808 Vector Methods in Inductance Measurements. 



evidently equal to the angle OCA in fig. 5. The triangles 

 QST (fig. 8) and CAO (fig. 5) are clearly similar, so 

 angle QTS (fig. 8) is equal to angle a above, and u + /3 = d. 



Hence, to find the total impedance to the transverse current 

 we must compound the two vectors p and p", the angle be- 

 tween them being 6. Now, this angle is always acute, and 

 the vector p" may be regarded as fixed by the condition that 

 Ti should be equal to I, so to obtain a minimum transverse 

 impedance we should make p as small as possible, i.e. K 

 should be large. 



Two rather interesting conclusions have here been reached. 



(1) Given r 1? R, and L, the transverse impedance de- 



pends solely on the value o£ K, and not on the 

 distribution of the resistance necessary for balan- 

 cing between the arms r 2 , r 3 , and r 4 . 



(2) Owing to the angle 6 being acute, no reduction in 



the transverse impedance by " tuning " is possible, 

 i. e. there is no value of K giving a minimum 

 impedance for a given frequency. 



Under the best conditions of working, r 1 = I = Lo) nearly 

 and cos 6 is small, so 



P = W2, 9'=^, and p » = ±jL } 

 so the total transverse impedance is not much greater than 



V2VKV 



+ LV). 



In most cases -^- will be greater than L&>. For example, 



XV ft) 



if a) = 3300, corresponding to the Note C in the treble clef, 

 and our largest available condenser is 1 microfarad, 



¥7 - =300 ohms. If L = 0'05 henry [which is a compara- 



tively large value for a coil without an iron core], 

 L&) = 165 ohms. Hence in most cases we should use the 

 largest available condenser, though little would be gained by 



the use of one whose capacity greatly exceeded -r — ^, even if 



it were available. 



It must be remembered that a large value for K entails a 

 small value for r 1 r 4 +-r 8 (r 1 +K). As before, it is a disad- 

 vantage to have r 2 and r 4 small, as this causes excessive 



