Theory  of  Pliasemeters.  101 
must  coincide  with  the  current  triangle,  and  for  moderate 
variations  from  balance  we  may  put : — 
K1=K(l  +  el\     A2  =  A(l  +  e2),     A3=A(l  +  e3), 
A0  =  A(l  +  £0), 
where  A  is  the  mean  value  of  the  load  currents,  so  that 
e1  +  e24-€3=0, 
and  where  we  may  regard  all  the  quantities  e  as  small 
fractions  whose  squares  and  products  may  be  neglected 
compared  with  unity. 
Since  when  the  load  currents  are  equal  <^1  =  ^2  =  <^3,  the 
above  assumptions  necessarily  imply  that  if 
</>i  =  X  +  #i>      </>2  =  %-r-#2,      </>3  =  %  +  #3, 
where  ^  is  the  mean  value  of  the  quantities  <fy,  so  that 
01+02+02  =  O, 
we  may  also  neglect  the  squares  and  products  of  the 
quantities  0. 
Now,  if  we  substitute  in  (9)  and  simplify,  neglecting 
squares  and  products  of  small  quantities,  we  easily  find 
3  cos  <£  =  3  cos  x  +  (*i  +  e2  +  e3)  cos  %  —  (81  +  02  +  #3)  &in  %•> 
or,  using  the  above  relations,  we  have 
COS  (f)  =  cos  %, 
or  <f>  is  the  mean  value  of  </>l5  <£2,  $3. 
Collecting  formulae  we  thus  have 
$l  =  </>  +  01,  </>2  =  <£  +  #2j  </>3=<£  +  #3;  1 
A^ACl  +  O,  A2  =  A(l  +  68),  A3  =  A(l  +  e3);    I 
A0=A(l+e0),         £1  +  62  +  63  =  0.  J 
Now  certain  relations  can  be  found  between  the  quantities 
e  and  6  for  the  current  triangle  shown  in  fig.  6,  by  equating 
the  ratios  of  the  sides  to  the  sines  of  the  opposite  angles  and 
reducing. 
These  relations  prove  to  be 
2>h  6i=  62  —  03,      —  3^  #!  =  e2  —  e;., 
3*  *,  =  . *,-*!,       -3*tf,  =  €,-€!,     j 
3^3  =  0.-^02,        -8^3  =  6,-60. 
