DIRECT-CURRENT MACHINERY 177 



flowing from bar 1 to the brush is I c + i and the drop of potential 

 is (I c + i) TI] the current from bar 2 to the brush is I c i and 

 the drop of potential is (/. i)r z . Since the resistance n is in- 

 creasing while r% is decreasing, the current from bar 2 will increase 

 while that from bar 1 will decrease and the current in the coil will 

 decrease. Neglecting the resistance and self-induction of the coil 

 the current flowing in the coil will be zero when r and r 2 are equal 

 and when therefore half of the time of commutation has passed. 

 Any further increase in the resistance n will cause part of the 

 current from coil b to flow through coil c in order to reach the 

 brush BI by the path of least resistance. As the resistance r\ 

 still increases, more and more current flows through c until r\ 

 becomes infinite as the brush breaks contact with bar 1 and the 

 total current I c from b flows through c. Commutation is then 

 complete. 



In Fig. 145, curve (1), the current in coil c is plotted on a time 

 base for half of one revolution; it is reversed in the time T, rep- 

 resented by OT, during which the coil moves across the brush BI, 

 and it must vary according to a straight line law. This can be 

 proved as follows: 



If Fig. 143 represents the condition i seconds after the beginning 

 of commutation, neglecting the resistance and self-inductance of 

 the coil, the drop of potential from the commutator to the brush 

 must be the same at both sides, or 



(I c + i) n = (I c - i) r 2 ; 

 but 



T T 



= r c ~~ and Tz = Tc > 



therefore, 



Solving for i this gives 



r c 7 ^- i =(I c -i)r c j. ... (203) 



(204) 



which is the equation of a straight line. 



T 



When t = , i = 0, and when t = T, i = I c . 



