Experiment No: 2
Open circuit and short circuit tests on single phase transformer 1 Aim • To understand the basic working principle of a transformer. • To obtain the equivalent circuit parameters from OC and SC tests, and to estimate eﬃciency & regulation at various loads.
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 = N1 ωΦm cos ωt, The r.m.s. value of this voltage is given by: E1 = 4.44F N1 Φm 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 N2 Φm where Φm 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
ω = 2πF
Note that in actual practice, φp = 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.
In a practical transformer a very large proportion of the primary and secondary ﬂux paths are common and leakage ﬂuxes are comparatively small. Therefore φp ≈ φs = φmutual and therefore Φm ≈ Φm . If in addition, winding resistances are neglected – being usually small in a practical transformer, then V 1 ≈ E1 Similarly, V 2 ≈ E2 Although the iron core is highly permeable, it is not possible to generate a magnetic ﬁeld in it without the application of a small m.m.f.(magneto-motive force), denoted by Mm . Thus even when the secondary winding is open circuited, a small magnetizing current (im ) is needed to maintain the magnetic ﬂux. The current of the primary circuit on no-load is of the order of 5% of full load current. Also, the pulsation of ﬂux in the core is productive of core loss, due to hysteresis and eddy currents. These losses are given by: 1.6 Ph = Kh Bmax F, 2 Pe = Ke Bmax F 2
and Pc = Ph + Pe
where Ph , Pe and Pc are hysteresis, eddy current and core losses respectively, Kh and Ke are constants which depend on the magnetic material, and Bmax is the maximum ﬂux density in the core. These losses will remain almost constant if the supply voltage and frequency are held constant. The continuous loss of energy in the core requires a continuous supply from the electrical source to which the primary is connected. Therefore, there must be a current component ic which accounts for these losses. It should be noted that magnetizing current (im ) and core loss component of current(ic ) are in phase quadrature. The resultant of these two currents is the no-load current io . Generally the magnitude of this current is very small compared to that of the rated current of the transformer ( may be of the order of 5% of the rated). This current makes a phase angle ζo of the order of (cos −1 (0.2)) with the applied voltage. If a load of ﬁnite impedance is connected across the second coil, a current i2 will ﬂow through it. This tends to alter the mmf and...
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