114 Dr Searle, Calculation of the electrical resistance, ete. 
-,§4. The currents in the cables and in the wires can now be 
easily found. Let the EMF. applied between A, and B, be 4, let 
the current along A,A, be a, and that along Ajm_,Am be %m. Let 
the current in the wire A,,B,, be y». Since the resistance of the 
system beyond A,B, 1s Rpm, we have 
SY = Din —i@aeris os eee (17) 
The “continuity ” of the current gives 
Lin lim: = Yi, ais one ee (18) 
From (17) and (18) we have, by (16), : 
eas _sinh(n—m+1)@ 
Din atk owe ll a Tsinh(n—m)0_ m-+1° 
Hence 
eUm+1 bri Lm, bed a ual 
sinh(n—m)@ sinh{n—(m—1)}@ ~~ sinhn@’ 
E _ _Sualn(@—mesp ye) 
so that GBrye ‘sinhind £y. 2 eee (19) 
But #,= £/R,, and hence, by (19) and (16), 
sinh (n —m+1)60 E 
“m= smh (n+1)0—sinh n° 8° 
Since 2 sinh? $0 = cosh @—1=r/s, we have 
f _sinh(n—m+1)0 £ 
- cosh(n+4)0 °V/2rs° 
Since, by (18), Ym =%m— mii, we have 
= ee {sinh (n — m + 1) @—sinh (nm — m) a 
2E cosh (n—m+4)@sinh $6 - 
Tre. cosh (n+ +) 0 <a 
_-cosh(n—m+4)@ E 
cosh(n+4)0 s— 
Um 
