156 THE MAGNETIC CIRCUIT [ART. 49 



curve; thenE t =M t v. Substituting the value of M t from eq. (81), 

 we have 



<=/ cos (<'+/?), ..... (83) 



where 



Ef =Q3Qk b mniv. ....... (84) 



a known quantity introduced for the sake of brevity. The 

 angle ^ in formula (83) is expressed through <' and /?, because, 

 from Fig. 40, 



#=' + /3 ........ (85) 



Another relation between E t and ft is obtained from the triangle 

 ODG, from which 



E t = Esinp ........ (86) 



A comparison of eqs. (83) and (86) gives that 

 Et'cos (<' +/?) = # sin/?. 

 Expanding and dividing throughout by cos /? we find the relation 



(87) 



from which the angle /? can be determined, and then E n calculated 

 byeq.(82). 



The next step is to take from the no-load saturation curve the 

 value M n of the net excitation necessary on the main poles in order 

 to induce the voltage E n . The real excitation Mf must be larger, 

 because part of it is neutralized by the direct armature raction Md. 

 We thus have 



...... (88) 



where M d is calculated from eq. (79), the angle <[> being known 

 from eq. (85). When the load is thrown off, the only excitation 

 left is Mf; let it correspond to a voltage e on the no-load satura- 

 tion curve. From e and e the per cent voltage regulation of the 

 machine is determined from its definition as the ratio (e e)/e. 



The same general method and the same equations apply in the 

 case of Fig. 41, when one is required to determine a point on one of 

 the phase characteristics of a synchronous motor. The beginner 

 must be careful with the sign minus in the case of the motor. 



