Question

Open circuit and short circuit test of single phase/ three phase transformer and obtain its equivalent circuit.

Answer 1

Theory

The physical basis of the transformer is mutual induction between two circuits linked by a common

magnetic field. Transformer is required to pass electrical energy from one circuit to another, via the

medium of the pulsating magnetic field, as efficiently and economically as possible. This could be

achieved using either iron or steel which serves as a good permeable path for the mutual magnetic

flux. An elementary linked circuit is shown in Fig.1. The principle of operation of this circuit can be

explained as follows:

Let an alternating voltage v1 be applied to a primary coil of N1 turns linking a suitable iron core. A

current flows in the coil, establishing a flux φp in the core. This flux induces an emf e1 in the coil to

counterbalance the applied voltage v1. This e.m.f. is

e1 = N1

dφp

dt .

Assuming sinusoidal time variation of the flux, let φp = Φm sin ωt. Then,

e1 = N1ωΦm cos ωt, where ω = 2πF

The r.m.s. value of this voltage is given by:

E1 = 4.44F N1Φm

Now if there is a secondary coil of N2 turns, wound on the same core, then by mutual induction an emf

e2 is developed therein. The r.m.s. value of this voltage is given by:

E2 = 4.44F N2Φ

0

m

where Φ0

m is the maximum value of the (sinusoidal) flux linking the secondary coil (φs).

If it is assumed that φp = φs then the primary and secondary e.m.f.’s bear the following ratio:

e1

e2

=

E2

E1

=

N2

N1

Note that in actual practice, φp 6= φs since some of the flux paths linking the primary coil do not

link the secondary coil and similarly some of the flux paths linking the secondary coil do not link the

primary coil. The fluxes which do not link both the coils are called the “leakage fluxes” of the primary

and secondary coil.