3/6 ELEMENTS OF ELECTRICAL ENGINEERING. 



under (a). Add these reluctances and multiply their sum by 

 <>. This gives the magnetomotive force which is required to over- 

 come the reluctance through which <3> flows. Add to this the mag- 

 netomotive force eP and we have the total magnetomotive force & 

 between the pole pieces. Then calculate < exactly as under (a). 



19. Magnetic reluctances in series and in parallel. Two por- 

 tions of a magnetic circuit through which the same flux flows 

 are said to be in series. The combined reluctance of two such 

 portions of a magnetic circuit is equal to the sum of their indi- 

 vidual reluctances. 



When two or more portions of a magnetic circuit are so 

 related that the total magnetic flux <l> through the circuit di- 

 vides and flows in part through each portion, the portions are 

 said to be in parallel. The combined reluctance of a number of 

 branches of a magnetic circuit which terminate in the same branch 

 points, is equal to the reciprocal of the sum of the reciprocals of 

 the individual reluctances. 



The total flux < divides up among a number of parallel 

 branches in exactly the same way that an electric current divides 

 up among a number of parallel branches of an electric circuit. 



The various leakage paths and the path of the useful flux 

 (across gap spaces and armature core) are all in parallel between 

 the pole pieces of a dynamo. 



20. Work required to magnetize iron. When an iron rod is 

 magnetized by sending an electric current through a coil of wire 

 surrounding the rod, an opposing electromotive force is induced 

 in the coil by the growing magnetism of the rod, and the work 

 done in forcing the current against this opposing electromotive 

 force, is the work expended in magnetizing the rod. 



The work W, in ergs, which is done in magnetizing V cubic 

 centimeters of iron from a given initial flux density IB' to a given 

 final flux density eB", is given by the equation : 



V f^" 



(14) 



