Page 63 - Solutions Manual to accompany Electric Machinery Fundamentals

N N  N
I  I  I  I  C I  SE C I
L
SE
C
C
N SE C N SE C
N
or I  SE I
C
L
N SE  N C
so the input voltage can be expressed in terms of the input current as:
N
V  I Z  SE I Z
eq
C
L
N SE  N C L eq
The input impedance of the autotransformer is defined as Z eq  V L I / L , so
 V N
Z  L  SE Z
eq
I L N SE  N C eq
This is the expression that we were trying to prove.
2-17. A 10-kVA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a
120-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.
(a) Sketch the transformer connection that will do the required job.
(b) Find the kilovoltampere rating of the transformer in the configuration.

(c) Find the maximum primary and secondary currents under these conditions.
SOLUTION (a) For this configuration, the common winding must be the smaller of the two windings, and
N SE  4N . The transformer connection is shown below:
C
+

N
SE

+
600 V
N
C
120 V

- -

(b) The kVA rating of the autotransformer can be found from the equation
N  N 4N  N
S  SE C S  C C  10 kVA  12.5 kVA
IO
N SE W 4N C
(c) The maximum primary current for this configuration will be
S 12,500 VA
I  P   20.83 A
V P 600 V

and the maximum secondary current is
S 12,500 VA
I    104 A
S
V S 120 V
2-18. A 10-kVA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a
480-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.

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