This preview shows page 1. Sign up to view the full content.
polar form
arg I
0
( 29
69.897

deg
=
I
0
5.962A
=
rectangular form
I
0
2.049
5.599j

A
=
I
0
I
fe
I
m
+
=
I
m
5.599j

A
=
I
m
V
prated
jX
mls
=
I
fe
2.049A
=
I
fe
V
prated
R
fels
=
b.
The primary side is the low side and the core parameters are given on the low side.
Using the cantilever circuit we can find the exciting current directly
Z
eq
0.026
0.038j
+
Ω
=
Z
eq
R
p
j X
p
⋅
+
( 29
R
s
j X
s
⋅
+
( 29
a
2
⋅
+
=
a.
The equivalent winding impedance referred to the low side is found by referring
everything to the primary side in this case:
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: a 0.433 = a V prated V srated = First find the turns ratio Solution X mls 37.15 Ω ⋅ = X s 0.10 Ω ⋅ = R s 0.07 Ω ⋅ = R fels 101.5 Ω ⋅ = X p 0.0189 Ω ⋅ = R p 0.0132 Ω ⋅ = V srated 480 volt ⋅ = V prated 208 V ⋅ = S rated 75 kVA ⋅ = Problem 512: Solution kVAR kW ≡ kVA kW ≡...
View
Full
Document
This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University.
 Spring '12
 Piller

Click to edit the document details