATION OF REGULATION FROM OPENCIRCUIT AND SHORT CIRCUIT TEST
As we know
%Regulation = (I2 R02 cosΦ - I2 X02 sinΦ)100
E2
Equivalent R02 and X02 can be easily determined from short circuit test , I2 is the secondary current and cosΦ is the power factor either leading or lagging. No load secondary terminal voltage is equal to emf induced E2.
Open circuit data are not required for determination of regulation
LOSSES AND EFFICIENCY
There are two type of losses in transformer.
1) Iron loss or core loss , fixed loss
2) Copper loss or ohmic loss
Iron losses:-Iron losses are due to alternating flux in the core and consists of hysteresis and eddy current losses.
a) Hysteresis Losses:- The core of a transformer is subjected to an alternating magnetizing force and for each cycle of e.m.f. A hysteresis loop is traced out. The hysteresis loss is given by
Ph = ń(ßmax) ˣ f v joules per second or watts
ń = hysteresis coefficient
ßmax = flux density maximum
f = frequency, The value x varies from 1.5 to 2.5 and it depends upon the material.
b) Eddy current losses:- We have seen whenever flux linkage with closed electric circuit changes an e.m.f induced in the circuit and current flows, If the magnetic circuit is made up of iron and if the flux in the circuit is , current will be induced by induction in iron circuit itself. All such currents are called eddy currents.
Pe = Ke (ßmax)² t² f² v watts

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COPPER LOSS:- These losses occur due to the ohmic resistance of the transformer winding. If I1 and I2 are the primary and secondary currents, R1 and R2 are primary and secondary winding resistance. =
In that case the copper losses are = I1² R1 + I2 ² R2
These losses vary in the square of load current, If the copper loss at full load is Pc than copper loss at ½ load and 1/3 load shall be
(½)² Pc = Pc/4,
(1/3)² = Pc/9
Copper losses can be determined by short circuit test.
TRANSFORMER EFFICIENCY :- The efficiency of transformer is defined as the ratio of out put power to the input power. Both are to be measured in same unit either in watts or Kw
Therefore ƞ = output output
Input input + losses
ƞ = output power
output power +losses
= V2 I2 Cosϕ2
V2 I2 Cosϕ2 + Pi + Pc
V2 Secondary voltage
I2 Secondary current
ϕ2 Power factor of load
Pi = Iron loss = eddy current loss + hysteresis loss
Pc = Copper loss = I2² Res
If x is the fraction of load than the efficiency at x
Ƞx = Xoutput
Xoutput + Pi + x² Pc
CONDITION FOR MAXIMUM EFFICIENCY :- The terminal voltage V2
is approximately constant, Thus for a given P.F efficiency depends on load current I2

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ƞ = V2 I2 Cosϕ2
V2 I2 Cosϕ2 + Pi + Pc
In the above expression numerator is constant and efficiency will be maximum when denominator is minimum . Thus the maximum condition is obtained by differentiating the quantity in the denominator w.r.t I2
d (V2 I2 Cosϕ2 + Pi/ I2 + I2 Res) = 0
d I2
0 = - Pi/ I2² + Res
Pi = I2²Res
Iron loss = copper loss
I2 =√ Pi/Res
If x is the fraction of full load KVA at which the efficiency of transformer is Maximum
Then copper loss = x²Pc
Iron loss = Pi
x²Pc = Pi
x = √Pi/Pc
therefore output KVA corresponding Maximum efficiency
= full load kva √iron loss/copper loss at FL

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ALL DAY EFFICIENCY
The load on certain transformer fluctuates throughout the day , The distribution transformers are energized for 24 hours, but they deliver very light loads for the major portion of the day. Thus iron loss occur for the whole day but copper loss occurs only when transformers are loaded. Hence transformer cannot be judged for the commercial efficiency.
All day efficiency is the ratio of Kwh output to Kwh input for 24 hours
Ƞ allday = output in kwh ( 24hrs)
Input in kwh
The find out all day efficiency , we must know the load cycle of the transformer.
OPEN CIRCUIT TEST
PURPOSE OF THIS TEST IS TO DETERMINE
•Core loss or Iron loss Or Magnetic loss (Pi)
•No load current (Iо)
•Shunt branch parameters R о and X о
One of the winding is kept open.
Rated voltage at rated frequency is applied to other(LV) winding.
A voltmeter, wattmeter, and an ammeter are connected in LV side of the transformer.
Ammeter > Reads No-Load Current, I о
Voltmeter > Reads Applied Voltage, V о
Wattmeter> Reads No-Load Input Power, W о or P о

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DETERMINATION OF EQUIVALENT CIRCUIT CONSTANTS THROUGH NO- LOAD TEST
No load power factor, CosΦо= W о / V о I о
Core loss component, Iw = I о CosΦ о
Magnetising component, Im = I о SinΦ о
Core Loss, Pi = No load power (W о)
Core loss resistance, R о = V о / Iw = V о / I о CosΦ о
Magnetising reactance, X о = V о / Im = V о / I о SinΦ о

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SHORT CIRCUIT / IMPEDANCE TEST ON TRANSFORMER
PURPOSE OF THIS TEST IS TO DETERMINE
•Zо₁ or Z о ₂ – Total impedance referred to either primary or secondary side
•R о ₁ or R о ₂ - Total resistance referred to either primary or secondary side
•X о ₁ or X ₂ - Total reactance referred to either primary or secondary side
•Full load cu loss I ₂ ² R о ₂
*In this test one of the winding is short circuited by thick conductor.
* Current rating of HV side is low compared with LV side.
* Power input gives total cu loss at rated load.
* Unity power factor wattmeter is used for measuring power in SC test.

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DETERMINATION OF EQUIVALENT CIRCUIT CONSTANTS THROUGH LOAD TEST
SC power factor, CosΦsc = Wsc / Vsc Isc
Resistance of transformer referred to primary side ,
Rо ₁ = Wsc / (Isc)2
Reactance of transformer referred to primary side ,
X о ₁ = Z о ₁ SinΦsc = 퐙 о ₁ ퟐ−퐑 о ₁ ퟐ
Impedance of transformer referred to primary side,
Z о ₁ = Z о ₁ Cos Φsc= Vsc / Isc

RGPV BE 104 Unit II

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RGPV BE 104 Unit II