THE SYNCHRONOUS MOTOR. 165 



we have 



A 2 sin (0 8) B 2 sin (<j> + 6) = 2^ sin (i) 



which determines the value of <?> tor which the efficiency of transmission is a maximum. 

 Geometrical construction of equation (i). Draw lines, Fig. 139, representing 

 A 2 and B l to scale, the angle between them being 20. About the point p as a 

 center describe a circle of which the radius represents 2AB sin 6. From the point 

 ^ draw a tangent to this circle. The angle between the line OS and this tangent 

 is the required value of 0. Two tangents can be drawn from the point q. One of 

 these tangents determines the value of (less than 180) for which the efficiency of 

 transmission is a maximum with machine A acting as a synchronous motor, and the 

 other tangent determines the value of (greater than 180) for which the efficiency 

 of transmission is a maximum with machine B acting as a synchronous motor. The 

 machine - A is distinguished as having the greater electromotive force. The angle $ 

 is the lag of B behind A. 



79. Value of B to give maximum intake of machine B with given cur- 



rent; A, X and R being given. Let /, Fig. 140, be the given current and 

 E( = I VR* + X*) the resultant electromotive force. In order that the intake of 

 B may be a maximum, BIcos(BI] or Bcos(BI] must be a maximum. Now 

 Bcos(BI] is the projection of B on the current line OL Describe a circle of 

 radius A about the point P, Fig. 140 ; then Ox is the greatest possible value of 



Fig. 140. 



B cos (BI} for the given current, and OC is the required value of B. From the 

 triangle OPC we have as the required value of B 



B VA* -\-E* 2.AE cos 



Remark. From Fig. 140 it is evident that A is in phase with T when B is 

 adjusted to give maximum P" with given /. It was shown in Article 78 that A 



