242 



Mr. Albert Griffiths : Some 



It may be of interest to examine, by means of an example, 

 what sort of results one would expect from reasoning on the 

 principles explained above, 



Let the E.M.F. acting at the extremities of the Wheat- 

 stone bridge be represented by (£ sin pt, and the back E.M.F. 

 acting in the bismuth spiral by ^- -6 sin (p£ — 45°). 



The adjoining figure represents a 

 Wheatstone bridge. 



The bismuth spiral is supposed to 

 be in the arm CD. 



CBA represents the bridge- wire 

 divided into 1000 parts, the divisions 

 counting from C to A. 



Let X=Y ; this involves the as- 

 sumption that the change in resistance 

 of the bismuth is only apparent. 



Let 



Q 



P + Q 



= r. 



then 



P + Q 



=1— r. 



The theoretical change in resistance of the bismuth spiral 

 as determined by the telephone will first be calculated. 

 Assuming that the current in BD is negligible, the current 

 through the telephone varies as 



X f (£ ") P 



iQrsinpt— g^sinCpt— 45°) > - p Q @sin^, 



X + Y 



or as 



-smpt — £-rsin(jo£ — 45°)— (1— r) s'mpt 



60 



Where 



= Rsin (pt — a) say. 



to make R 2 a minimum we have 



\/2 



r =2 + 



120 



Reading = 511*8 approx. 

 Resistance with telephone 



= '5118 approx. 



511-8 



Resistance with steady current 488*2 



= 1*048. 



