86 
Dr.  W.  E.  Sumpner  on  the 
the  current  vectors,  we  nave  from  elementary  vector  con- 
siderations the  following  relations  between  scalar  products : 
YA^VpA^,     YA2  =  V^A2,     VA3=VpA3. 
We  also  have 
COS  0  —  COS  a.  COS  0p, 
where  0  is  the  phase-angle  between  V  and  any  current  A, 
0P  is  the  phase-angle  between  Vp  and  the  same  current  A, 
and    a  is  the  angle  between  the  vectors  V  and  Vp. 
Hence,  it  will  be  apparent  that  in  any  such  equation  as  (3) 
or  (5)  we  can  regard  the  vectors  V  and  Vp  as  interchangeable, 
and  also  that  if  we  may  neglect  the  square  of  the  small  angle  a, 
we  can  regard  cos  0  and  cos  0P  as  identical. 
Now  the  vector  figure  will  be  as  represented  in  fig.  1. 
In  the  case  considered  the  load  currents  are  equal,  hence 
the  vectors  A  will  be  of  equal  magnitude,  and  it  follows  that 
any  two  will  differ  in  phase  by  120°.  If  therefore  0  is  the 
angle  between  the  vectors  V  and  A1}  and  if  we  denote  by 
Fig.l. 
v 
V  and  A  the  magnitudes  of  the  voltage  and  current  vectors, 
we  have 
V^A^YAcos^), 
VA2  =VA  cos  (120  +  0)  =  -VA  cos  (60-0), 
VA3"  =VAcos  (240 +  0)  =  --VA  cos  (60  +  0). 
Substituting  in  (3)  we  have  for  balance  the  condition  : 
F1cos0  =  F2cos(6O-0)+F3cos(6O  +  0).     .     .     (6) 
If  now  in  the  direct-current  test  we  make  the  steady  currents 
such  that 
-A!=Aj+Ai, 
and 
A_2  =  cos  (60-0)  =  1  +  3*  tan  ft 
A~3      cos  (60  +  0)      1— 3*tan0'      "     * 
(?) 
