802 Dr. H. H. Poole on Vector Methods for 



the telephone. We must also consider the effect of a small 

 change in one resistance, say ~R lt 



(1) Let L x become L x + 8Li, all the other quantities re- 

 maining as before. Neglecting the telephone current, the 

 vector diagram of fig. 1 now becomes as shown in fig. 7. 

 The points 0, A 2 , P 2 , and B are unchanged from their 

 original positions, but A 1 and P x have separated as shown 

 from A 2 and P 2 . 



Fig. 7. 



It is evident that as the angle OPB is always right, 

 P moves on a semicircle on OB as diameter. Also, since 

 the resistances are unchanged, the triangle OA x A 2 is clearly 

 similar to 0PiP 2 , so that A x lies on the semicircle OA 2 Q on 

 OQ as diameter, where 



OQ 



n 



OB r 1 + R 1 =r 2 + H 2 * 



For a very small motion AjA 2 is a tangent to this circle, so 



that it is evident that the angle OA 2 A x is equal to -x —a. 



z 



Let BE be the P.D. (AjAg on the figure) across the 

 telephone terminals due to the increment SL. 



Then 



s-CT CirxSu -, . 



oJi/ = =Cir 1 cosao(tana). 



cos a v ' 



