Page 37 - Solutions Manual to accompany Electric Machinery Fundamentals

R  eq 0.20  R  C 300 

X  0.80  X  100 
eq M
SOLUTION The equivalent circuit of this transformer is shown below. (Since no particular equivalent
circuit was specified, we are using the approximate equivalent circuit referred to the primary side.)

The secondary voltage and current are
282.8


V   0 V   200 0 V
S
2
7.07

I S    36.87 A   5  36.87 A
2
The secondary voltage referred to the primary side is



S  V aV S  100 0 V
The secondary current referred to the primary side is
 I I S  10  36.87 A

S
a
The primary circuit voltage is given by
 
 V  V I  R  jX 
P S S eq eq


 V 100 0 V   10  36.87 A 0.20   j 0.80   106.5 2.8 V  
P
The excitation current of this transformer is




106.5 2.8 V 106.5 2.8 V


 I  I I   0.355 2.8   1.065   87.2
EX C M 300  j 100 

 I 1.12   68.8 A
EX
Therefore, the total primary current of this transformer is


 I  I I 10  36.87  1.12  68.8  11.0  40.0 A
P S EX
The voltage regulation of the transformer at this load is
V  aV 106.5 100

VR  P S  100%   100% 6.5%

aV S 100
The input power to this transformer is
31