State the principle of potentiometer. With the help of circuit diagram, describe a method to find the internal resistance of a primary cell.

Explanation

Measurement of internal resistance of a cell using potentiometer is shown in figure. The cell of emf, E is connected across a resistance box (R) through key `K_(2)`.
When, `K_(2)` is open, balance length is obtained at length `AN_(1) = l_(1)`
`E = rho l_(1)" "...(i)`
When `K_(2)` is closed :
Let V be the terminal potential difference of cell and the balance is obtained at `AN_(2) = l_(2)`
`therefore" "V = rho l_(2)" "(ii)`
From equations (i) andd (ii), we get
`(E)/(V)=(l_(1))/(l_(2))" "...(iii)`
`E = I(r + R)`
`V = IR`
`(E)/(V) = (r+R)/(R)" "...(iv)`
From (iii) and (iv), we get `(R+r)/(R) = (l_(1))/(l_(2))`
`(E)/(V)=(l_(1))/(l_(2))`
`r = R((E)/(V)-1) rArr r = R((l_(1))/(l_(2))-1)`
We know `l_(1), l_(2) and R`, so we can calculate r.