Showing posts with label Transformer. Show all posts
Showing posts with label Transformer. Show all posts

Parallel Operation of Transformers

The transformers are connected in parallel when the load on them is more than the rating of the individual transformers. Before going to discuss about parallel operation of single phase transformers  A small question for you all that is 'why parallel operation of transformers is required?'.Answer is simple we have some advantages over operating single transformer units so we go for parallel operation of transformers.
Parallel Operation of Transformers

Advantages of Parallel Operation of Transformers

Parallel operation is nothing but connecting primary windings of two or more transformers to supply and secondary windings to common loads.To supply more than  the rating of existing transformer parallel connections will be employed.It is very economical to operate small no.of transformer units in parallel to supply rated output than a big rated transformer.

Parallel operation of transformer because of following advantages 

  1. Maximum power system efficiency: Power system loads always varies to maintain its efficiency we operate transformers in parallel because transformer gives the maximum efficiency at full load,if we use single large transformer,load on the large transformer always varies so it will operate with low efficiency. In other hand if we operate small units in parallel we can switch on/off as per load demand and we can maintain high efficiency.
  2. Ease of electrical power switching: Scheduling the power with single large transformer will become difficult because of it's high rating and distribution system will be connected ,controlled by a single transformer we can avoid this problem using parallel operation of transformers.This also improves power system reliability,flexibility.
  3. Economical Issues /Non-availability of large transformer: If a large transformer of required rating is unavailable we can go for small rated transformers which can perform better than single large unit. And large transformer operating costs will be more than parallel operation of single phase transformers.
  4. Easier transportation: Transportation will be very easy for small transformers: If installation location is far away from transformer manufacturing/selling point.So parallel operation of transformers is to be encouraged. 

Conditions for Parallel Operation of Transformers

We can't connect any two or more transformers in parallel blindly.When two or more transformers need to operate in parallel, they must met some conditions for efficient performance.Major conditions for parallel operation of transformers are listed below.
  • The voltage ratings and voltage ratios of the transformers should be the same.
  • Transformers should be properly connected with regard to their polarities.
  • The per unit or percentage impedance of the transformers should be equal.
  • The reactance/resistance ratios of the transformers should be the same.

Same Voltage Rating and Voltage Ratio

The voltage ratios of two transformers must be same but why?
Reason is simple if voltage ratio of two transformers is different  and they put  in parallel with same primary supply voltage, there will be a difference in secondary voltages. As secondaries of transformers completes a closed loop there will be circulating  currents.In this case, considerable amount of current is drawn by the transformers even without load.As the internal impedance of transformer is small, a small voltage difference may cause sufficiently high circulating  current  causing unnecessary extra I²R loss.
Parallel Operation of Transformers
Circulating Currents=IC= (EA-EB)/ZA+ZB

Connections with regard to Polarity

Polarity of all connected transformers must be same in order to avoid short circuit.Polarity of a transformer taken w.r.t dot notation.Dots of all transformers connected together on primary and secondary separately. If polarity is opposite to each other, huge circulating current flows.In the below diagram right and wrong parallel operation connections are shown take a look.
Parallel Operation of Transformers

The phase sequence must be identical of all parallel transformers.

This condition is can be applied only to poly-phase transformers. If the phase sequences are not same, then transformers can not be connected in parallel.During the cycle, each pair of phases will be short circuited.

Equal Percentage Impedance

The per unit (pu) impedance of each transformer on its own base must be same.This condition is also desirable for proper parallel operation of single phase  transformers. If this condition is not met, the transformers will not share the load according to their kVA ratings.

Sometimes this condition is not fulfilled by the design of the transformers. In that case, it can be corrected by inserting proper amount of resistance or reactance or both in series with either primary or secondary circuits of the transformers where the impedance is below the value required to fulfill this condition.That is why per unit impedance of the connected transformers must be same.

Incoming Queries:
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Open and Short Circuit Test on Transformer

We conduct open circuit and short circuit test on single phase transformer to determine the  efficiency and regulation of a transformer on any load condition and at any power factor.Open circuit & short circuit test are also called as OC test & SC test on transformer.This method of finding the parameters of a transformer is called as an indirect loading method.

Open Circuit(OC) and Short Circuit(SC) Test on Transformer

What is the need of OC & SC test on transformers?

Open circuit test and short circuit test on transformer are very economical and convenient because they are performed without actually loading of the transformer.because they furnish the required information without actually loading the transformer. In fact, the testing of very large a.c. machinery consists of running two tests similar to the open and short-circuit tests of a transformer.
Open-circuit or No-load Test On Single Phase Transformers

The circuit diagram for open circuit test is shown in the figure. A voltmeter(V), wattmeter(W), and an ammeter(A) are connected in LV side of the transformer.Usually high voltage (HV) winding is kept open and the low voltage (LV) winding is connected to its normal supply.Because we are going to find the max output voltage of transformer which is present at HV side.This is the reason why HV winding is open circuited.

Procedure of Open Circuit(OC) Test

  1. Switch on single phase AC supply,with the help of variac,increase the voltage until the voltmeter raeds the rated voltage of the LV side.
  2. At the instant of  rated LV voltage, Note down the readings of all three instruments.(Voltmeter, Ammeter and Wattmeter readings) are recorded.

The ammeter reading gives the no load current I0.As no load current (I0) very small,voltage drop due to this current also neglected.The input power supplied to the transformer is indicated by the watt meter (W).Output power of transformer in open circuit test is zero because  other side of transformer is open circuited.That means input supply is to just compensate the  core losses and copper losses.By neglecting some voltage drop due to small no-load current,Watt meter (W) reading gives the core losses of the transformer.

 Here,Wo = Pi  = Iron losses
       Calculations : We know that,
             Wo = VoIo cos Φ(watt meter reading)
             cos Φo = Wo /(VoIo) = no load power factor
Once cos Φo is known we can obtain,
              Ic  = Io cos Φo 
 and        Im = Io sin Φo
       Once Ic  and Im are known we can determine exciting circuit parameters as,
              Ro = Vo /Ic   Ω 
          Xo = Vo /Im   Ω

where, X0,R0 are equivalent exciting reactance,resistance of transformer.

Key Point : We should use LPF (low power factor)watt meter to get error free results.Beacuse cos Φo is very low in the above case.The above values are calculated by taking LV side of transformer as reference.If the meters are connected on secondary and primary is kept open then from O.C. test we get Ro' and Xo' with which we can obtain Ro and Xo knowing the transformation ratio K.
The equivalent circuit derived by the OC test is shown below.

Hence,we can conclude that open circuit test on transformer gives gives core losses of transformer and shunt parameters of the equivalent circuit.

Short Circuit Or Impedance Test On Transformer

The connection diagram for short circuit test on transformer is shown in the figure. A voltmeter(V), wattmeter(W), and an ammeter(A) are connected in HV side of the transformer as shown.The secondary winding is short circuited with the help of thick copper wire or solid link. As high voltage side is always low current side, it is convenient to connect high voltage side to supply and shorting the low voltage side.

Procedure of Short Circuit (OC) Test

  1. Apply voltage to HV side.
  2. Increase the voltage from the zero until the ammeter reading equals the rated current.
  3. At the instant of  rated HV current, Note down the readings of all three instruments.(Voltmeter, Ammeter and wattmeter readings) are recorded.

Now the current flowing through the windings of transformer is rated current. Hence the total copper loss will be full load copper loss.Iron losses are neglected due to small fraction of voltage supllied. Hence the wattmeter reading shows the power loss which is equal to full load copper losses as iron losses are neglected.

Wsc = (Pcu) F.L. = Full load copper loss
      Calculations : From S.C. test readings we can write,
              Wsc = Vsc Isc cos Φsc
...            cos Φsc = Vsc Isc /Wsc = short circuit power factor
               Wsc = Isc2 R1e = copper loss
...             R1e =Wsc /Isc2
while        Z1e =Vsc /Isc = √(R1e2 + X1e2)
                X1e = √(Z1e2 - R1e2)

Thus we get the equivalent circuit parameters R1e, X1e and Z1e. Knowing the transformation ratio K, the equivalent circuit parameters referred to secondary also can be obtained.

We conduct open circuit and short circuit test on single phase transformer to determine the  efficiency and regulation of a transformer on any load condition and at any power factor.Open circuit & short circuit test are also called as OC test & SC test on transformer.This method of finding the parameters of a transformer is called as an indirect loading method.

Calculation of Efficiency of transformer from O.C. and S.C. Tests

Efficiency, η = Power output in KW/ Power input in KW

= Power output in KW/ (Power output in KW + Losses)

= Power output in KW/ (Power output in KW + Copper loss + Core loss)

Consider that the KVA rating of the transformer is S, a fraction of the load is x and the power factor of the load is Cos Φ. Then
The output power in KW = xSCos Φ
Suppose the copper loss at full load is Pcu (since x =1),
Then copper loss at x per unit loading = x2Pcu
Therefore the efficiency of the transformer is
Efficiency, η = xSCos Φ / (xS Cos Φ + x2 Pxcu + Pxcore)
In the above efficiency equation, the core or iron losses and full load copper losses are found by OC and SC tests.

Calculation of Regulation

Percentage voltage regulation, %R = ((E2 – V2)/ V2 )×100
The expression of voltage regulation in terms voltage drops is given as
%R = ((I1R01 cos Φ +/- I1X01 sin Φ) / V1) ×100
%R = ((I2R02 cos Φ +/- I2X02 sin Φ) / V2) ×100
The above two equations are used based on the parameters are referred to primary or secondary sides. Hence, from the SC test data we can find out the regulation of a transformer. The positive sign is used for lagging power factor and negative sign is used for leading power factor.


Losses In Transformers ; Hysteresis Loss, Eddy Current Loss,Efficiency

Hysteresis Eddy Current(Iron) or Core Losses and Copper Loss in Transformer

We have discussed about electrical transformer construction,working.When it comes to efficiency of the transformer it is above 95%.In modern power transformers we can get above 98% of efficiency.Electrical transformer is a static device,so no rotating losses or frictional,windage losses.So no mechanical losses occur in transformer.We consider only electrical losses in transformer.Now read efficiency of transformer and losses in transformer below.

Losses in Transformer

As we no machine in this world is ideal.Transformer is also not an exception for this.There are two types of losses in transformers they are
(i) Core Losses Or Iron Losses
(ii) Copper Loss In Transformer 

(i) Core Losses Or Iron Losses in Transformer

Eddy current,Hysteresis losses are considered as core losses of transformer.Core losses of transformer almost constant for a transformer after it is built for certain and frequency.Because eddy current loss and hysteresis loss depends on the magnetic properties,volume of the core which is used for the construction.As volume is fixed we can say core losses or iron losses strictly depends only on frequency.

Hysteresis Loss in transformer

Hysteresis losses occurs due to reversal of magnetization in the transformer core.The magnetizing and demonstration curve of any material will not be same.Some loss happens due to cohesive force between magnetic atoms. 
The hysteresis loss in transformer depends on the volume and grade of the iron, frequency of magnetic reversals and value of flux density. It can be given by, Steinmetz formula:
Wh= ηBmax1.6fV (watts)
where,   η = Steinmetz hysteresis constant
             V = volume of the core in m3

How to reduce Hysteresis Loss in Transformer?

We can reduce hysteresis losses by using Cold Rolled Grain Oriented (CRGO) Silicon Steel.In earlier days we used silicon steel as transformer core after that CRGO steel found as less hysteresis constant.

Eddy current loss in transformer

Transformer works only on alternating current,when AC current is supplied to the primary winding of transformer,it sets up alternating magnetizing flux. When this flux cuts the secondary winding,emf will induced in it.This main flux also cuts the core of the transformer,As we know whenever flux cuts the magnetic material emf will induce in it.which will result in small circulating currents in them due to closed path of the core. This induced current is called as eddy current. Due to eddy currents, some energy will be dissipated in the form of heat.

 We= ηB²f²T (watts)

W∝ T

where,   η = Steinmetz hysteresis constant
             T = Thickness of lamination in m

How to reduce Eddy current loss in transformer?

By making the core into laminations.In the above relation you can find that eddy current losses are directly proportional with square of it's thickness.As the lamination thickness is much smaller than the depth of penetration of the field, the eddy current loss can be reduced by reducing the thickness of the lamination. Present day we use lamination thickness of 0.25mm operated at 2 Tesla.By decreasing lamination thickness we can  reduce the eddy current losses in the core.This loss also remains constant until the frequency of operation is constant.

(ii) Copper Loss In Transformer 

Copper loss in transformer also known as ohmic losses.This loss due to resistance of copper winding.We know that when current flows through a conductor definitely there will be I²R loss in the form of heat.

1.Primary copper losses in transformer takes place due to the flow of current in the primary winding of transformer.

2.Secondary copper losses takes place due to the flow of current in the secondary winding of transformer.
Total copper losses=I²R=(I1)²R1+(I2)²R2

where,I1,I2 and R1,R2 are current and resistances of primary ,secondary winding respectively.

The primary and secondary resistances differ from their d.c. values due to skin effect and the temperature rise of the winding.

Apart from these major losses we have dielectric loss in transformer.As it's name says dielectric losses takes place in the insulation coating of the transformer due to the large electric stress.It is constant and in the case of low rating transformers it is neglected.

Efficiency Of Transformer

Like all machines efficiency of a transformer can be defined as the output power divided by the input power.

Efficiency = output / input 

As a transformer being highly efficient[95-99%], output and input are having nearly same value,So it is impractical to measure the efficiency of transformer by using output / input.A practical method to find efficiency of a transformer is using, efficiency = (input - losses) / input = 1 - (losses / input).

Condition For Maximum Efficiency

Copper loss = (I1)²R1
Iron loss = Wi
Hence, efficiency of a transformer will be maximum when copper loss and iron losses are equal.That is Copper loss = Iron loss.

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Buchholz Relay Construction, Working In Transformer

What is a Buchholz relay?

Buchholz relay is one of the protective device for electrical transformer from internal faults.It works with all gas and oil immersed transformer.Buchholz relay in transformer is an oil contained arrangement. Buchholz relay connected in between pipe from main tank to conservator tank and it is filled with oil throughout transformer operation.It is used for all power transformers,but not economical in the case of low rated transformers.Mostly buchholz relay employed for gas and oil operated transformer(rated >500 kVA).

buchholz relay diagram

Why Buchholz relay is used in transformers?

The buchholz relay works on the principle of gas vaporization from insulating oil in the fault conditions.Buchholz relay protects power transformer from all internal faults such as breakdown of the insulating oil, insulation failure of winding,over heat generated in the coils due to failure of insulation etc.

Working,Construction of Buchholz Relay

Buchholz relay function is based on very simple mechanical phenomenon.It consists of two elements mounted in a metallic chamber located in the pipe connecting the conservator and the transformer tank. When minor fault occur, heat is produced due to current leakage, some of the oil in transformer tank evaporates and some of the vapor gets collected at the top of the chamber while passing it to the conservator.

When a predetermined amount of vapor accumulates at the top of the chamber, the oil level falls, the mercury switch attached to a float is tilted so it closes the alarm circuit and rings the bell. A release cock is provided at the top of the chamber, so that after operation, the pressure in the chamber can be released and the gas emitted allows the chamber to refill with oil.

When severe faults occur a large volume of gas is evaporated, so that the lower elements containing a mercury switch mounted on a hinged type flap is tilted and the trip coil is energized. A drain is provided at the bottom of the chamber to allow air to be pumped into the chamber for test purposes.

The Buchholz relay operation may be actuated without any fault in the transformer. For instance, when oil is added to a transformer, air may get in together with oil, accumulated under the relay cover and thus cause a false Buchholz relay operation.

Buchholz relay is employed for the following types faults, 

1. Core bolt insulation failure
2. Local-overheating
3. Entrance of air into the oil
4. Less of oil due to leakage
5. Short circuited lamination due to burns.

Merits & Demerits of Buchholz Relay

Advantages of  Buchholtz relay

1. The advantage of using a Buchholtz relay is that they indicate incipient faults. For example, It indicates the faults between turns or heating of core which enables us to take the transformer out of service before severe damage occurs.

2. Buchholtz relay gives an alarm when the oil level reduces below a certain level due to leakage of oil from transformer.

3. Buchholtz relay gives an audible warning which informs the operator that there is some fault in the transformer.
4.Type of insulation failure of the transformer can also be detected by testing the gas. 

Drawbacks of  Buchholtz relay

1. This type of relay can be used only for transformers with conservators.
2. The relay does not operate when the faults are above the oil level.
3. Buchholz relay does not protect the cables associated with the transformers below. A separate protection has to be employed for the cables.
4. Due to economic considerations, this relay is not employed for transformers above 500 kVA.
5. Setting of the mercury switch cannot be too sensitive, otherwise there can be a false operation due to vibrations,earthquakes. mechanical shocks to the pipe.
6. The operation of this new is unsatisfactory due to slow operating time. (Nearly 0.2 to 0.1 seconds)

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EMF Equation of Transformer & Voltage Transformation Ratio

 Transformer EMF Equation

In the previous post we are discussed about working principle of transformer.In this article the amount of emf induced in the secondary winding due to the current supplied to the primary winding can be calculated by EMF equation of the transformer.This is very essential in the designing process of any transformer.Because amount of emf induced due to mutual induction also depends on number of turns in the coil.Transformer emf equation gives the output voltage value,so that we can design a transformer as per our requirements.
EMF Equation of Transformer

Derivation : EMF Equation of Transformer


N1 = Number of turns in primary winding

N2 = Number of turns in secondary winding

Φm = Maximum flux in the core (in Webers) = (Bm x A)

f = frequency of the AC supply (in Hz).

Flux propagates in the sinusoidal wave form.If the time period of the flux wave is 'T' sec.Then it reaches it's maximum value Φm in T/4 sec i.e.,quarter of the cycle.We know that T=1/f (f is the frequency of the sine wave).

Average rate of change of flux = Φ/(T/4)    = Φm /(1/4f)
Average rate of change of flux = 4f Φm       ....... (Webers/s).
Induced emf per in the one turn = Rate of change of flux per in the one turn
So, average emf per turn = 4f Φm   ..........(Volts).
Since,Form factor = RMS value / Average value.
RMS value of emf per turn = Form factor X Average emf per turn.
Form factor of sine wave is 1.11
Therefore, RMS value of emf per turn =  1.11 x 4f Φm = 4.44f Φm Volts.
The above equation for one turn.If there are N number of turns in the coil then emf equation of the transformer becomes,

E=4.44f N Φm Volts

f = frequency of the AC supply (in Hz).
Φm = Maximum flux in the core (in Webers) = (Bm x A)
N = Number of turns in winding.[it can be primary or secondary]

Induced EMF Equation of Transformer In Primary,Secondary winding

RMS value of induced emf in the primary winding (Say E1) = RMS value of emf per turn X Number of turns in the primary winding

          E1 = 4.44f N1 Φm          ............................. (1)

Same as above, RMS induced emf in the secondary winding (Say E2) can be given as

          E2 = 4.44f N2 Φm.          ............................ (2)

By dividing eqs (1) & (2).

The above relation between primary and secondary induced voltages  is called the emf equation of transformer.We can observe emf / number of turns is same for both primary and secondary winding of a transformer.

Voltage Transformation Ratio (K) of Transformer 

From equation (3) E1/N1=E2/N2=K
Where, K = constant
This constant K is known as voltage transformation ratio.
  1. If N2 > N1, i.e. K > 1, then the transformer is called step-up transformer.
  2. If N2 < N1, i.e. K < 1, then the transformer is called step-down transformer
  3. Based on the requirement we can use any transformer as step-up or step down by changing the supply.

Electrical Transformer Construction, Working,EMF Equation

In this post we are going to learn about electrical transformer.we can say transformer as static induction motor.You know why we are using AC power instead of DC power in our power system,Because of so many advantages of AC power over DC power.One of main advantages of AC supply is transformation [changing one level to another level ].This transformation is possible by a static electrical machine called transformer.

Definition of Transformer: 

Transformer is an electrical device which transfers power from one circuit to another circuit without electrical coupling at constant frequency.In the transformer the total power supply is constant with negligible losses so,it is also called as constant power device.Transformer is also called as constant frequency device.

To maintain constant power at input and output side voltage and current values will changed in inverse proportional manner.i.e.,For example at input side you have 200 V,10 A.If you want 400 V at output side then current must be 5 A.So,overall power in the transformer will become constant.This is the answer you can tell who asks you "what is transformer ?".

Construction of Transformer

Transformer consists of two coils,based on power supply primary winding and secondary winding.Based on actual construction HV(high voltage)  winding and LV(low voltage) winding.This two windings are placed on a magnetic core made up of CRGO steel(latest transformers) or Silicon steel (old transformers).CRGO steel is used to reduce iron losses.Magnetic core is laminated to reduce eddy current losses.In the next post we will post a detailed information on losses in transformer. 

Constructional Parts of Transformer

There are three major parts in the transformer.They are,
1.Primary Winding of transformer :- This is connected to power sulppy.It produces magnetic flux.
2.Magnetic Core of transformer :- This acts as a pipe line to carry the flux through it and delivers to the secondary winding.
3.Secondary Winding of transformer :- the flux, produced by primary winding, passes through the core, and cuts the secondary winding.So emf will induce in the secondary winding of transformer.

Working Principle of Transformer

The working principle of transformer is based on Faraday's laws of electromagnetic induction.According to Faraday's law,whenever there is a relative space (or) time variation between magnetic field and set of conductors an emf will be induced in the conductors,vice-versa.

Another main principle of transformer working is mutual induction. According to mutual induction,current in the one coil is responsible of emf induced in another coil.

Let us understand in more easy way,assume there are two coils.Say first and second.Now we are supplying voltage to first coil,then according to Faraday's law an alternating flux surrounding that coil.If suppose any coil present in the range of alternating flux,then this alternating flux cuts the coil.Hence emf induce in the second coil.This is the fundamental working of any transformer.The flux linkage between two coils is always constant.So transformer is also called as constant flux device.

Based on number of turns in the coil emf will induce in the second coil.The transformer which is discussed above will not work practically.Because in practical applications we have to consider efficiency of transformer,losses,economical aspects.In the above figure you can find that so much flux wasted and not linked with the second coil.This is called leakage flux in the transformer.So to reduce leakage flux we use a magnetic core which gives low reluctance path for flux.We can find reduced leakage flux after placing a magnetic core.

Mathematics of Transformer

  VP  -  is the primary voltage
  VS  -  is the secondary voltage
  NP  -  is the number turns of primary winding
  NS  -  is the number of secondary winding
  Φ (phi)  -  is the flux linkage

We heard, transformer can increase or decrease the voltage but up to what level it can increase or decrease ?.This depends on turns ratio (or) transformation ratio(denoted with n or k) of transformer.This can be defined as the ratio of number turns of primary winding to number of secondary winding (or) ratio of primary voltage to secondary voltage.
Based on the turns ratio we can classify step-up transformer or step-down transformer.n<1 for step-up,n>1 for step down transformer.

Transformer EMF Equation

  ƒ  -  is the flux frequency in Hertz,  = ω/2π
  Ν  -  is the number of coil windings.
  Φ  -  is the flux density in webers.

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Sumpner's Test Or Back-To-Back Test On Transformer

Sumpner's Test Or Back-To-Back Test On Transformer

Sumpner's test provides data for finding the regulation, efficiency and heating under load conditions and is employed only when two similar transformers are available.The O.C. and S.C. tests give us the equivalent circuit parameters but ca not give heating information under various load conditions.The Sumpner's test requires two identical transformers.One transformer is loaded on the other and both are connected to supply. The power taken from the supply is that necessary for supplying the losses of both transformers and the negligibly small loss in the control circuit. Sumpner's test  is also called as back to back test on transformers because of connections are in parallel at input side and connections are in series at output side.

Procedure Sumpner's Test on Transformers :

As shown in Fig, primaries of the two transformers are connected in parallel across the same a.c. supply. With switch S open, the wattmeter W1 reads the core loss for the two transformers.

Finding Losses With Sumpner's Test On Transformer

The secondaries are so connected that their potentials are in opposition to each other. This would so if VAB = VCD and A is joined to C whilst B is joined to D. In that case, there would be no secondary current flowing around the loop formed by the two secondaries. T is an auxiliary low-voltage transformer[Regulation Transformer].which can be adjusted to give a variable voltage and hence current in the secondary loop circuit. By proper adjustment of T, full-load secondary current I2 can be made to flow as shown. It is seen, that I2 flows from D to C and then from A to B. Flow of I1 is confined to the loop FEJLGHMF and it does not pass through W1. Hence, W1 continues to read the core loss and W2 measures full-load Cu loss (or at any other load current value I2). Obviously, the power taken in is twice the losses of a single transformer.

i.e. copper loss per transformer PCu = W2/2.

i.e. iron loss per transformer Pi = W1/2.

From results of sumpner's test, the full load efficiency of each transformer can be given as

Advantages of Sumpner's Test 

1.Low  power is required to conduct this test. Because no external load is connecting 
2.Full load copper losses and iron losses of both transformers determined.
3.Increase in  transformer temperature can be found.