A potentiometer wire has a length 400 cm and a resistance of 20`Omega`. It is connected in series with an external resistance R and a cell of e.m.f. 1.5 V and internal resistance 1`Omega`. A source of e.m.f.6 mV is balanced against a length of 30 cm of the potentiometer wire. Find R. According to the principle of potentiometer `Vpropl`.
i.e., V = ilp where l'is the balancing length & p the resistance per cm of the potentiometer wire.

Explanation

Data supplied
Potentiometer resistance ,`P=20Omega" "r=1Omega" "E=1.5V," "p=(20)/(400)Omega.cm^(-1)`
`i=(E)/(P+R+r)" " thereforeV=(Elp)/(P+R+r): " "V=6xx10^(-3)V`
`P+R+r==(Elp)/(V)=(1.5xx30xx20)/(6xx10^(-3)xx400)=375`
Substitution for P and r ,we get ,20+R+1=375
`therefore` R=375-21=354`Omega`