Tests on a transformer:-

• Tests on a transformer are done to calculate efficiency and regulation of a transformer at any load and at any power factor . Depending upon how the tests are performed the tests on a transformer are divided into two categories. They are

1. Direct loading test:-

• In this test load is directly connected to the secondary of the transformer and various meters are connected to primary and secondary windings of a transformer. Tha advantage of this method is results are accurate. The disadvantage of this method is for large rating transformers suitable load is difficult to obtain in laboratories. Power losses are large during the test. Hence this method is suitable for small transformers and rarely used.

1. Indirect loading test.

• In this method the actual load is not used on transformer. The equivalent circuit parameters of a transformer are determined by conducting two tests on a transformer. They are

1. Open circuit test(O.C Test)
2. Short circuit test(S.C Test)

• Tha advantage of this method is without much power loss the tests are conducted and results are effective.

Open circuit test(O.C Test):-

• In this test primary winding of the transformer is connected to a.c supply through ammeter & wattmeter. Secondary winding of the transformer is kept open. In this test low votage side is used as primary and high votage side is used as secondary. The circuit diagram of O.C. test is as follows.

• When the primary is connected to the rated voltage then the wattmeter measures input power i.e. P0, ammeter measures input current i.e. I0 and the voltmeter measures primary voltage i.e. V1.
• As secondary is open, I2=0 and its reflected current on primary i.e I2' is also zero. Thus primary current I1=I0.
• As I2=0, secondary copper losses are zero. As the transformer no load current is very small and as I1=I0, the primary current is also low which causes low copper losses on primary. Thus total copper losses in O.C. test are negligible.
• When input voltage is applied at rated frequency then there will be maximum flux density in the core which causes iron losses. As output power is zero and copper losses are negligible the total input power is used to suppy iron losses. Hence P0=Pi=Iron losses.
• We know

$P_o = V_o I_o \cos \phi$

• Thus no load power factor is given by

$\cos \phi =\dfrac { P_o} { V_o I_o} $

• After finding no load power factor then the two components of load current is caluculated as follows

$I_c = I_o \cos\phi_o \\ I_m= I_o \sin\phi_o$

• After finding  Ic & Im then the circuit parameters i.e Ro & Xo are caluculated as follows

$R_o = \dfrac{V_o }{I_c} \Omega \\ X_o = \dfrac{V_o }{I_m} \Omega$

• If the meters are connected on secondary and primary is kept open then O.C. test gives Ro' and Xo' . After knowing transformer ratio K we can calulate Ro and Xo.