82 Tin: wKECT-crKKKvr MMT<H; OH. iv 



Taking now the sum of the two torques. \ve find that 

 the total torque becomes 



" i!"^ (*>) 



In the example illustrated in the diagram the speed at 

 this point is (i2(i r.p.m., c a is 48 amperes, c 6 is 72 amperes, 

 the total current is 120 amperes, and the combined torque 

 812 inch-pounds. 



When the total torque is less than that given by 

 Equation 40 the dynamo with the larger induction factor 

 will do less work than the other, until at a certain load it 

 will do no work at all, and all the work will be done by 

 the motor with the smaller induction factor. 



The torque for which one motor is doing all the 

 work may be found by considering the fact that at this load 

 the induced tension of the other motor is equal to that of 

 the line. Hence we can write : 



l ' : = --71^" ...(41). 



M,, V,, M,; 2 



From which we find the required torque to be 



'-"ifgc*-".) (> 



In our example this torque is 207 inch-pounds. 



It has been assumed that when one of two coupled 

 motors is generating a current, the other is in danger of 

 being burnt out at heavy loads, on the supposition that 

 it would have to do all the useful work and also drive 

 the second motor as a generator. In practice it will be 

 found that it is only at light loads that one of the two 

 motors will act as a generator, the effect of an increase 



