Page 10 - Solutions Manual to accompany Electric Machinery Fundamentals

l 0.37 m

R  5   56.1 kA t/Wb
 A 5  0 1500 4  10  H/m 0.07 m  0.05 m
5  7 
 r
The total reluctance is
 R    R R R 168 152 168 108 



 R  R 1 2 3 4  56.1  204 kA t/Wb
TOT 5  R  R  R R 168 152 168 108



1 2 3 4
The total flux in the core is equal to the flux in the center leg:
    F  300 t 1.0 A   0.00147 Wb

center
TOT
R 204 kA t/Wb
TOT
The fluxes in the left and right legs can be found by the “flux divider rule”, which is analogous to the
current divider rule.
  R 168 108 R 
 left  3 4  TOT   0.00147 Wb  0.00068 Wb



1  R 2  R 3  R R 4 168 152 168 108
  R 168 152 R 
 right  1 2  TOT   0.00147 Wb  0.00079 Wb



1  R 2  R 3  R R 4 168 152 168 108
The flux density in the air gaps can be determined from the equation   BA:
 0.00068 Wb
B left  left   0.185 T

eff  A 0.07 cm  0.05 cm 1.05 
 0.00079 Wb
B right  right   0.215 T

eff  A 0.07 cm  0.05 cm 1.05 
1-7. A two-legged core is shown in Figure P1-4. The winding on the left leg of the core (N 1) has 600 turns,
and the winding on the right (N 2) has 200 turns. The coils are wound in the directions shown in the
figure. If the dimensions are as shown, then what flux would be produced by currents i 1 = 0.5 A and i 2 =
1.0 A? Assume  = 1200 and constant.
r

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