Question

When a resistor of `5 Omega` is connected across a cell, its terminal potential differnce is balanced by 140 cm of potentiometer wire and when a resistance of `8 Omega` is connected across the cell, the terminal potential difference is balanced by 160 cm of the potentiometer wire. Find the internal resistance of the cell.

Answer 1

Here, `R_(1) = 5 Omega , l_(1) = 140 cm` ,
`R_(2) = 8 Omega , l_(2) = 160 cm`
Let emf of the cell be balanced by length l of the potnetiometer wire. Then as per question,
In the first case,
`r = R_(1)((l_(1)-l_(1))/(l_(1))) or (r l_(1))/R_(1) = l-l_(1)`
In the second case,
`r = R_(2)((l_(1)-l_(2))/(l_(2))) or (r l_(2))/R_(2) = l-l_(2)`
Subtracting (ii) from (i), we get
`r[(l_(1))/(R_(1)) - l_(2)/R_(2)]=(l - l_(1)) - (l - l_(2))= l_(2) - l_(1)`
or `r=(l_(2) - l_(1))/(l_(1)/R_(1) - l_(2)/R_(2)) = (160-140)/(140/5 - 160/8) = 2.5Omega`