0195136047.pdf

11.6 AUTOTRANSFORMERS 493

(b) Determine the kVA rating as an autotransformer, and find how much of that is the

conducted kVA.

(c) Based on the data given for the two-winding transformer in Example 11.4.2, compute the

efficiency of the autotransformer corresponding to full load, 0.8 power factor lagging,

and compare it with the efficiency calculated for the two-winding transformer in part (a)

of Example 11.4.2.

+

−

+ VH

−

VX

IX

IC

240 V

2400 V

IH Figure E11.6.1

Solution

(a) The two windings are connected in series so that the polarities are additive. Neglecting

the leakage-impedance voltage drops,

VH= 2400 + 240 =2640 V

VX=2400 V

The full-load rated current of the 240-V winding, based on the rating of 50 kVA as a two-

winding transformer, is 50, 000 / 240 = 208 .33 A. Since the 240-V winding is in series

with the high-voltage circuit, the full-load current of this winding is the rated current on

the high-voltage side of the autotransformer,

IH= 208 .33 A

Neglecting the exciting current, the mmf produced by the 2400-V winding must be

equal and opposite to that of the 240-V winding,

IC= 208. 33

(

240

2400

)

= 20 .83 A

in the direction shown in the figure. Then the current on the low-voltage side of the

autotransformer is given by

IX=IH+IC= 208. 33 + 20. 33 = 229 .16 A

(b) The kVA rating as an autotransformer is