Electrical Machines I
6
Prof. Krishna Vasudevan, Prof. G. Sridhara Rao, Prof. P. Sasidhara Rao
Phasor diagrams
r1
I1
r’2
jxl1
jx’l2
Io
Ic
Im
Rc
V1
jXm
V’2
Z’L
R
jX
(a)
I1
Ic
V1 Rc
r1
I’2
jxl1
r’2
jx’l2
I1
I’2
R=r1+r’2
Io
Im
Z’L
jxm
V’2 x=xl1+x’l2
V’2
V1
I1=I’2
(b)
(c)
Figure 16: Exact,approximate and simplified equivalent circuits
The resulting equivalent circuit as shown in Fig. 16 is known as the exact
equivalent circuit. This circuit can be used for the analysis of the behavior of the transformers. As the no-load current is less than 1% of the load current a simplified circuit known
as ‘approximate’ equivalent circuit (see Fig. 16(b)) is usually used, which may be further
41
Indian Institute of Technology Madras
Electrical Machines I
Prof. Krishna Vasudevan, Prof. G. Sridhara Rao, Prof. P. Sasidhara Rao
simplified to the one shown in Fig. 16(c).
On similar lines to the ideal transformer the phasor diagram of operation can be
drawn for a practical transformer also. The positions of the current and induced emf phasor
are not known uniquely if we start from the phasor V1 . Hence it is assumed that the phasor
φ is known. The E1 and E2 phasor are then uniquely known. Now, the magnetizing and loss
components of the currents can be easily represented. Once I0 is known, the drop that takes
place in the primary resistance and series reactance can be obtained which when added to
E1 gives uniquely the position of V1 which satisfies all other parameters. This is represented
in Fig. 17(a) as phasor diagram on no-load.
Next we proceed to draw the phasor diagram corresponding to a loaded transformer.
The position of the E2 vector is known from the flux phasor. Magnitude of I2 and the load
power factor angle θ2 are assumed to be known. But the angle θ2 is defined with respect
to the terminal voltage V2 and not E2 . By trial and error the position of I2 and V2 are
determined. V2 should also satisfy the Kirchoff’s equation for the secondary. Rest of the
construction of the phasor diagram then becomes routine. The equivalent primary current
′
I2 is added vectorially to I0 to yield I1 . I1 (r1 + jxl1 )is added to E1 to yield V1 . This is shown
in fig. 17(b) as phasor diagram for a loaded transformer.
42
Indian Institute of Technology Madras
Electrical Machines I
Prof. Krishna Vasudevan, Prof. G. Sridhara Rao, Prof. P. Sasidhara Rao
V1
IoX l1
Ior1
E1
E2
Io
φ
Il
Im
φ
(a)No-load
V1
I1X l1
E1
I1r1
E2
I2x2 I r
2 2
I2
V2
I’2
Il
Io
φ
(b)On-load
Figure 17: Phasor Diagram of a Practical Transformer
43
Indian Institute of Technology Madras
φ