0195136047.pdf

478 MAGNETIC CIRCUITS AND TRANSFORMERS

Solution

The net cross-sectional area of the core is 2. 52 × 10 −^4 × 0. 9 = 0. 5625 × 10 −^3 m^2. Note that the

stacking factor of 0.9 is applied for the laminated core. It does not, however, apply for the air-gap

portion,

BC=

φ

net area

=

0. 625 × 10 −^3

0. 5625 × 10 −^3

= 1 .11 T

The correspondingHCfrom Figure 11.1.2 for M-19 is 130 A/m. Hence,


C=HClC= 130 × 0. 1 =13 At

The cross-sectional area of the air gap, corrected for fringing, is given by

Ag=( 2. 5 + 0. 01 )( 2. 5 + 0. 01 ) 10 −^4 = 0. 63 × 10 −^3 m^2

Bg=

φ

Ag

=

0. 625 × 10 −^3

0. 63 × 10 −^3

= 0 .99 T

Hg=

Bg

μ 0

=

0. 99

4 π× 10 −^7

= 0. 788 × 106 A/m

Hence,


g=Hglg= 0. 788 × 106 × 0. 1 × 10 −^3 = 78 .8At

For the entire magnetic circuit (see Figure E11.2.1)

NI=
TOTAL=
C+
g= 13 + 78. 8 = 91 .8At

Thus, the coil current

I=

NI

N

=

91. 8

100

= 0 .92 A

+

−

φ

Figure E11.2.1Equivalent magnetic circuit.

EXAMPLE 11.2.2

In the magnetic circuit shown in Figure E11.2.2(a) the coil of 500 turns carries a current of 4 A.

The air-gap lengths areg 1 =g 2 = 0 .25 cm andg 3 = 0 .4 cm. The cross-sectional areas are related

such thatA 1 =A 2 = 0. 5 A 3. The permeability of iron may be assumed to be infinite. Determine

the flux densitiesB 1 ,B 2 , andB 3 in the gapsg 1 ,g 2 , andg 3 , respectively. Neglect leakage and

fringing.

Solution

Noting that the reluctance of the iron is negligible, the equivalent magnetic circuit is shown in

Figure E11.2.2(b).