Transformador de potencia

1

PROBLEM SOLUTIONS: Chapter 1
Problem 1.1
Part (a):
Rc =

lc lc =
=0
µAc µr µ0 Ac

A/Wb

Rg =

g
= 1.017 × 106 µ0 Ac

A/Wb

part (b):
Φ=

NI
= 1.224 × 10−4
Rc + Rg

Wb

part (c): λ = N Φ = 1.016 × 10−2

Wb

part (d):
L=

λ
= 6.775 mH
I

Problem 1.2 part (a):
Rc =

lc lc =
= 1.591 × 105 µAc µr µ0 Ac

Rg =

g
= 1.017 × 106 µ0 Ac

A/Wb

A/Wb

part (b):
Φ=

NI
= 1.059 × 10−4
Rc + Rg

Wb

part (c): λ = N Φ = 8.787 × 10−3 part (d):
L=

λ
= 5.858 mH
I

Wb

2
Problem 1.3 part (a):
Lg
= 110 turns µ0 Ac

N= part (b):
I=

Bcore
= 16.6 A µ0 N/g

Problem 1.4 part (a):
N=

L(g + lc µ0 /µ)
=
µ0 Ac

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

part (b):
I=

Bcore
= 18.2 A µ0 N/(g + lc µ0 /µ)

Problem 1.5 part (a):

part (b): µr = 1 +

I=B

3499
1 + 0.047(2.2)7.8 g + µ0 lc /µ µ0 N

= 730

= 65.8 A

3 part (c):

Problem 1.6 part (a):
Hg =

NI
;
2g

Bc =

Ag
Ac

Bg = Bg 1 −

x
X0

part (b): Equations
2gHg + Hc lc = N I ;

Bg Ag = Bc Ac

and
Bg = µ0 Hg ;

Bc = µHc

can be combined to give


NI
=
Bg = 
Ag
(lc + lp )
2g + µ0
2g + µ Ac


NI
µ0 µ 1−

Problem 1.7 part (a):

I = B

g+

µ0 µ (lc + lp )

µ0 N


 = 2.15 A

part (b):
1199
µ = µ0 1 + √
1 + 0.05B 8

I = B

g+

µ0 µ (lc + lp )

µ0 N

= 1012 µ0

 = 3.02 A

 x X0

(lc + lp )

4 part (c):

Problem 1.8 g= µ0 N 2 Ac
L

µ0 µ −

lc = 0.353 mm

Problem 1.9 part (a): lc = 2π (Ro − Ri ) − g = 3.57 cm;

Ac = (Ro − Ri )h = 1.2 cm2

part (b):
Rg =

g
= 1.33 × 107 µ0 Ac

A/Wb;

Rc = 0

A/Wb;

part (c):
L=

N2
= 0.319 mH
Rg + Rg

part (d):
I=

Bg (Rc + Rg )Ac
= 33.1 A
N

part (e): λ = N Bg Ac = 10.5 mWb
Problem 1.10 part (a): Same as Problem 1.9 part (b):
Rg =

g
= 1.33 × 107 µ0 Ac

A/Wb;

Rc =

lc
= 3.16 × 105 µAc A/Wb

5