A potentiometer , with a wire of length 10 m, is connected to an accululator of steady voltage. A lenchlanche cell gives a null point at 7.5 m If the length of the potentiometer wire is increaseed by 1 m, find the new posistion of the null point.

Explanation

Use `epsi = (l)/(L)V_(s)`
In the first case ` epsi = (750 cm)/(10 m ) . V_(s) " "…….(i)`
In the second case ` epsi = (l)/((10+ 1)m),V_(s) " "…..(ii)`
Divide to get ` 1 = (750 xx 11)/(10 l) or l = (750 xx 11)/(10) = 825 cm`
In order to reduce the potential drop across the wire other than increasing the length of the potentiometer wire, the most commoly used parctice is to add resistance R by the side of the supply voltage as shown below.

Problem sloving technique : In this type of problem treat the entire circuit in two parts. The lower portion in which the unknow EMF `epsi` is balanced against some balancing lenght l is treated just as to calculate the potential drop V across the wire AB using method `(epsi)/(V_(s)) = (l)/(L)`

Now treat separately the branch in which supply voltage `V_(S)` and potential difference V connected as follows .
Now , if current l flows in this loop and the resistance of the potentiometer wire is .i. then you can use the following equations depending on which quantity has been asked .

`I = (V_(s))/(R+r)" "......(i)`
`V = ir" "......(ii)`
`V_(s) - V = IR " "......(iii)`