Explain method to measure internal resistance of cell by using potentiometer.

Explanation

`rArr` As shown in figure between point A and C, positive terminal of battery (B) variable resistance R and key (K) are connected.
`rArr` Cell whose internal resistance (r) is to be measured is connected in parallel with key `(K_(2))` and resistance box (R).
`rArr` Positive of cell `(epsilon)` is connected with A and negative terminal of cell is connected with galvanometer Another end of galvanometer is connected with jockey key and moved on wire AC.
When key `K_(2)` is open the null point obtained for zero deflection of galvanometer be `N_(1)`.
Assume `AN_(1) = l_(1)`
`therefore epsilon = phi l_(1) " "` ...(1)
`" "` (For open circuit condition )
`rArr` When switch `K_(2)` is closed then current will pass through resistance box R and cell `(epsilon)`. Let null point obtained in this condition be `N_(2)`.
Let `AN_(2) = l_(2)`
`therefore ` If V is terminal voltage of cell `.epsilon.` then,
`V = phi l_(2) ` .... (2)
By taldng ratio of (1) and (2),
`(epsilon)/(V) = (l_(1))/(l_(2))`
but `epsilon ` = I (R + r) and V = IR
`therefore (I (R + r))/(IR ) = (l_(1))/(l_(2))`
`therefore (R + r)/(R) = (l_(1))/(l_(2))`
`therefore 1 + (r)/(R) = (l_(1))/(l_(2))`
`therefore (r)/(R) = (l_(1))/(l_(2)) - 1 `
`therefore r = R [ (l_(1) - l_(2))/(l_(2)) ] " or " r = R [ (l_(1))/(l_(2)) - 2 ] `
Advantage of potentiometer is suoh that cµrrent do not flow through cell whose internal ressistance is to be measured. Hence, it do not affect internal resistance of the cell.