In this topic, you study Transformer on Load.
Let the equivalent circuit shown in Fig. 3.15 represent the actual transformer. When the primary winding of a transformer is connected across the ac supply of voltage with no load on the secondary side (i.e. with secondary winding open circuited), as already mentioned earlier, its primary winding draws a small current from the supply (Fig. 3.13). This current has two components. The magnetizing component sets up mutual flux which is in time phase with it. This flux induces emfs and in the primary and secondary windings. The other component of i.e. core-loss component provides the core losses.
Now, if the load is put on the transformer by connecting an impedance of finite value across its secondary winding, a current flows through the secondary winding causing voltage drops in its resistance and leakage reactance . Hence the secondary terminal voltage under this condition is given by
E, – 12 E, -12 Z,
(3.13)
The phase relationship between the secondary current and the secondary terminal voltage is determined by the characteristics of the load. To neutralize the demagnetizing effect of secondary current , primary draws additional current in anti-phase with such that,
Thus, the primary current Il on load can be considered as having two components 12 and . Hence,
This primary current Il causes voltage drops in the resistance and leakage reactance of the primary winding. The primary impressed voltage is therefore given by the phasor sum of (part of the applied voltage balancing ) and the drops and i.e. The cosine of the angle between and gives the power factor on the primary side of the transformer. The phasor diagram for the transformer on load can therefore be constructed as follows
Assume first the load to be inductive in nature and having a power factor of . Starting with voltage , (i.e. taking it as reference phasor), is drawn lagging behind by The magnitude of is of course decided by the load impedance. The phasor is drawn in phase with the current and the phasor leading it by 90o. The phasor sum of V2, and is . The emf is in time phase with . The load component of the primary current to neutralize the demagnetizing effect of is drawn opposite to . The total current Fig. 3.16 : Phasor diagram of a transformer with inductive load in the primary is obtained by the phasor sum of the no-load currents and . The current
is drawn such that its magnitude is equal to the current taken by the primary when the transformer is on no load. The angle of lag of behind (i.e. ) is assumed to be equal to the angle between the primary impressed voltage and the no-load primary current (i.e. when is zero). The Resistance and reactance drops and drawn in phase with the current and leading by 90o respectively. Finally, the primary impressed voltage is obtained from the phasor sum of , and . Fig. 3.16 shows the complete phasor diagram of a transformer. If is the phasor difference between and then obviously the power factor of the primary side of the transformer is . The same method can be used for constructing the phasor diagram for the secondary load of any nature.
Figs. 3.17 (a) and (b) show the phasor diagrams for the resistive (with unity power factor)
and capacitive (with leading power factor) loads.
Fig. 3.17 : Phasor diagrams of a transformer with
(a) Resistive load, (b) Capacitive load
Incidentally, it should be noted that the no-load current is of small magnitude in comparison with the total primary current on full load. Therefore, if it be neglected,
Then from the Equation (3.14), we have
or,
i.e. the primary and secondary currents are inversely proportional to the respective turns.
