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Open-Circuit and Short-Circuit Tests in Transformers

Join our engineering community sign-in with:, a transformer could be tested under no-load and full-load conditions to determine its turns ratio, regulation, and efficiency. however, without fully loading the transformer, it is possible to perform two tests (open-circuit and short-circuit) from which all the important data can be derived. in this article, learn how to analyze the results of open-circuit and short-circuit transformer tests to determine the component values of the transformer equivalent circuit..

To discuss transformer theory accurately, we must take into account some losses associated with transformers. Generally, these losses can be divided into two categories: magnetic, or core, losses, and I 2 R, or coil loss. The core losses are relatively constant, resulting from the magnetic circuit that does not change much as transformer current changes. Currents cause the I 2 R, or coil loss, and because currents change, so do the coil losses.

All the losses associated with the core of the transformer are magnetic in nature and are relatively constant. The effect of hysteresis and eddy currents do not change much because of current flow; they result from core material and design.

A transformer  is just like any electrical circuit or device. When current flows, it always creates a magnetic field and heat. The heat produced by the current flow in a transformer is referred to as the coil, or copper, loss. The formula for calculating this loss is the same as the formula for wattage: I 2 R.

The important thing to remember about coil loss is that the amount of coil loss is directly related to the current flow through the transformer  and manifests itself as heat. To reduce this heat loss, transformer coils may be wound with copper conductors, which have less resistance than aluminum conductors of the same size.

Two tests (open-circuit and short-circuit) are performed to determine core and copper losses in the transformer. These losses are further used to compute the transformer’s efficiency. 

\[Efficiency=\frac{Output\,Power}{Input\,Power}\times100=\frac{Output\,Power}{Output+\Sigma(Losses)}\times100\]

\[Efficiency=\frac{Output\,Power}{Output\,Power+Core\,Losses+Copper\,Losses}\times100\]

Open-Circuit Test

Figure 1(a) shows the circuit for the transformer open-circuit test. The alternating input voltage is set to the normal primary level for the transformer , and the voltage at the open-circuited output terminals is monitored on a voltmeter, as illustrated. The wattmeter measures the input power, and the ammeter measures the primary current. Because the secondary is open-circuited, the primary current is very small, and the voltage drops across the ammeter and wattmeter can be assumed to be negligible. In this case, the input voltage can be taken as the transformer's primary voltage; thus, the ratio of the voltmeter readings gives the turns ratio.

\[\frac{E_{P}}{E_{S}}=\frac{N_{P}}{N_{S}}\]

Figure 1(a).  A transformer open-circuit test is performed by measuring the (no-load) secondary voltage, the primary current, and the input voltage and power. Image used courtesy of Amna Ahmad

With a very small primary current, and near-zero secondary current (i.e., the voltmeter current), the copper loss in the windings can be assumed negligible. The input power measured on the wattmeter is then the total transformer core losses, and the ammeter indicates the no-load primary current (I o ) (see Figure 1(b)). From the measured values of input voltage, current, and power, the components of the no-load equivalent circuit can be determined.

Figure 1(b).  Secondary components referred to the primary. Image used courtesy of Amna Ahmad

True Power,

\[P=\frac{E^{2}_{P}}{R_{o}}\]

\(R_{o}=\frac{E^{2}_{P}}{P}\)   (1)

Apparent Power, 

\[S=E_{P}I_{o}\]

\[S=\sqrt{(true\,power)^{2}+(reactive\,power)^{2}}=\sqrt{P^{2}+Q^{2}}\]

\[Q=\sqrt{S^{2}-P^{2}}\]

\(Q=\sqrt{(E_{P}I_{o})^{2}-P^{2}}\)   (2)

The reactive power can be determined by,

\[Q=\frac{E^{2}_{P}}{X_{o}}\]

\(X_{o}=\frac{E^{2}_{P}}{Q}\)   (3)

An open-circuit test on a transformer produced the following measurements: E P =115 V, E S =57.5 V, P =9.5 W, and I o =180 mA. Determine the transformer turns ratio and the values of R o  and X o .

\[\frac{N_{S}}{N_{P}}=\frac{E_{S}}{E_{P}}=\frac{57.5V}{115V}=\frac{1}{2}\]

From Equation 1,

\[R_{o}=\frac{E^{2}_{P}}{P}=\frac{(115\,V)^{2}}{9.5\,W}=1.39k\Omega\]

From Equation 2,

\[Q=\sqrt{(E_{P}I_{o})^{2}-P^{2}}=\sqrt{(115V\times180mA)^{2}-(9.5W)^{2}}==18.39vars\]

From Equation 3,

\[X_{o}=\frac{E^{2}_{P}}{Q}=\frac{(115V)^{2}}{18.39\,vars}=719\Omega\]

Short-Circuit Test

The transformer  short-circuit test is performed with the secondary terminals short-circuited, as illustrated in Figure 2(a). Note that the primary voltage (E P ) is measured right at the transformer primary terminals to avoid error due to the voltage drops across the ammeter and wattmeter. The input voltage is increased from zero until the ammeter in the primary circuit indicates normal full-load primary current. When this occurs, the normal full-load secondary current is circulating in the secondary winding. Because the secondary terminals are short-circuited, the input voltage required to produce full-load primary and secondary currents is around 3% of the normal input voltage level. With such a low input voltage level, the core losses are so small that they can be neglected. However, the windings are carrying normal full-load current, and so the input is supplying the normal full-load copper losses.

Figure 2(a).  To perform a transformer short-circuit test, the secondary is shorted, and the primary current is adjusted to the normal full-load level. The primary voltage and input power are measured. Image used courtesy of Amna Ahmad

The output power (to the short-circuit) is zero, so the wattmeter measuring true input power indicates the full-load copper losses. The product of the ammeter and voltmeter readings gives the apparent input power. From these quantities, calculations may be made of the resistive and reactive components of the full-load equivalent circuit referred to the primary (see Figure 2(b)).

Figure 2(b).  Equivalent Transformer Circuit simplified by neglecting R o  and X o . Image used courtesy of Amna Ahmad

\[P=I^{2}_{P}R_{e}\]

\(R_{e}=\frac{P}{I^{2}_{P}}\)   (4)

Apparent Power,

\[S=E_{P}I_{P}\]

Reactive Power,

\(Q=\sqrt{(E_{P}I_{P})^{2}-P^{2}}\)   (5)

\[Q=I^{2}_{P}X_{e}\]

\(X_{e}=\frac{Q}{I^{2}_{P}}\)   (6)

Determine R e  and X e  for the transformer in Example 1 when the following measurements were made on a short-circuit test: E P(SC)  =5.5 V, I P =1 A, and P=5.25 W.

From Equation 4,

\[R_{e}=\frac{P}{I^{2}_{P}}=\frac{5.25W}{(1A)^{2}}=5.25\Omega\]

From Equation 5,

\[Q=\sqrt{(E_{P}I_{P})^{2}-P^{2}}=\sqrt{(5.5V\times1A)^{2}}-(5.25W)^{2}=1.64\,var\]

From Equation 6,

\[X_{e}=\frac{Q}{I^{2}_{P}}=\frac{1.64\,var}{(1A)^{2}}=1.64\Omega\]

Key Takeaways of Open-Circuit and Short-Circuit Transformer Tests 

The performance of a transformer is described in terms of its voltage regulation and efficiency, and can be predicted from the results of two tests: the open-circuit test and the short-circuit test. An open-circuit test is accomplished by measuring the (no-load) secondary voltage, the primary current, and the input voltage and power. In order to conduct a short-circuit test, the secondary side is shorted, and the primary current is adjusted to the normal full-load level. The primary voltage and input power are measured.

Featured image used courtesy of Adobe Stock

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• Transformer Inspection and Testing
• Basic Inductance Principles in Transformers
• Calculating Voltage Regulation in Transformers
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• open circuit

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Open and Short Circuit Test of Transformer

Open and short circuit tests are performed on a transformer to determine the:

Open Circuit Test on Transformer

The ammeter reading gives the no load current I e . As no load current I e is quite small compared to rated current of the transformer , the voltage drops due to this current that can be taken as negligible.

Short Circuit Test on Transformer

These values are referred to the HV side of the transformer as the test is conducted on the HV side of the transformer. These values could easily be converted to the LV side by dividing these values with the square of transformation ratio.

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Home > Electrical Machines > Transformer > Short Circuit Test and Open Circuit Test of Transformer

Short Circuit Test and Open Circuit Test of Transformer

Open circuit (no load) and short circuit (on-load) test of a transformer.

As we have seen in the equivalent circuit of transformer , there are four main parameters;

• Equivalent resistance (R 01 )
• Equivalent reactance (X 01 )
• Core-loss resistance (R 0 )
• Magnetizing reactance (X 0 )

The open circuit and short circuit tests are performed to find circuit parameters, regulation, and efficiency of a transformer . These tests are performed without the actual loading of a transformer. Therefore, these tests are considered as an indirect method of testing.

These tests give more accurate results compared to the test performed on a fully loaded transformer (direct method). Also, these tests are more economical as the power consumption is very less. Two tests considered an indirect method of testing;

• Open-circuit Test
• Short-circuit Test

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Open Circuit Test (No-Load Test)

The open-circuit test (aka No-load test ) is performed to determine the losses in a transformer such as core loss (iron loss), no-load current (I 0 ), and no-load equivalent circuit parameters (R 0 and X 0 ). This test is performed either on primary winding or secondary winding. But in most cases, this test is performed on low-voltage winding. Because it is difficult to obtain high voltage in laboratories and the current that passes through the high voltage winding is very small. So, it may be difficult to measure the accurate readings.

Therefore, the open circuit test is performed on low voltage winding. The experimental connection diagram of the open-circuit test on a single-phase transformer is shown in the figure below.

As shown in the above figure, the primary winding (low voltage winding) is supplied by rated voltage and frequency (commonly, a single phase supply from autotransformer ) . And the secondary winding is kept open. Now, a voltmeter V 0 , an ammeter I 0 , and a wattmeter W 0 are connected in the primary winding.

The secondary winding is kept open-circuited. Therefore, the current that passes through the secondary winding is zero. And the load is not connected. Hence, the current that passes through the primary winding is no-load current I0. The current that passes through the primary winding is measured by an ammeter that gives the value of no-load current.

The supply voltage given to the primary winding is rated voltage. So, the flux produced in the core of a transformer is normal. And this flux is the same for all loading conditions. The iron loss produced in the transformer depends on the supply voltage and frequency. In this test, we have given rated supply voltage and frequency. Hence, the iron loss or core loss produced in this test is the same for all loads.

The current that passes through the secondary winding is supplied to the iron loss and copper loss in the primary winding. No-load current passes through the primary winding that is very small (2 to 5 percent of full-load current). Therefore, we can neglect the copper loss. And the primary current is supplied for the core loss.

A wattmeter is connected to the primary winding that measures the supplied power. So, the wattmeter indicates the power loss occurred in the transformer core. In an open-circuit test, the reading of instruments is as below;

Ammeter: no-load Current I 0

Voltmeter: rated supply voltage V 1

Wattmeter: iron or core loss P i

Observation Table

The observation table of an open circuit test is shown below.

Now, we can find the circuit parameter (R0 and X0) using the no-load current.

No-load power W 0 = V 1 I 0 Cos ϕ 0 = Iron loss

Working component of no-load current;

I W = I 0 Cos ϕ 0

Magnetizing component of no-load current;

I M = I 0 Sin ϕ 0

Now, from the working component and magnetizing component, we can find the no-load resistance and reactance as follows;

No-load Resistance;

No-load Reactance;

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Short Circuit Test (On-Load Test)

The short circuit test (aka On-load test ) is performed on the high-voltage side and the low-voltage side is short-circuited. This test could be conducted on the low voltage side, but this test required hardly 5 to 7 percent of rated voltage. On the low voltage side, this voltage is quite small and has chances of a measurement error. Also, reduced voltage (5 to 7 percent) of the high voltage side is easily available in the laboratory. Therefore, it is convenient to perform short circuit tests on the high voltage side.

The schematic diagram of the short circuit test is shown in the figure below.

Usually, a low voltage winding is short-circuited using a thick wire. But in some cases, an ammeter is connected to measure the rated load current. An ammeter, a voltmeter, and a wattmeter are connected in the high voltage side as shown in the above figure. Here, we have considered primary winding as high voltage winding and secondary winding as low voltage winding.

The high voltage winding is supplied by the reduced input voltage from a variable supply source. The supply voltage gradually increases until full-load primary current flows through the primary winding. When full-load current passes through the primary winding, by transformer action, the current flows through the secondary winding are full-load secondary current.

So, the ammeter connected in the high voltage side measures the full-load primary current. The voltmeter measures the supplied voltage when full-load current flows through the primary winding. In this condition, the supplied voltage is hardly 5 to 10 percent of full load voltage. Due to low input voltage, the flux produced in the core is very low. And core loss is proportional to the square of flux. Hence, the core loss is very small that can be neglected.

Also, the current that passes through the windings is a full-load current. So, a copper loss that occurs during a test is a normal full-load copper loss. And the wattmeter indicates the full-load copper loss. The secondary winding is short-circuited. So, the secondary voltage (output voltage) is zero. Therefore, the entire primary voltage is used to supply the voltage drop in total impedance referred to as the primary side.

The approximate equivalent circuit of the transformer under the short circuit test is shown in the figure below.

Observation Table:

The reading of instruments in short-circuit test are as follows;

• Ammeter: Full-load primary current (I SC )
• Voltmeter: Supplied voltage (V SC )
• Wattmeter: Full-load copper loss (P C )

Full-load copper loss;

W SC = I 2 SC R 01

The equivalent resistance of transformer referred to primary;

Equivalent impedance referred to primary;

Equivalent reactance referred to primary;

Power factor;

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Open circuit test & short circuit test in transformers.

Open circuit test and short circuit test are conducted to determine the core loss, copper loss, and equivalent circuit parameters of a transformer. In this article, in addition to the tests mentioned in the title, we will also discuss the polarity test and Sumpner’s test.

Why is it necessary to conduct OC and SC tests?

It is necessary to know the impedance of the transformer in order to calculate its voltage regulation and efficiency . The impedance and other circuit parameters can be determined by conducting simple no-load tests. No-load tests have a minimal power loss when compared to that during direct load tests.

Open circuit test

The open circuit test is performed to determine the no load losses or core losses as well as the turns ratio, no load currents, magnetizing components and core loss components of the transformer.

Circuit for open circuit test

For convenience, the supply is connected to the LV side of the transformer and the HV side of the transformer is left open. Voltmeters, ammeters and watt meter are connected as shown in the figure below.

 Open circuit test Procedure

Calculation of core losses and magnetizing components.

Neglecting the copper loss, we calculate the core losses and the core loss components.

Short Circuit test

The purpose of conducting a short circuit test is to determine the winding resistance, reactance, and the copper loss of the transformer.

Circuit for short circuit test

Short circuit test procedure.

It can be noted that the applied voltage, V sc , required to circulate current I sc is very small compared to the rated voltage of the winding (typically 5% of rated voltage). Therefore, the excitation current required is too small and can be neglected.

The power input to the transformer measured by P sc corresponds to copper loss. Therefore

The resistance offered by the coil,

The Susceptance of the core,

The attained values of R and X are referred to the HV side of the transformer from which the test is conducted. If can be referred to the other side using the operator a 2 (square of turns ratio).

Calculation of transformer efficiency

If P 0 and P sc are the core loss and the copper loss of a transformer respectively, the efficiency of the transformer can be calculated using the following formula:

Polarity test

Polarity test is conducted to find the polarity of the transformer windings. Those ends of the two transformers that acquire positive or negative polarity simultaneously are said to be at similar polarity.  It is necessary to know the polarity of the winding in order to connect it properly.

Connection and procedure for polarity test

Method 2:  dc flashing method.

When the switch is closed, the end of the LV winding that is at the same polarity as that of the HV winding to which the positive terminal of the battery connected becomes positive (This is indicated by the deflection of the DC voltmeter). When the switch is opened, the similar polarity end acquired negative potential.

Sumpner’s test

Sumpner’s test, otherwise known as “back-to-back test”, is a method to determine the steady-state temperature rise of the transformer. This test is conducted by connecting two similarly rated transformers back to back.

Circuit for Sumpner’s test

Since the secondaries are connected in phase opposition, no current shall flow through it even if they are short-circuited and the current measured by the ammeter A 1 corresponds to the core loss component.

Similarly when the primaries are short-circuited, full current flows through the circuit. The current measured by W 2 corresponds to the copper loss of the transformer.

Thus by conducting Sumpner’s test, we can calculate the iron and copper losses without even loading the transformer. The total power input to the transformer is (2P o + 2P c ).

The steady-state temperature rise of the transformer can be monitored by conducting this test for a longer period and monitoring the temperature rise using a thermometer.

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Open Circuit and Short Circuit Test on Transformer

• April 4, 2024
• By Ravi Teja

In this tutorial, we will learn what is an Open Circuit Test and Short Circuit Test in the context of Transformers, how to perform Open Circuit and Short Circuit Test on Transformer, calculate the efficiency of these open circuit and short tests and also calculate the regulation.

Overview of Open Circuit and Short Circuit Test

It is possible to predict the performance of a transformer at various levels of load by knowing all the equivalent circuit parameters. These circuit parameters are supplied in terms Open Circuit (OC) and Short Circuit (SC) test data of a transformer. Without actually loading the transformer, these two assessed tests give the test results, which are used to determine the equivalent circuit parameters.

By these parameters, we can easily predetermine the efficiency and regulation of the transformer at any power factor condition as well as at any load condition. This method of finding the parameters of a transformer is called as an Indirect Loading Method.

This tutorial enumerates how to perform these tests, how determine the equivalent parameters from test data and significance HV or LV side in which the calculation to be performed.

Open Circuit or No Load Test on Transformer

This test is performed to find out the shunt or no load branch parameters of equivalent circuit of a transformer. This test results the iron losses and no load current values, thereby we can determine the no load branch parameters with simple calculations.

As the name itself indicates, secondary side load terminals of the transformer are kept open and the input voltage is applied on the primary side. Since this test is carried out without placing any load, this test is also named as No Load Test.

How to Perform Open Circuit Test?

The open circuit (OC) test is carried out by connecting LV side (as primary) of the transformer to the AC supply through variac, ammeter, voltmeter and wattmeter instruments. The secondary side or HV side terminals are left open and in some cases a voltmeter is connected across it to measure the secondary voltage.

The primary side voltmeter reads the applied voltage to the transformer, ammeter reads the no load current, wattmeter gives the input power and the variac used to vary the voltage applied to transformer so that rated voltage is applied at rated frequency. The OC test arrangement of a transformer is shown in below figure:

When a single phase supply is given to the transformer, the rated value of the primary voltage is adjusted by varying the variac. At this rated voltage, the ammeter and wattmeter readings are to be taken. From this test, we get rated voltage V O , input or no load current I O and input power W O .

We know that, when the transformer is on no load, the no load current or primary current is very small, typically 3 to 5 percent of the rated current value. Thus, the copper loss in the primary winding is negligible.

In OC test, transformer is operated at rated voltage at rated frequency so the maximum loses will be the flux in the core. Since the iron or core losses are at rated voltage, the power input is drawn to supply the iron losses by the transformer under no load.

W O = Iron losses

The no load shunt parameters are calculated from the OC test as

Once the power factor is obtained, the no load component currents are determined as:

When the transformer is operating on no load, the current drawn by the shunt or parallel parameters is very small, about 2 to 5 percent of the rated current. Thus, a low current will flow through the circuit during OC test. In order to be readable by the instruments, the measurements of voltage, current and power must be performed in the low voltage side.

And also, low range current coils and low range ammeter must be selected. The power factor of the transformer on no load is too low. which is typically below 0.5 . So, in order work with this low value, a LPF watt meter is selected. The equivalent circuit obtained by the OC test is shown below:

Short Circuit Test on Transformer

This test is performed to find series branch parameters of an equivalent circuit such as equivalent impedance (Z o1 or Z o2 ), total winding resistance (R o1 or R o2 ) and total leakage reactance (X o1 or X o2 ). Also, it is possible to determine copper losses at any desired load and total voltage drop of the transformer referred to primary or secondary. In this test, usually LV winding is shorted by a thick wire. And the test is conducted on the other side, i.e. HV side (as primary).

How to Perform Short Circuit Test?

In Short Circuit (SC) test, the primary or HV winding is connected to the AC supply source through voltmeter, ammeter, wattmeter and a variac as shown in figure. This test is also called as Reduced Voltage Test or Low Voltage Test. As the secondary winding is short circuited, at rated voltage, the transformer draws a very large current due to its very small winding resistance.

Such high current can cause overheating and also burning of the transformer. Thus, to limit the high current, the primary winding must be energized with a low voltage, which is just enough to produce the rated current in the primary of the transformer.

The SC test is conducted on HV side due to the two main reasons. The first one is, the SC test conducted by applying rated current and the rated current of the HV side is much less than that of the LV side. Therefore, the rated current is easily achieved at HV side (due to the low current value) as compared to the LV side.

On the other hand, if we short the HV terminals by connecting measuring instrument on LV side, voltage in the secondary is zero. Therefore, the current flow through HV side is very high (as VA rating is constant) compared to the LV side and hence it will cause to burn the transformer.

During this test, by varying the variac slowly, we apply a low voltage to the primary typically 5 to 10 percent of the rated voltage to cause a rated current to flow in both primary and secondary windings that we can observe on ammeter reading (in some cases, the secondary is shorted through an ammeter). At this rated current, we have to record the voltmeter (V sc ), ammeter (I sc ) and wattmeter (W sc ) readings.

In this test, the current flow is rated value and hence no load current is very small and is 3 to 5% of the rated current. In other words, the voltage applied to the primary winding is very low, thereby the flux level in the core is very small. In turn there is negligible core loss. Therefore, the no load shunt branch is considered as absent in equivalent circuit of this test as core loss is negligible.

As the iron or core losses are function of voltage, these losses are very small. Therefore, the wattmeter reading shows the power loss or I 2 R loss equal to the full load copper losses of the whole transformer.

W sc = Full load copper losses

Form the test results we determine the series branch parameters of an equivalent circuit as

The equivalent circuit obtained from this test is shown below.

It should be noted that, before calculation of parameters, you must be aware in which side (primary or secondary) the test reading being recorded. Suppose if the transformer is step-up transformer, then we carry out the SC test on secondary side (HV side) while primary or low voltage side is shorted. In such case, we get the parameters referred to the secondary from calculations such as R02, X02 and Z02.

If it is a step-down transformer, we get the parameter values as R01, X01 and Z01 because the meters are connected to the HV side of the primary.

From the OC test we get, shunt branch parameters referred to the LV side and from SC test we get series branch parameters referred to HV side. Therefore, for a meaningful equivalent circuit, all the parameters must be referred to the one particular side. The explanation regarding this transformation is explained in equivalent circuit of the transformer topic in our earlier articles.

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

As we have seen that, the practical transformer has two types of major losses namely copper and core losses. The temperature of the transformer rises due to these losses which are dissipated as heat. Due to these losses, input power drawn by the primary no longer equal to the output delivered at secondary. Therefore, the efficiency of the transformer is given as

We have discussed that, the core loss Pcore remains constant from no load to full load as the flux in the core remains constant. And the copper losses are depend on the square of the current. As the winding current varies from no load to full load, copper losses are also get varied.

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

Therefore the efficiency of the transformer is

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

For a fixed voltage in the primary, the secondary terminal voltage will not be maintained constant from no load to full load. This is due to the voltage drop across leakage impedance which magnitude depends on both degree of loading and the power factor.

So the regulation gives change in secondary voltage from no load to full load at a given power factor. It is defined as the change in the secondary voltage when the transformer is operating at full load of specified power factor supplied at rated voltage to no load with primary voltage held constant.

The expression of voltage regulation in terms voltage drops is given as

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.

A beginner’s guide on Open Circuit and Short Circuit Test of a Transformer. You learned how to perform Open Circuit and Short Circuit Test on Transformer, calculate the equivalent circuit parameters, calculate the efficiency and percent of regulation.

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5 Responses

It is very helpful article for me so thank you very much.??

Very useful article. The open-circuit power is the voltage difference measured between two terminals when no current is supplied. The short circuit current is the current that flows when the stations are forced to have zero voltage change.

This is a good article on Transformer tests.

the explanatoin was ok .only that i expected better elaborations through some worked examples

It was very explanation. Thank you.

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Open Circuit and Short Circuit Test on Transformer – Experimental Set-up & Procedure

The Open Circuit and Short Circuit Test on Transformer can be used to determine the efficiency and regulation of a transformer at any predetermined load for that we do not have to actually load the transformer. The parameters of the equivalent circuit of a transformer can be obtained by using the open circuit (O.C.) test and the short circuit (S.C.) test on the transformer.

Open Circuit ( O.C.)  Test on Transformer

Fig. 1 : Set-up for Open Circuit Test on Transformer.

This test is performed so as to calculate the no load losses (core losses) of a transformer and the values of I 0 , R 0 and X 0 of the equivalent circuit. The set up for O.C. test of a transformer is shown in Fig. 1. The primary or secondary winding of the transformer is connected to the rated ac voltage by means of using a variac. The other winding is left open. Generally the high voltage winding is open circuited, and the voltage is applied to the low voltage winding. Assume that this is a step up transformer. A voltmeter is connected across the primary winding to measure the primary voltage. An ammeter is used for measuring the no load primary current (I 0 ) and the wattmeter is connected to measure the input power. The secondary is open circuited because it is an open circuit (O.C.) test. Sometimes a voltmeter is connected across the secondary to measure V 2 = E 2 . Note that the ac supply voltage is applied generally to the low voltage side and the higher voltage side is used as secondary.

• Connect the circuit as shown in Fig. 1.
• Keep the variac at its minimum voltage position.
• Switch on the ac power supply and adjust the variac to get the rated primary voltage as measured by voltmeter V across the primary.
• Now measure the primary current (I 0 ) and power (W 0 ) using the ammeter and wattmeter respectively.
• The ammeter reads the no load primary current I 0 whereas the wattmeter measures the no load input power W 0 .
• The observation table for the O.C. test is as follows.

• The two components of no load current I 0 are,

\[{{I}_{m}}={{I}_{0}}\sin {{\phi }_{0}}\]

\[{{I}_{c}}={{I}_{0}}\sin {{\phi }_{0}}\]

• The no load power factor is given by cosϕ 0 and the input power at no load is given by,

\[{{W}_{0}}={{V}_{1}}{{I}_{0}}\cos {{\phi }_{0}}\]

• The phasor diagram on no load showing the two components of I 0 is shown in Fig. 2.

Fig. 2 : Phasor Diagram for Open Circuit Test on Transformer.

• The no load current I 0 is very small as compared to the full load primary current. The no load current I 0 is about 3 to 5 % of the full load value.
• As I 2 is zero, the secondary copper loss is zero. The primary copper loss will be negligible because I 0 is small.
• Therefore the total copper loss is very small and can be assumed to be equal to zero. Hence the wattmeter reading W 0 represents the iron losses.

\[{{W}_{0}}={{P}_{i}}=\text{ Iron losses}\]

Calculation of parameters:

The two parameters which can be calculated from the open circuit test are R 0 and X 0 . They are calculated as follows.

Step 1: Calculate no load power factor cosϕ 0,

The wattmeter reads the real input power.

\[\cos {{\phi }_{0}}=\frac{{{W}_{0}}}{{{V}_{1}}{{I}_{0}}}\]

Calculate ϕ 0 from this.

Step 2: Calculate I m and I c,

\[{{I}_{c}}={{I}_{0}}\cos {{\phi }_{0}}\]

So calculate I m and I c from the above equations.

Step 3: Calculate R 0 and X 0,

\[{{R}_{0}}=\frac{{{V}_{1}}}{{{I}_{c}}}\Omega \]

\[{{X}_{0}}=\frac{{{V}_{1}}}{{{I}_{m}}}\Omega \]

The value of cosϕ 0 is very small. Therefore it is necessary to use the low power factor type wattmeter to avoid any possibility of error in measurements.

Short Circuit (S.C.) Test on Transformer

Fig. 3 : Set-up for short Circuit Test on Transformer.

The set up for carrying out the shown circuit (SC) test on a transformer is shown in Fig. 3. Generally the high voltage side is connected to the ac supply and the low voltage high current side is shorted. Variac is used to adjust the input voltage precisely to the rated voltage. We assume that the transformer used here is a step down transformer. Hence the secondary is shorted and primary is connected to the variac. The voltmeter is connected to measure the primary voltage The ammeter measure the short circuit rated primary current I sc and the wattmeter measures the short circuit input power. The secondary is short circuited with the help of thick copper

• Connect the circuit as shown in Fig. 3.
• Shown circuit the secondary which is a low voltage high current, low resistance winding.
• Keep the variac at its minimum voltage position and switch on the ac supply voltage.
• Increase the primary voltage very gradually: and adjust it to get the primary current equal to the rated value I sc . Do not increase the primary voltage further.
• Note down the wattmeter, voltmeter and ammeter readings. The observation table is as shown below in Table.

Parameter calculations:

The primary and secondary currents are the rated currents. Therefore the total copper loss is the full load copper loss. If we adjust the primary current to half the full load current then we get the copper loss at half load. The iron losses are a function of applied voltage. As the applied voltage in S.C. test is small, the iron losses will be negligibly small. Hence the wattmeter reading W sc corresponds almost entirely to the full load copper loss.

\[{{W}_{SC}}=\text{ Full load copper loss}\]

\[={{P}_{cu(FL)}}\]

We can calculate the parameters R IT , X IT and Z IT of the equivalent circuit from the short circuit (S.C.) test.

We know that

\[{{W}_{SC}}={{\text{V}}_{SC}}{{I}_{SC}}\cos {{\phi }_{SC}}\]

Hence the short circuit power factor is given by,

\[\cos {{\phi }_{SC}}=\frac{{{W}_{SC}}}{{{V}_{SC}}{{I}_{SC}}}\]

But the wattmeter reading W sc indicates the full load copper loss.

\[{{W}_{SC}}=\text{Copper loss}\] \[=I_{SC}^{2}\times {{R}_{1T}}\]

\[{{R}_{1T}}=\frac{{{W}_{SC}}}{I_{SC}^{2}}\]

\[{{Z}_{1T}}=\frac{{{V}_{SC}}}{I_{SC}^{{}}}=\sqrt{R_{1T}^{2}+X_{1T}^{2}}\]

\[{{X}_{1T}}=\sqrt{Z_{1T}^{2}+R_{1T}^{2}}\]

In this way the parameters R IT , X IT and Z IT can be calculated from the S.C. test. If the transformation ratio K, it is possible to obtain the parameters referred to the secondary side.

Efficiency Calculation from O.C. and S.C. Test:

The expression for full load efficiency IS given by,

\[{{\eta }_{FL}}=\frac{{{V}_{2}}{{I}_{2(FL)}}\cos \phi }{{{V}_{2}}{{I}_{2(FL)}}\cos \phi +{{P}_{i}}+{{P}_{cu(FL)}}}\]

The iron loss P i can be obtained from the O.C. test because,

\[ {{W}_{0}}={{P}_{i}}=\text{ Total iron loss}\],

And the full load copper loss is obtained from the S.C. test because,

\[{{\text{W}}_{SC}}={{P}_{cu(FL)}}=\text{ Copper loss}\]

Voltage Regulation Calculation from O.C. and S.C. Test:

The percentage regulation (%R) is given by,

\[\%R=\frac{{{I}_{2}}{{R}_{2T}}\cos \phi \pm {{I}_{2}}{{X}_{2T}}\sin \phi }{{{V}_{2}}}\times 100\]

\[\%R=\frac{{{I}_{1}}{{R}_{1T}}\cos \phi \pm {{I}_{1}}{{X}_{1T}}\sin \phi }{{{V}_{1}}}\times 100\]

We can obtain the parameters such as R IT , X IT , R 2T , X 2T from the S.C. test of the transformer. Whereas the rated voltages V 1 ,V 2 and the rated currents I 1 and I 2 in the above expressions are known from the given transformer

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ECE Undergraduate Laboratories

Ece 342 energy conversion.

Experiment 3: Power Transformer Open and Short Circuit Tests

• To conduct standard open and short circuit tests in order to find the parameters of the equivalent circuit of a transformer.
• Evaluate the regulation and efficiency of the transformer at a given load.
• Fluke meter from the stockroom.
• Two scope leads from the stockroom.
• 3-phase AC Variac.
• One four-winding single phase transformer. (Model # T-1000)
• A. Fitzgerald, C. Kinsley, Jr., S. Umans, Electric Machinery, Ch. 1, 6th Edition, McGraw-Hill Inc., 2005.
• P.C SEN, Principles of Electric Machines and Power Electronics , 3rd Edition, John Wiley, 2013

A power transformer is usually employed for the purpose of converting power, at a fixed frequency, from one voltage to another. If it is used for converting power from a high voltage to a low voltage, it is called a step-down transformer. The conversion efficiency of a power transformer is extremely high and almost all of the input power is supplied as output power at the secondary winding.

Consider a magnetic core as shown in figure 3.1, carrying primary and secondary windings having N 1 and N 2 turns, respectively. When a sinusoidal voltage is applied to the primary winding, a flux Φ will exist in the core which links both the primary and secondary windings, inducing the RMS voltages

The transformer is said to have a transformation ratio

Equivalent Circuit

The transformer may be represented by the equivalent circuit shown in figure 3.2. The parameters may be referred to either the primary or the secondary side. The series resistances R 1 and R 2 represent the copper loss in the resistance of the two windings. The series reactances X 1 and X 2 are leakage inductances and account for the fact that some of the flux established by one of the windings does not fully couple the other winding. These reactances would be zero if there were perfect coupling between the two transformer windings.

The shunt resistance R p accounts for the core losses (due to hysteresis and eddy currents) of the transformer. The shunt inductance X p is representative of the inductances of the two windings and would be infinite in an ideal transformer if the number of turns of the two windings were to be infinite.

A knowledge of the equivalent circuit parameters permits the calculation of transformer efficiency and of voltage regulation without the need to conduct actual load tests. But experimental data must first be obtained in order to determine those parameters.

It will be confirmed at the conclusion of the first two parts of this experiment that the impedances of the series branch of the transformer equivalent circuit are substantially smaller than the impedances of the parallel branch. Because of this large discrepancy in the magnitudes of the elements we can redraw the equivalent circuit shown in figure 3.2 into that shown in figure 3.3. The errors introduced into calculations using figure 3.3 in place of figure 3.2 are quite insignificant. Furthermore, the large difference in the magnitudes of the transformer parameters allows for the determination of the elements in the series branch using one set of measurements and the elements in the parallel branch using another set of measurements.

Open Circuit Test

The open circuit test is used to determine the values of the shunt branch of the equivalent circuit R p and X p . We can see from figure 3.3 that with the secondary winding left open, the only part of the equivalent circuit that affects our measurement is the parallel branch. The impedance of the parallel branch is usually very high but appears lower when referred to the low voltage side. This test is therefore performed on the low voltage side of the transformer terminals 1 − 1' in figure 3.3) to increase the current drawn by the parallel branch to a readily measurable level. Besides, the rated voltage on the low voltage side is lower and therefore more manageable.

T-1000 transformer has four windings. Create a 1:2 ratio step up transformer by connecting the two primary windings in series and the two secondary windings in series.

This transformer will also be used in the next part of the experiment, so leave the connections intact when the present part is finished.

This transformer is rated at 1.0 KVA. The rated current is 1000 VA/240 V = 4.16A on the 240 V side and 1000 VA/120 V = 8.32A on the 120 V side.

Instructions

• Connect the circuit as shown in figure 3.4. Make sure that the low voltage side of the transformer corresponds to the left side of the connection diagram. A low power factor wattmeter should be used.
• Connect the Power Quality Meter to the left side (primary) of the transformer. If a low Power Factor Wattmeter is used, it should be connected also to primary and DVM’s connected to allow measurement of phase voltage (V 1 ) and primary current (I P )
• Connect the power supply from the bench panel to the INPUT of the three-phase variac and connect the OUTPUT of the variac to the circuit.
• Vary the input voltage starting at 0 V in 20 V increments to go up to 120 V. At each step change, record  I p , W 0 and V 1   in table 3.1.
• Turn off the variac.
• Complete table 3.1
• Compute the parameters R p and X p   at the rated voltage by using

These parameters are referred to the low voltage side.

• Find the value s of   R p and X p as referred to the high voltage side.
• Plot the no-load current  I p , magnetizing current   I m , core loss W 0  and no-load power factor cos Φ , against the applied voltage   V 1  on the same graph paper.

Short Circuit Test

The short circuit test is used to determine the values R s and X s of the series branch of the equivalent circuit. These impedances are usually very low, but appear higher in value when referred to the high voltage side. This test is consequently performed on high voltage side of the transformer (terminals 2 − 2ʹ in figure 3.3) in order to keep the current drawn by these impedances at a manageable level.

• Using the 2:1 ratio transformer of the previous part connect the circuit as shown in figure 3.5. Make sure that the high voltage side of the transformer corresponds to the left side (primary) of the connection diagram. Use the voltage terminals ± and 150 V of the standard AC wattmeter, if used.
• Make sure that the variac is turned all the way down before starting this experiment. Turn on the variac.
• Turn the variac up slowly until the current I s (consult figure 3.5) is at the rated value (about 4 amps). Record I s , V s and W s in table 3.2.
• Repeat the previous step by reducing the current I s in 0.5 A and record all values in table 3.2.
• Plot the copper losses W s against the current I s .
• Compute the equivalent circuit parameter R s   and X s   at the rated high voltage winding current by first calculating

• Calculate the values of R s and X s referred to low voltage side.
• Now that we have all the parameters for the transformer equivalent circuit, compute the voltage regulation at the rated power and at a lagging power factor of 0.8.
• Calculate the per unit efficiency at the rated power and at a lagging power factor of 0.8.
• Calculate the value of the maximum efficiency of the Transformer and determine the current at which it occurs.
• Using the laboratory data, determine percent efficiency of the Transformer at half rated power and 0.8 lagging power factor.

ELECTRICAL ENGINEERING

Open circuit and short circuit test on Single-Phase  transformer.

Experiment No.: – 08

Aim of the Experiment: –

To perform the open circuit and short circuit test of a single phase transformer and to draw the equivalent circuit after determining its constants.

Apparatus Required: –

Circuit Diagram: –

Theory: –

The performance of a transformer can be calculated on the basis of its equivalent circuit which contains four main parameters, the equivalent resistance R 01 as referred to primary( or secondary R 02 ), the equivalent leakage reactance X 01 as referred to primary, the core-loss conductance G 0 and the magnetizing susceptance B 0 . These constants or parameters can be easily determined by two test i.e. Open circuit test and short circuit test. These are very economical and convenient, because they furnish the required information without actually loading the transformer.

The purpose of O.C. test is to determine no load loss or core loss and no load current I 0 which is helpful in finding X 0 and R 0 . One winding of the transformer whichever is convenient but usually high voltage winding is left open and the other is connected to its supply of normal voltage and frequency. A wattmeter (W), Voltmeter (V) and ammeter (A) are connected in the low voltage winding i.e. primary winding in the present case. With normal voltage applied to the primary, normal flux will be setup in the core, hence normal iron loss will occur which are recorded by the wattmeter. As the primary no load current I 0 is small, Cu loss is negligibly small in primary and nil in secondary. Hence, the wattmeter reading represents practically the core loss under no load condition.

For short circuit test, one winding usually the low voltage winding, is solidly short circuited by a thick conductor (or through an ammeter which may serve the additional purpose of indicating rated load current). A low voltage (usually 5 to 10% of normal primary voltage) at correct frequency (though for Cu losses it is not essential) is applied to the primary and is continuously increased till full- load current is flowing both in primary and secondary (as indicated by the respective ammeters). Since, in this test, the applied voltage is a small percentage of the normal voltage, the mutual flux ɸ produced is also a small percentage of its normal value. Hence, core loss is very small with the result that the wattmeter reading represents the full load Cu loss or I 2 R loss for the whole transformer i.e. both primary Cu loss and secondary Cu loss. If Vsc is the voltage required to circulate rated load current, then Z 01 = Vsc/I 1 . A two winding transformer can be represented by means of an equivalent circuit as shown below

Procedures: –

Open Circuit Test

• Connect circuit as shown in the circuit diagram. Open circuit the secondary and apply full load voltage to the primary through a Variac. The cupper loss is negligible since there is only no load current is flowing. Hence power consumed is the core losses of the core.
• Note voltmeter, ammeter and wattmeter readings.

Short Circuit Test

• Connect as shown in the circuit diagram. Short circuit the secondary and apply a low voltage to the primary through an auto transformer. The iron losses are negligible since the flux will be very low on account of the primary and secondary.
• Increase the voltage gradually till full load current flows in the primary.
• Note voltmeter and ammeter and wattmeter reading.

Observation Table for Open Circuit Test: –

Calculations:

See the no load phasor diagram below

Calculation:

Let the total equivalent resistance of primary and secondary referred to primary side be R 1 ohms and the total equivalent leakage reactance referred to primary side be X 1 ohms.

ohms and the total equivalent leakage reactance referred to primary side be X 1 ohms.

Precautions: –

• Don’t switch on power supply without concerning teachers.
• Single Phase Auto transformer must be kept at minimum potential point. Before switch on the experiment.

Conclusion: –

To be written by Student.

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Lab 3 Transformers short and open circuit test

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1. Three-Phase Transformer. Short Circuit Test

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1. THREE-PHASE TRANSFORMER . SHORT CIRCUIT TEST

1.1 INTRODUCTION. DESCRIPTION OF THE EXPERIMENT The short-circuit test consists of measuring the input quantities of the transformer when its secondary winding is short-circuited and the primary winding is supplied with a suitably decre- ased voltage , so that the currents in both windings are equal to the rated currents.

The input power of the transformer in short-circuit operation is coincident with the copper losses in the transformer. In fact, the supply voltage is completely used to overcome the volta- ge drops of the windings and the only flux generated is the leakage flux, whose path is almost exclusively developed in air. Being affected by a virtually null flux (main flux), the core does not give rise to any loss.

In this experiment, students will measure the value of the short-circuit voltage VSC and of the

power factor cosφSC. These values are essential for the calculation of the voltage drops under any load condition. They are useful to define the conditions of load division in case of parallel operation with other transformers .

Objectives By performing this experiment, the students will study the short circuit operation of a three- phase transformer while reaching the following main objectives: ¾¾ To understand the schematic diagram corresponding to the short circuit test of a three- phase transformer. ¾¾ To perform the three-phase transformer wiring connections, in order to run the short circuit test. ¾ ¾ To obtain the characteristic curve related to the short circuit test (VSC - short-circuit voltage,

ISC - short circuit current):

VSC = f(ISC) and cosφSC = f(ISC)

1.2 COMPONENTS LIST The modules required for this experiment are: ¾¾ DL 1080 Three-Phase Transformer ¾¾ DL 10065N Electrical Power Digital Measuring Unit ¾¾ DL 1013M2 Power Supply Module

1.3 PROCEDURE OUTLINE Schematic diagram

The values of Psc and Isc are those corresponding to an input current that is normally made equal to the rated current. It is preferable to perform the test under different current values,

to draw the graphs of the individual quantities as a function of the short-circuit currentsc I . The advantage is obtained that the measurement errors are reduced through the graphical inter- pretation of possible anomalous results. This is rather easy because the curves have a compul- sory and foreseeable behavior. When selecting the test currents, it is worth to stick to a wide range around the rated value

5 of the input winding. This does not have to be exceeded by over the 10 ÷ 15%, in order not to significantly heat the windings. The test results are affected by the temperature of the windings and the latter must be provi- ded to have an accurate meaning. It is, therefore, suggested to perform the test very fast and to start the measurements from the highest current values.

The circuit diagram for testing the three-phase transformer in short circuit is shown in figure 1. Supply the transformer through the high voltage side from a variable power supply, to prevent too high currents from flowing through the measuring circuit.

Figure 1. Circuit diagram for the short ciruit test of a three-phase transformer The LV side of the transformer is short circuited and the wattmeter (W), the voltmeter (V) and the ammeter (A) are connected to the HV side of the transformer.

The test can be performed by selecting at will the input winding, because neither the scP va- lue nor the Vsc change. Since in the short-circuit operation the behavior of the transformer is perfectly balanced and no reasons exist of wave deformation, also the connection of the input winding is perfectly free and can be indifferently selected as a function of supply and measu- rement convenience.

Characteristics curves The short circuit test of the three-phase transformer refers to the following curves presented in figure 2. The behavior of the test diagrams can be justified by analyzing:

1. The short-circuit voltage is expressed by the formula:

VSC = ZS ISC where ZS represents the equivalent impedance of the transformer, referred to the input side. This impedance is composed of the equivalent resistance and the leakage inductance of the windings.

6 Figure 2. The characteristic curve for the short ciruit test of a three-phase transformer

Both of them have no reason to change when the current Isc is varied, because: ¾ ¾ the Re could be modified only as a consequence of a winding temperature variation. The test must prevent significant heating from being generated. ¾ ¾ the Xe is only generated by the core flux. Therefore, its value is surely constant.

Being Ze constant, the graph Vsc = f(Isc) will have to be a straight line crossing the axes origin.

2. The power Psc measured during the test represents the total copper losses. It includes the measured Joule effect losses due to the winding resistances and the additional losses due to the eddy currents induced, by the leakage flux, in the mass of both the windings and the sur- rounding conductive materials.

3. The function cosφSC = f(Isc) must be a constant. In fact, being constant the equivalent para-

meters RS and XS of the transformer, also will be constant when Isc is varied:

cosφSC = RS / ZS

Setup and connection diagram

Figure 3 shows the schematic diagram of the short circuit test, where the three-phase tran- sformer is supplied from the AC three-phase variable section of the power supply DL 1013M2 (0÷240V/8A). The transformer parameters are measured with the measuring module DL 10065N. The schematic diagram from figure 3 is close to the student’s theore- tical knowledge (it contains the classical electrical symbols). We invi- te you to use this diagram while performing the experiment, having the wiring diagram from figure 4 as a reference.

7 Figure 3. Schematic diagram of the short circuit test of the three-phase transformer Follow the diagram below to connect the power cables:

Figure 4. Wiring diagram of the short circuit test for the three-phase transformer Before starting any wiring activity, check all the power connections: all switches must be OFF.

Do not forget to connect the ground terminal! As shown in the dia- gram with specific symbols, all the equipment is connected to the protective network with a dedicated connector and cable. Experimental procedure and learning plan Before starting the experiment, connect all the modules to the main power supply using the supply cables. Perform the circuit configuration shown in the wiring diagram in figure 4. Power ON the DL 10065N measuring device.

8 Follow the next steps to enable and prepare the DL 1013M2 power supply for use:

¾¾ Raise up all the switches on the power supply.

¾¾ Turn the key clockwise. from position 0 to 1. ¾¾ Switch the selector "a0b" to position "b". We will use the AC part (0÷240V/8A) of the power supply. This action is necessary in order to start the power supply ¾¾ Press the green “start” button on the power supply module.

Before using the power supply, make sure that the safety connector (dongle) K1 is installed on the DL 1013M2 module (see figure 4).

Supply voltage to the three-phase transformer (DL1080) using the DL 1013M2 power supply.

Make sure that the knob of the power sup- ply is turned counterclockwise at “0” posi- tion and the main switch to “b”. Switch the selector of DL 1013M2, corresponding to the variable AC voltage "L1L2L3/ 0÷240V•8A", from off (O) to the on (I) position.

Perform the measurements starting from the high current values, with a certain speed between measurements, to avoid a possible thermal jump due to the nature of the test, so that the temperature remains approximately constant throughout the experiment.

Gradually increase the voltage and, while adjusting the knob, read the voltmeter V, using the DL 10065N module. For each voltage, measure the corresponding input currents and powers through the ammeter A and the wattmeter W (use the arrows from the front panel of the DL 10065N module to switch between voltage, current and power).

9 Table 1. Measured short-circuit values of the three-phase transformer Calculate the power factor using the following formula and compare the result with the power factor measured using the DL 10065N module:

When the experiment is completed, turn off the power supply and switch all the selectors to “off”, the “a0b” to position zero and turn the knobs fully-counterclockwise to the zero position.

1.4 QUESTIONS Answer the following questions related to the experiment.

1. Explain about the short circuit test on transformers. 2. Why is the transformer rated in kVA and not in kW? 3. How are the copper losses determined in a transformer?

10 1.5 CONCLUSIONS From the no load and the short circuit transformer tests, it can be seen that the copper loss of a transformer depends on the current, while the iron loss depends on the voltage. Thus, the total transformer loss depends on volt-ampere (VA). It does not depend on the phase angle between voltage and current (the transformer loss is independent on the load power factor). This is the reason why transformers are rated in kVA.

The short circuit test is used to determine the values Re and Xe of the series branch of the equi- valent circuit. These impedances are usually very low, but they appear higher in value when referred to the high voltage side. This test is consequently performed on the high voltage side of the transfor- mer in order to keep the current drawn by these impedances at an acceptable level.

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Open circuit and Short circuit Test on transformer

These two transformer tests are performed to find the parameters of equivalent circuit of transformer and losses of the transformer . Open circuit test and short circuit test on transformer are very economical and convenient because they are performed without actually loading of the transformer.

Open circuit or No load test on Transformer

Open circuit test or no load test on a transformer is performed to determine 'no load loss (core loss)' and 'no load current I 0 '. The circuit diagram for open circuit test is shown in the figure below.

Usually high voltage (HV) winding is kept open and the low voltage (LV) winding is connected to its normal supply. A wattmeter (W), ammeter (A) and voltmeter (V) are connected to the LV winding as shown in the figure. Now, applied voltage is slowly increased from zero to normal rated value of the LV side with the help of a variac. When the applied voltage reaches to the rated value of the LV winding, readings from all the three instruments are taken.

The ammeter reading gives the no load current I 0 . As I 0 itself is very small, the voltage drops due to this current can be neglected.

The input power is indicated by the wattmeter (W). And as the other side of transformer is open circuited, there is no output power. Hence, this input power only consists of core losses and copper losses. As described above, no-load current is so small that these copper losses can be neglected. Hence, now the input power is almost equal to the core losses. Thus, the wattmeter reading gives the core losses of the transformer.

Sometimes, a high resistance voltmeter is connected across the HV winding. Though, a voltmeter is connected, HV winding can be treated as open circuit as the current through the voltmeter is negligibly small. This helps in to find voltage transformation ratio (K) .

The two components of no load current can be given as,

From this, shunt parameters of equivalent circuit of transformer (X 0 and R 0 ) can be calculated as

Short circuit or Impedance test on Transformer

The connection diagram for short circuit test or impedance test on transformer is as shown in the figure below. The LV side of transformer is short circuited and wattmeter (W), voltmere (V) and ammeter (A) are connected on the HV side of the transformer. Voltage is applied to the HV side and increased from the zero until the ammeter reading equals the rated current. All the readings are taken at this rated current.

The voltage applied for full load current is very small as compared to rated voltage. Hence, core loss due to small applied voltage can be neglected. Thus, the wattmeter reading can be taken as copper loss in the transformer.

Therefore, equivalent reactance of transformer can be calculated from the formula  Z eq 2 = R eq 2 + X eq 2 .

Why Transformers are rated in kVA?

From the above transformer tests, it can be seen that Cu loss of a transformer depends on current, and iron loss depends on voltage. Thus, total transformer loss depends on volt-ampere (VA). It does not depend on the phase angle between voltage and current, i.e. transformer loss is independent of load power factor. This is the reason that transformers are rated in kVA .

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Open and Short Circuit Test of Transformer

In this article, we will go through the Open and Short circuit tests of a Transformer. Open and short circuit tests are performed on a Transformer to evaluate its performance characteristics and parameters which mainly determine the Voltage regulation of the transformer, the Transformer’s efficiency, and the Equivalent circuit of the Transformer. They determine Iron and copper losses in the transformer. In this tutorial, we will go through the open and short circuit tests on a transformer with their circuit diagrams and an example and also we will tabulate the differences between the two.

Basic Terminologies

• Short circuit : An electrical connection in which current flows along a path that has very little resistance. This may result in high current flow, damage to the equipment, heating, etc.,
• Open circuit: It is an electrical connection in which there is a break in the path in which electric current flows. This may result in High resistance, no current flow, etc.,
• Transformer: An electrical device that is used to transfer electrical energy from one circuit to another without physical connection through electromagnetic induction. Now, let us learn open circuit tests and short circuit tests step by step with an example for each.

Open Circuit Test of Transformer

It is a method in electrical engineering used to determine the iron or core loss and it determines the exciting current of the transformer when it operates at rated voltage. The circuit diagram of this is shown below. In this, the secondary winding or the high voltage side is left open and the low voltage side is used to perform the test completely. Using this test we calculate no-load circuit parameters(X 0, R 0 ) and iron loss.

Open CIrcuit Transformer

In the circuit diagram, the voltmeter(V), ammeter (A) and wattmeter(W) were all connected on the low-voltage side of the transformer, which is supplied at rated voltage(V1). The secondary winding side is left open, so which a small amount of current(I 0 ) is flowing in primary winding. Here, I 0 is called as no – load current. The reading of the wattmeter gives the iron loss.

The iron loss of transformer is given by (P i ),

where (P i ) = V 1 I 0 cosΦ 0 (where, cosΦ 0 is called as no – load power factor. )

True power(P) is given as , ( P) = E 2 p / R 0

so, no – load resistance (R 0 ) = E 2 P /P = V 1 / I 0

Apparent power(S) is given as , (S) = E P I 0

also, S = sqrt(P 2 + Q 2 ) (where P=true power, Q=reactive power )

so, Q = sqrt((E P I 0 ) 2 – P 2 ——— eq1)

no- load reactance (X 0 ) can be calculated as,

(X 0 ) = E 2 P / Q

Example: An open circuit test is conducted on a transformer and noted the following recordings E S = 57V, E P = 110V , I 0 = 150mA , P= 9.0W, Calculate the no load resistance(R 0 ) and no load reactance(X 0 )

Given, R 0 ,I 0 ,E p , E s , P now, according to the formula R 0 = E 2 P / P so, R 0 = (110V) 2 / 9 => 1.35 kΩ now, on substituting the required values in eq1 , we have sqrt((110V*150mA) 2 -(9W) 2 = 14.41vars and, X 0 = E 2 P / Q = (110v) 2 / 14.41vars = 840Ω

Short Circuit Test of Transformer

It is a method in electrical engineering used to determine the copper or winding loss and also it determines the impedance of the transformer . The circuit diagram of this is shown below. In this, the secondary winding (low voltage side is shorted by a thick conductor) and on primary side(high voltage side) Ammeter(A), Voltmeter(V) and wattmeter(W) are connected.

Transformer

so, generally short circuit test is performed on high winding side and short circuited on low winding side. It will read copper losses because of high current passing through windings.

we have Power losses ( W s. c ) = I 2 s. c * R e Hence, R e = W s c / I 2 s.c we know that, Impedance( Z e ) = V s .c / I s. c Reactance(X e ) = sqrt((impedance) 2 – (resistance) 2 )

A short circuit test is conducted on a transformer and noted the following recordings V s .c = 7V , I s .c = 1.5mA , W s .c = 4W, Calculate the Reactance(X e ) , resistance(R e ) ?

Given that, V s. c = 7V , I s. c = 1.5mA , W s . c = 4W now, resistance ( R e ) = W s. c / I 2 s. c so, Resistance (R e ) = 1.8Ω Impedance(Z e ) = V s .c / I s. c => 7/1.5 Now, Reactance( X e ) = sqrt((impedance) 2 – (resistance) 2 ) X e = 1.46Ω

Differences Between Open Circuit Test and Short Circuit Test

Both the open circuit test and short circuit test are done to measure the efficiency of the transformer through various parameters in each case, voltage regulation. In this article, we have covered this topic in detail. We have covered Open CIrcuit Test and Short Circuit Test both separately with proper example.

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open circuit and short circuit test on single phase transformer

To conduct parameters and losses in a single phase transformer by open circuit and short circuit test on single phase transformer

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Mfad-rtdetr: a multi-frequency aggregate diffusion feature flow composite model for printed circuit board defect detection.

1. Introduction

2. methodology, 2.1. rtdetr network, 2.2. the proposed mfad-rtdetr model, 2.2.1. the dfr module, 2.2.2. vss block, 2.2.3. multi-frequency fusion module, 2.2.4. dattention module, 2.2.5. drbc3 module, 2.2.6. mfad-feature composite paradigm, 2.2.7. the loss function wiou, 3. experiment and result analysis, 3.1. experimental environment and data preprocessing, 3.2. evaluation metrics, 3.3. experimental result analysis, 3.3.1. comparison experimental analysis, 3.3.2. ablation experiments, 3.3.3. visual analysis of experimental results, 4. conclusions, author contributions, data availability statement, acknowledgments, conflicts of interest.

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Xie, Z.; Zou, X. MFAD-RTDETR: A Multi-Frequency Aggregate Diffusion Feature Flow Composite Model for Printed Circuit Board Defect Detection. Electronics 2024 , 13 , 3557. https://doi.org/10.3390/electronics13173557

Xie Z, Zou X. MFAD-RTDETR: A Multi-Frequency Aggregate Diffusion Feature Flow Composite Model for Printed Circuit Board Defect Detection. Electronics . 2024; 13(17):3557. https://doi.org/10.3390/electronics13173557

Xie, Zhihua, and Xiaowei Zou. 2024. "MFAD-RTDETR: A Multi-Frequency Aggregate Diffusion Feature Flow Composite Model for Printed Circuit Board Defect Detection" Electronics 13, no. 17: 3557. https://doi.org/10.3390/electronics13173557

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