Electric Machinery (7th Edition, 2014 – Fitzgerald, Kingsley & Umans) | Solutions Manual with Step-by-Step Answers PDF

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SOLUTION MANUAL

, 1

PROBLEM SOLUTIONS: Chapter 1


Problem 1.1
Part (a):
lc lc
Rc = = =0 A/Wb
µAc µr µ0 Ac

g
Rg = = 1.017 × 106 A/Wb
µ0 Ac

part (b):

NI
Φ= = 1.224 × 10−4 Wb
Rc + Rg

part (c):

λ = N Φ = 1.016 × 10−2 Wb

part (d):

λ
L= = 6.775 mH
I
Problem 1.2
part (a):
lc lc
Rc = = = 1.591 × 105 A/Wb
µAc µr µ0 Ac

g
Rg = = 1.017 × 106 A/Wb
µ0 Ac

part (b):

NI
Φ= = 1.059 × 10−4 Wb
Rc + Rg

part (c):

λ = N Φ = 8.787 × 10−3 Wb

part (d):

λ
L= = 5.858 mH
I

,2

Problem 1.3
part (a):


Lg
N= = 110 turns
µ0 Ac

part (b):

Bcore
I= = 16.6 A
µ0 N/g

Problem 1.4
part (a):



L(g + lc µ0 /µ) L(g + lc µ0 /(µr µ0 ))
N= = = 121 turns
µ0 Ac µ0 Ac

part (b):

Bcore
I= = 18.2 A
µ0 N/(g + lc µ0 /µ)

Problem 1.5
part (a):




part (b):
3499
µr = 1 +  = 730
1 + 0.047(2.2)7.8

 
g + µ0 lc /µ
I=B = 65.8 A
µ0 N

, 3

part (c):




Problem 1.6
part (a):
   
NI Ag x
Hg = ; Bc = Bg = Bg 1 −
2g Ac X0
part (b): Equations

2gHg + Hc lc = N I; Bg Ag = Bc Ac
and
Bg = µ0 Hg ; Bc = µHc
can be combined to give
   
N I N I
Bg =     =    
Ag
2g + µµ0 Ac (l c + l p ) 2g + µ0
µ 1 − x
X0 (l c + l p )

Problem 1.7
part (a):
   
µ0
g+ µ (lc + lp )
I = B  = 2.15 A
µ0 N

part (b):
 
1199
µ = µ0 1 + √ = 1012 µ0
1 + 0.05B 8
   
µ0
g+ µ (lc + lp )
I = B  = 3.02 A
µ0 N