Open circuit and short circuit test of a single phase transformer


To find the equivalent circuit parameters and different losses of a single phase transformer.



This test is used to determine the core (or iron or excitation) loss, Pi, and no-load current I0, and the shunt branch parameters R0 and X0 of the equivalent circuit.

In this test the high voltage winding is kept open and the rated voltage at rated frequency is applied to the other winding as shown in fig.1. The wattmeter connected in the primary (LV) will then read the total hysteresis and eddy current loss (W0) in the single phase transformer. Since the primary no load current is small, the copper losses due to it can be neglected. No copper loss will occur at secondary because the secondary current is zero. If the supply voltage is V1 volts,

V1 I0 cosØ0 = W0

The open circuit test gives enough data to compute the equivalent circuit constants R0, X0, no load power factor cosØ0 , no load current I0, and the iron losses of a transformer.

Iron loss, Pi = Input power on no load= W0 (watt)

No load current = I0 amperes

Applied voltage to primary = V1 volts.

Angle of lag Ø0 = cos-1

No load working component of current, Ie = I0 cosØ0 =

No load magnetizing component of current, Im =

Equivalent circuit parameter, R0 =

Equivalent circuit parameter, X0 =

Circuit Diagram:



The purpose of this test is to determine full load copper loss and the equivalent resistance and reactance referred to metering side.

In this test, the terminals of secondary winding (Low voltage side) are short circuited through an ammeter and a variable low voltage is applied to the primary side through a variac, as shown in fig-2. The transformer now becomes equivalent to a coil having an impedance equal to impedance of both the windings.

The applied voltage, V1 to the primary is gradually increased till the ammeter A indicates the full load (rated) current of the metering side. Since applied voltage is very low (5-8% of the rated voltage) so the flux linking with the core is very small and therefore, iron losses are so small that these can be neglected. Thus, the power input gives total copper loss at rated load, the output being nil. Let the readings of voltmeter, ammeter and wattmeter be V1, I1 and W1 respectively.

Full load copper loss, Pc= Req= W1

Equivalent resistance, Req = W1/I12

Equivalent impedance, Zeq =V1/I1

Equivalent reactance, Xeq(Zeq2Req2)

Circuit Diagram:

Devices under test:

Apparatus Required:


(Open circuit test):

  • The circuit is connected as shown in the circuit diagram (fig. 1).
  • Maintaining the supply frequency constant, the applied voltage is varied to the rated normal value by means of a Variac and the reading of primary current, input power, voltage impressed and secondary voltage are noted.

(Short Circuit Test):

  • The circuit is connected as shown in the circuit diagram (fig. 2). Here the LV winding is short circuited by a connection having negligible resistance and good contacts, preferably by an ammeter.
  • Maintaining the supply frequency constant, the applied voltage is varied so that full load rated current flows and the readings of current, power, voltages are noted.

 Experimental Data:


  • The open circuit test is performed at rated voltage and frequency.
  • For measuring power at no load, a wattmeter of low power factor should be used.
  • During short circuit test, the supply voltage should be applied through a variac and voltage is increased very slowly from its low value, so that rated current flows through the secondary. The current should not exceed the rated value otherwise it would damage windings.
  • All connection must be clean and tight.




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