Page 858 - Basic Electrical Engineering

the loss component, I of the primary current, I which corresponds to the
0
c
losses in the core of the transformer. For the purpose of analysis, we will
draw the simplified phasor diagram of a transformer having the core flux, φ
as the reference axis as shown in Fig. 11.31.

Figure 11.31 Simplified phasor diagram of a CT

Actually, nI must be equal to I as in Fig. 11.31. However, due to the
2
1
presence of the no-load current, I , I is somewhat away from nI . (Note that
1
2
0
in Fig. 11.31, for clarity I has been shown enlarged. Infact, I is only 3 to 5
0
0
per cent of I .)
1
The vector sum of nI and I will give the value of I as shown. In the
1
2
0
figure, OC is equal to I . We extend nI as shown by the dotted line. We draw
2
1
a perpendicular from C on this extended line to meet at point B. OA is equal
to nI and ac is equal to I . The vertical line drawn downwards in E . E lags
2
2
0
2
the flux, ϕ by 90°.
Current flowing through the secondary winding due to E is I . I lags E 2
2
2
2
by an angle δ.
The angle between nI and I is θ.
2
1
The angle θ actually is very small, and hence OC can be considered equal
to OB.

Therefore