66 



ELECTRICAL ENGINEERING 



the flux through section (3) is 3> 3 = sr ; 



the total flux through the ring is 



< = $1 -f- $2 + $3 



_ M M M_ 



Q I CO I CV7 



o/V c/V o/V 



(96) 



But the flux is equal to the m.m.f. divided by the reluctance of 

 the ring, or 



M 



and therefore 



-* 



- 





(97) 



d! 



(6) Fig. 55 shows a ring of iron with a piece set in made up of 

 three parts of different permeabilities. The lengths, sections and 



FIG. 55. Series parallel magnetic circuit. 



permeabilities are indicated and the reluctances of the various parts 



, and %&4 = - . Deter- 



are oKi = 



mine the reluctance of the circuit. 



Let M = the m.m.f. applied to the ring, 



< = flux through the ring, 

 MI = m.m.f. consumed in section (1), 

 M 2 = m.m.f. consumed in section ab, 

 then MI = ^oftj and M 2 = 



A4M4* 



