The O.C test and S.C. test of single phase transformer is done in order to find all the parameter of transformer. It is also done to find the iron loss and cu loss occuring in x mer at any load.

** **

**Open Circuit Test (O.C. Test)**

The experimental circuit to conduct O.C test is shown in the figure.

Fig 1. Experimental circuit for O.C. test |

_{2 }= E

_{2 }when primary is supplied with rated voltage. As voltmeter resistance is very high, though voltmeter is connected, secondary is treated to be open circuit as voltmeter current is always negligibly small.

The observation table is as follows

V_{o }= Rated voltage

W_{o }= Input power

I_{o }= Input current = no load current

_{o}. The two components of this no load current are,

I_{m }= I_{o }sin Φ_{o}

I_{c }= I_{o }cos Φ_{o}

where cos Φ_{o} = No load power factor

And hence power input can be written as,

W_{o }= V_{o }I_{o }cos Φ_{o}

The phasor diagram is shown in the Fig. 2.

Fig. 2 |

_{2 }= 0. Thus its reflected current on primary is also zero. So we have primary current I

_{1 }=I

_{o}. The transformer no load current is always very small, hardly 2 to 4 % of its full load value. As I

_{2 }= 0, secondary copper losses are zero. And I

_{1 }= I

_{o }is very low hence copper losses on primary are also very very low. Thus the total copper losses in O.C. test are negligibly small. As against this the input voltage is rated at rated frequency hence flux density in the core is at its maximum value. Hence iron losses are at rated voltage. As output power is zero and copper losses are very low, the total input power is used to supply iron losses. This power is measured by the wattmeter i.e. W

_{o}. Hence the wattmeter in O.C. test gives iron losses which remain constant for all the loads.

**. ^{.}. **W

_{o }= P

_{i}

_{ }= Iron losses

Calculations : We know that,

W

_{o }= V

_{o }I

_{o }cos Φ

cos Φ

_{o}= W

_{o }/(V

_{o}I

_{o }) = no load power factor

Once cos Φ

_{o}is known we can obtain,

I

_{c }= I

_{o }cos Φ

_{o}

and I

_{m }= I

_{o }sin Φ

_{o}

Once I

_{c }and I

_{m }are known we can determine exciting circuit parameters as,

R

_{o }= V

_{o }/I

_{c }Ω

and X

_{o }= V

_{o }/I

_{m }Ω

**Key Point**: The no load power factor cos Φ

_{o}is very low hence wattmeter used must be low power factor type otherwise there might be error in the results. If the meters are connected on secondary and primary is kept open then from O.C. test we get R

_{o}

**‘**and X

_{o}

**‘**with which we can obtain R

_{o }and X

_{o }knowing the transformation ratio K.

**Short Circuit Test (S.C. Test)**

In this test, primary is connected to a.c. supply through variac, ammeter and voltmeter as shown in the Fig. 3.

Fig. 3 Fig 1. Experimental circuit for O.C. test |

**. ^{.}. **W

_{sc }= (P

_{cu}) F.L. = Full load copper loss

Calculations : From S.C. test readings we can write,

W

_{sc }= V

_{sc }I

_{sc }cos Φ

_{sc }

**.**cos Φsc = V

^{.}._{sc }I

_{sc }/W

_{sc }= short circuit power factor

W

_{sc }= I

_{sc}

^{2 }R

_{1e }= copper loss

**.**R

^{.}._{1e }=W

_{sc }/I

_{sc}

^{2}

while Z

_{1e }=V

_{sc }/I

_{sc }= √(R

_{1e}

^{2}

_{ }+ X

_{1e}

^{2})

**.**X

^{.}._{1e}

^{ }= √(Z

_{1e}

^{2}– R

_{1e}

^{2})

_{1e}, X

_{1e}

^{ }and Z

_{1e}. Knowing the transformation ratio K, the equivalent circuit parameters referred to secondary also can be obtained.

**Important Note**: If the transformer is step up transformer, its primary is L.V. while secondary is H.V. winding. In S.C. test, supply is given to H.V. winding and L.V is shorted. In such case we connect meters on H.V. side which is transformer secondary through for S.C. test purpose H.V side acts as primary. In such case the parameters calculated from S.C. test readings are referred to secondary which are R

_{2e}, Z

_{2e }and X

_{2e}. So before doing calculations it is necessary to find out where the readings are recorded on transformer primary or secondary and accordingly the parameters are to be determined. In step down transformer, primary is high voltage itself to which supply is given in S.C. test. So in such case test results give us parameters referred to primary i.e. R

_{1e}, Z

_{1e }and X

_{1e}.

**Key**

**point**: In short, if meters are connected to primary of transformer in S.C. test, calculations give us R

_{1e }and Z

_{1e}if meters are connected to secondary of transformer in S.C. test calculations give us R

_{2e}and Z

_{2e}.

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

We know that,

From O.C. test, W_{o }= P_{i }

From S.C. test, W_{sc }= (P_{cu}) F.L.

Thus for any p.f. cos Φ_{2} the efficiency can be predetermined. Similarly at any load which is fraction of full load then also efficiency can be predetermined as,

where n = fraction of full load

where I_{2}= n (I_{2}) F.L.

**Calculation of Regulation**

From S.C. test we get the equivalent circuit parameters referred to primary or secondary.

_{1}, V

_{2}and rated currents (I

_{1}) F.L. and (I

_{2}) F.L. are known for the given transformer. Hence the regulation can be determined as,

where I_{1}, I_{2} are rated currents for full load regulation.

For any other load the currents I_{1}, I_{2} must be changed by fraction n.

**. ^{.}. **I

_{1}, I

_{2}at any other load = n (I

_{1}) F.L., n (I

_{2}) F.L.

**Key**

**Point**: Thus regulation at any load and any power factor can be predetermined, without actually loading the transformer

thnx for giving me information about oc and sc test..

ur most welcome