Page 554 - Basic Electrical Engineering

shown as equal. Show two phasors E and E lagging flux ϕ by 90 °. Draw –
1
2
E in opposite direction to E , as has been shown in Fig. 6.16. Draw I 2
1
1
lagging E by a certain angle. Draw I′ opposite to I . Draw I lagging (–E )
1
o
1
2
1
by a large angle. Add I′ and I vectorially to get I . Now we have to locate
1
1
o
V and V . We apply Kirchhoff’s law in the primary winding circuit and
2
1
write V – I R – jI X – E = 0
1
1
1
1
1
1
or,

V = E + I R + jI X (i)
1
1
1
1
1
1
Similarly, for the secondary circuit we write E – I R – jI X – V = 0
2
2
2
2
2
2

V = E − I I − jI X (ii)
2
2 2
2
2
2
Using equation (i) develop the phasor diagram step by step.

Figure 6.15 Approximate equivalent circuit of a transformer with secondary parameters referred to
the primary side