CH. I T1IK INDUCTION FACTOR 5 



is the total number of lines of force entering the cylindrical 

 surface from one pole. Hence the couple or torque /, due 

 to the forces on both sides, tending to turn the armature* 



is ylLV10~ l . The whole current, c, passing into or out of 



7T 



the armature will be twice that in each conductor, so we 

 have . 



t= -AcNlQ' 1 (2). 



-7T 



We shall find it convenient to express this in inch- 

 pounds. Remembering that a weight of one pound is equal 

 to 1 15 x 10 5 dynes, we get 



t=l'4lAcNlQ~* inch-pounds (3). 



We have for simplicity supposed that the dynamo has 

 only two poles, and that iV lines enter the armature from 

 one pole, and pass from the armature into the other pole. 

 If the dynamo had four poles with N lines per pole and 

 A surface conductors/jVand A being the same as in the 

 two-pole machine, and if the current per conductor remained 

 unaltered, the total force on all the conductors would be 

 doubled. But if the four polar divisions of the armature 

 are connected in parallel, a current of c amperes entering 

 the armature will be divided into four instead of into two 

 parts, as with the two-pole machine ; hence, if c is the 

 same as before, the current per conductor in the four-pole 

 machine is half what it was in the two-pole machine, so 

 that the force and the torque remain unaltered. 



Equation 3 is then a general equation, applicable to 

 machines having any number of poles, provided that N is 

 the number of lines per pole, and that there are as many 

 surliHv o. mint-till-.- in NTU-S bet \\rni tin- main terminals of 



