Theory  of  Phasemeters. 
99 
load  currents  are  no  longer  balanced,  we  see  that  instead  of 
equation  (6)  we  shall  have  as  the  condition  for  balance  : 
Fx  kx  cos  fa  =  F2A2  cos  (60 -  </>2)  +  F3 A3  cos  (60  +  <£3) .    (20) 
Now  the  vectors  of  fig.  5  are  the  same  as  those  of  fig.  6,  in 
which  latter  the  current  vectors,  and  also  the  voltage  vectors, 
Fur.  6. 
are  drawn  so  as  to  form  closed  triangles.     Bearing  in  mind 
that  the  voltage  triangle  is  equilateral,  it  is  readily  seen  that 
A1cos^)1=A2cos(60— <£2)+A3cos(60-f-$3) ;  .      (21) 
and  as  we  have,  whatever  0  may  be, 
cos  0  =  cos  (60-0)  +  cos  (60  +  0), 
we  see  that  we  can  always   determine  an  angle  0  from  the 
consistent  equations 
A    _  AT  cos  <f>x  _  A2cos(60-cfe)  _  Agcos(60  +  (ftg) 
0  costf  cos  (60-0)  cos(6O  +  0)      '{     > 
and  that  for  such  value  of  0  we  have  from  (20) 
Fi  cos  0  =  F2  cos  (60  -  0)  +  F3  cos  (60  +  0).     .     (23) 
If  we  compare  this  equation  with  (6),  we  see  that  if  the 
instrument  has  been  correctly  calibrated  for  power-factor  on 
balanced  loads,  the  reading  of  the  instrument  will,  for  the 
H2 
