

the Synchronous Motor. 197 



We have 



. A Sc S 2irnL ,_, 



tan0=z -Wc = -R = --ET> ■ • ■ (9) 



that is, 6 is independent of the current and OT is a fixed 

 direction relative to Oc so long as the speeds of the machines 

 are kept constant, and L is considered constant. 



16. In fig. 3, OS (or PM) is proportional to the current, 

 [ = 27rnLc] ; 



OM is the component of e directly opposing c, 

 OR is the E.M.F. required to overcome resistance, 

 [=E C ]; 

 and ON is the component of E in the direction of c; 



hence, rectangle PSOM is proportional to the output of the 



motor [p] ; 

 rectangle OSTR is proportional to the c 2 R losses ; 

 and rectangle OSQN „ „ output of 



generator [cE cos -v^]. 



From this, and the equation 



p + c 2 R = cE cos yjr, 

 it follows that the efficiency of transformation 



= OM_ OM 



~ ON ~ MR* 



17. If the output of the motor is kept constant, we have 



rectangle PSOM = constant, 



and the locus of P is a rectangular Iryperbola having OM and 

 OS as asymptotes (fig. 4). 



Take any point P on this hyperbola. We have seen (9) 

 that OT has a fixed direction relative to Oe ; and the point T 

 (fig. 3) on this direction is found by drawing through P a 

 line parallel to Oc. Again, e lies on the line through P 

 parallel to OS and eT = E, in magnitude. Let the E.M.F. of 

 the generator be kept constant and equal to E. With centre 

 T and radius E describe a circle cutting PM in e and d ; 

 then the corresponding counter E.M.F. of the motor may be 

 either Oe or 0/, and the current is represented in magnitude 

 by PM : that is corresponding to given values of E and c 

 there are two values of e. The relative phases in the two 

 cases are shown in the parallelograms OeTE and O^TE' 

 (fig. 4). 



18. To find the point P' on the hyperbola corresponding to 

 minimum current we have to bring the points e and e' into 



