Method of Comparing Inductance with Capacity. 41 

 The current in that branch is 



^o / 17 n 



TL'(l + ripC)-r 2 ipC' ^ J 



and, as the resistance is non-inductive, the potential difference 

 between any two points along PB is proportional and parallel 

 to the same vector. Hence the voltage across YB, say, lags 

 behind the applied e.m.f. by an angle 



tan-' ^'-^ (lg) 



At the same time the voltage across QB lags behind the 

 applied e.m.f. by 



tan-'^ (19) 



Both of these tangents are simply directed proportional to 

 the frequency, and consequently if they are equal for one 

 frequency they are equal for all. If then we equalize 

 potentials, not, as in Rimington's method, between Q and a 

 point on AP, but between Q and a point on PB, we can 

 ensure that the balance obtained is continuous for all fre- 

 quencies, while we still avoid the troublesome double adjust- 

 ment of the Maxwell method. 

 Writing 



R 1 = r 1 + r 2 = AP + PY^ 



R 2 = YB L . . . . (20) 



R 3 = AQ and R 4 = QB 



the condition for phase equality becomes 



Lp (Ri-r,)(R 8 + r,)j?C 



R 3 + R 4 - R 1 + R 2 ' * ' * { } 



while the condition that voltage amplitudes are equal then 

 becomes 



R!R 4 =R 2 R 3 (22) 



Hence, for a balance, 



£=|?r,(R 2 + r 2 ) (23) 



The same result is easily obtained for the conjugate 

 position, where Y and Q are the current points and the 

 indicator is across AB. 



Putting r 2 = we get Maxwell's case; while if the arms 

 of the bridge are all made equal, (23) becomes 



5 =r 1 (R-hr 2 ), (24) 



