Formula' used in Inductance Measurements. 



799 



evidently as shown in the figure. The current in BO is the 

 vector sum of c 2 and c 3 , so in finding the components of BO 



Fig. 5. 



c, r, 

 Vector Diagram 



Circuit. Diagram 



along and perpendicular to OP we may treat each current 

 separately. Resolving along OP we have 



o s 



and 



c i r i = ^=c 2 r 2 cos <j) 

 C X R] = c 2 i\ cos <f> ; 

 J{ = r~> tne condition for steady current balance. 



Resolving perpendicular to OP, we have 

 C3 r 3 = c 2 r 2 sin $ 

 and cJjco = c s r 3 + c 3 r 4 -f c 2 r 4 sin 0, 



= Jr 3 + n+ ^l; 

 L- r* J 



or since 



Cz = c 1 r 1 K(o and r l i\ = r 2 R y we get 

 L = K[fir 4 + r i (r l +R)]. 



If we can obtain a balance with r 3 = 0, i.e. the points 

 A 2 and B coincident, we have the simpler formula L = Kr 1 ? , 4 . 

 This method is, however, less convenient unless either Lor K 

 is infinitely variable. 



