96 THE ROYAL SOCIETY OF CANADA 



This does not however take into account the fact that two of the coils 

 are always shunted by the vernier in actual use. A further correction 

 is thus introduced into the correction for the main slider. In a manner 

 similar to that employed before, we may obtain 



Ri -\~ R2 . , , , - , 



y— =1+C (15) 



where c is the known small quantity, but we know that 



Rx + R* = 2000 + ai + a 2 - 



mean slider ohms, hence 



101 



2 2 a, 



101 



2 2 a» 



1 



101 



7 = 2000+ (a,+a 2 ^ 2000 



= 2000+e (16) 



mean slider ohms, where e represents the known part. When the 

 slider reading is n the coils shunted are R n+i and R n +2- The equivalent 

 resistance, therefore, in terms of the mean slider ohm is, 



101 



1000+1 (a w+ i+a„+2-2 -i__ + g ) 



For any reading n of the slider we require for our calibration to 

 find the correction to be added to n to give the true ratio of the first 

 n coils to the whole resistance of the slide, the coils R n +i and R n +i 

 being shunted by the vernier. When shunted the total resistance 

 in terms of mean slider ohms is not 100,000 but 



101 

 / 2 2 a n v 



101000-i? w+ ,-i? w+2 +1000 + i (a n +i + an+2 j0j— + ej 



which equals 



101 

 2 a» 

 100000+ig-f ^a„ +1 +a„+2-2-^ r j (18) 



101 

 2a M 

 \e — f ( a n+ i + a„-f 2 — 2 * ) is thus the correction due to the 



shunting when the slide is at n — call this correction v n . 



Now, in most cases it is desirable to express a reading in terms of 

 the whole box as 100,000 ohms. Equation (14) gives us the correc- 

 tions when the vernier is absent and the total resistance is assumed 

 to be 101,000 ohms. In order to express the reading in terms of the 

 slider as 100,000 ohms when any two coils are shunted, we must 



