30 



THE MAGNETIC CIRCUIT 



[ART. 13 



is shunted through the cast-iron part of the circuit. At low saturations a 

 considerable part of the total flux is shunted through the cast-iron part, 

 but as the flux density increases the cast iron becomes saturated, and a 

 larger and larger portion of the flux is deflected into the air-gap. What 

 percentages of the total flux in the yoke are shunted through the cast iron 

 when the flux density in the air-gap is 1 kl/sq. cm. and 7 kl/sq. cm. 

 respectively? Solution : When the flux density in the air-gap is 1 kilo- 

 line per sq. cm. the m.m.f. across the gap is 1000X0.8 X 0.5 = 400 ampere- 

 turns. The flux density in the steel poles is 2 kl/sq. cm., and the required 

 m.m.f. in them is about 16 ampere-turns. Therefore, the total m.m.f. 

 across AC and consequently across the cast-iron part is 416 ampere-turns, 



Sq. Cm. 



35 Sq. Cm. 

 Cas.t Steel 



FIG. 6. A complex magnetic circuit. 



or H = 24.5 ampere-turns per centimeter of length of the path in the cast 

 iron. Thisvalue of //corresponds on the magnetization curve to B = 6kl/sq. 

 cm. ; hence, the total flux in the cast iron is 72 kl. The flux in the yoke is 

 60 + 72 = 132 kl., and the percentage in the cast-iron shunt is 72/132 or 

 about 55 per cent. Similarly, it is found that, when the flux density in the 

 air-gap is 7/kl sq. cm., about 25 per cent of the flux is shunted through 

 the cast-iron part. The foregoing arrangement illustrates the principle 

 used in some practical cases, when it is desired to modify the relation 

 between the flux and the magnetomotive force, by providing a highly 

 saturated magnetic path in parallel with a feebly saturated one. 



Ans. 55 per cent and 25 per cent approximately. 

 Prob. 17. Indicate how the preceding problem can be solved if the 

 cast-iron part were provided with a small clearance of say 1 mm. Hint: 

 See the second solution to problem 1 1 . 



