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Solution Manual - Electric Machinery 6th Edition Fitzgerald Kingsley Uman
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Solution Manual - Electric Machinery 6th Edition Fitzgerald Kingsley Uman
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1PROBLEM SOLUTIONS: Chapter 1
Problem 1.1
Part (a):
Rc=lc
µA c=lc
µrµ0Ac=0 A / W b
Rg=g
µ0Ac=1.017×106A/Wb
part (b):
Φ=NI
Rc+Rg=1.224×10−4Wb
part (c):
λ=NΦ=1.016×10−2Wb
part (d):
L=λ
I=6.775 mH
Problem 1.2
part (a):
Rc=lc
µA c=lc
µrµ0Ac=1.591×105A/Wb
Rg=g
µ0Ac=1.017×106A/Wb
part (b):
Φ=NI
Rc+Rg=1.059×10−4Wb
part (c):
λ=NΦ=8.787×10−3Wb
part (d):
L=λ
I=5.858 mH 2
Problem 1.3
part (a):
N=/radicalBigg
Lg
µ0Ac= 110 turns
part (b):
I=Bcore
µ0N/g=1 6.6A
Problem 1.4
part (a):
N=/radicalBigg
L(g+lcµ0/µ)
µ0Ac=/radicalBigg
L(g+lcµ0/(µrµ0))
µ0Ac= 121 turns
part (b):
I=Bcore
µ0N/(g+lcµ0/µ)=1 8.2A
Problem 1.5
part (a):
part (b):
µr=1+3499
/radicalbig
1+0.047(2.2)7.8=7 30
I=B/parenleftbiggg+µ0lc/µ
µ0N/parenrightbigg
=6 5.8A 3
part (c):
Problem 1.6
part (a):
Hg=NI
2g;Bc=/parenleftbiggAg
Ac/parenrightbigg
Bg=Bg/parenleftbigg
1−x
X0/parenrightbigg
part (b): Equations
2gH g+Hclc=NI;BgAg=BcAc
and
Bg=µ0Hg;Bc=µH c
can be combined to give
Bg=
NI
2g+/parenleftBig
µ0
µ/parenrightBig/parenleftBig
Ag
Ac/parenrightBig
(lc+lp)
=
NI
2g+/parenleftBig
µ0
µ/parenrightBig/parenleftBig
1−x
X0/parenrightBig
(lc+lp)
Problem 1.7
part (a):
I=B
g+/parenleftBig
µ0
µ/parenrightBig
(lc+lp)
µ0N
=2.15 A
part (b):
µ=µ0/parenleftbigg
1+1199
√
1+0.05B8/parenrightbigg
= 1012µ0
I=B
g+/parenleftBig
µ0
µ/parenrightBig
(lc+lp)
µ0N
=3.02 A 4
part (c):
Problem 1.8
g=/parenleftbiggµ0N2Ac
L/parenrightbigg
−/parenleftbiggµ0
µ/parenrightbigg
lc=0.353 mm
Problem 1.9
part (a):
lc=2π(Ro−Ri)−g=3.57 cm; Ac=(Ro−Ri)h=1.2c m2
part (b):
Rg=g
µ0Ac=1.33×107A/Wb; Rc=0 A / W b ;
part (c):
L=N2
Rg+Rg=0.319 mH
part (d):
I=Bg(Rc+Rg)Ac
N=33.1A
part (e):
λ=NB gAc=1 0.5m W b
Problem 1.10
part (a): Same as Problem 1.9
part (b):
Rg=g
µ0Ac=1.33×107A/Wb; Rc=lc
µA c=3.16×105A/Wb
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