149 TREATISE OX ALTERNATING CURRENTS. 



Solving equation (16) we get 



_ IE . \r*E* - 4J*rw 

 = ~ ~~ 



(18) 



This gives the maximum and minimum values of c when E and 

 w are given. 



If we consider w as variable, it also gives the greatest possible 

 value of e for a given value of the generator E.M.F. ; this occurs 

 when w = 0, and the greatest possible value of c is given by 



where 



IE 

 e = = E sec (19) 



n 



tan 6 = 

 r 



We have noticed before (87) that the generator E.M.F. is 

 not necessarily greater than the counter E.M.F. of the motor. We 

 are now in a position to say that, neglecting losses due to friction, 

 hysteresis, etc., a generator of E.M.F., E can drive as a motor any 

 alternating- current machine whose counter E.M.F. lies between 

 the values and E sec 0, where is defined by the equation 



tan 6 = - 

 r 



S being the total reactance, and r the total resistance of the two 

 armatures and the line. 



It will be well to remember that for the above statement to 

 be true, e and E are not the E.M.F.s on an open circuit, but the 

 E.M F.s due to rotation in the resultant fields determined by the 

 exciting currents, together with the armature currents. 



Equation (18) also shows that the maximum output of the 

 motor is given by 



E* 



W = -r~ 



4r 



where E is now the total E.M.F. generated by the driving machine, 

 and r is the total resistance of the two armatures and line. 



So long, therefore, that the output lies between the values 



