Page 43 - Solutions Manual to accompany Electric Machinery Fundamentals

SHORT CIRCUIT TEST (referred to the high-voltage or primary side):
17.1 V

 Z  R jX   1.97
EQ EQ EQ
8.7 A
P 38.1 W
  cos  1 SC  cos  1  75.2

VI  17.1 V 8.7 A 
SC SC

Z EQ  R EQ  jX EQ  1.97 75.2    0.503 j  1.905
R EQ  0.503 
X EQ  j 1.905 
To convert the equivalent circuit to the secondary side, divide each series impedance by the square of the
turns ratio (a = 230/115 = 2). Note that the excitation branch elements are already on the secondary side.
The resulting equivalent circuit is shown below:

R EQ,S  0.126  X EQ,S  j 0.476 
R CS  3383  X MS  1099 
,
,
(b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred
to the secondary side. The rated secondary current is
1000 VA
I   . 8 70 A
S
115 V
We will now calculate the primary voltage referred to the secondary side and use the voltage regulation
equation for each power factor.

(1) 0.8 PF Lagging:
  36 87 A

P V S  V Z EQ S  115 0 V   I  j  0.126 0.476  8.7   .



P  V 118.4 1.3 V
118.4-115

VR   100% 2.96%
115
(2) 1.0 PF:
  8.7 0.0 A

P V S  V Z EQ S  115 0 V   I  j  0.126 0.476   



P  V 116.2 2.04 V
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