Page 579 - Basic Electrical Engineering

The circuit diagram and the phasor diagram showing the currents with
reference to supply voltage, V have been shown in Fig. 6.25. I is the no-
o
1
load current making an angle of lag ϕ with V where cos ϕ = 0.2 or ϕ = 78°
0
o
1
0
(lagging). The load power factor, cos ϕ = 0.8 or, ϕ = 37°. I is the load
2
2
2
current. I′ is the additional current drawn by the primary to balance the load
1
current I such that
2

I′ N = I N 2
2
1 1
substituting values

As shown in Fig. 6.25, it is observed that the phasor sum of I and I′ is the
o
1
primary current when the transformer is loaded. The angle between I and I′ 1
o
is ϕ − ϕ , i.e., (78° − 37°) = 41°.
0
2
Using law of parallelogram,

substituting values

Example 6.9 A 400/200 V, 50 Hz, 10 kVA transformer has primary and
secondary winding resistances of 2.5 Ω and 0.5 Ω and winding leakage
reactances of 5 Ω and 1 Ω, respectively. Calculate the equivalent resistance

and reactance of the transformer referred to the secondary side. What amount
of power will be lost in the windings?