The potentiometer wire of length 100cm has a resistance of `10Omega`. It is connected in series with a resistance of `5Omega` and an accumulaot or emf 3V having negligible resistance. A source 1.2V is balanced against a length 'L' of the potentionmeter wire. find the value of L.

Explanation

Current flowing through the wire A to B is `I_(AB)=(3)/(5+10)=(3)/(15)=(1)/(5)=0.2A` and `R_(AB)=10Omega`
`V_(AB)=10xx0.2=2` volt so Potential gradient `(V)/(l)=(2)/(100)=0.02` volt/cm So L`=(1.2)/("Potential gradient")=(1.2)/(0.02)= 60cm`