In potentiometer experiment, if `l_(1)` is the balancing length for e.m.f. of the cell of internal resistance r and `l_(2)` is the balancing length for its terminal potential difference when shunted with resistance R then :

Options

A. `l_(1)= l_(2)((R+r)/(R ))`
B. `l_(1)=l_(2)((R )/(R+r))`
C. `l_(1)=l_(2)((R )/(R-r))`
D. `l_(1)=l_(2)((R-r)/(R ))`

Answer

A

Explanation

For the potentiometer, internal resistance,
`r=R((l_(1)-l_(2))/(l_(2)))`
`therefore r l_(2)=R l_(1)-R l_(2)`
`therefore R l_(1)= R l_(2)+ r l_(2)= l_(2) (R+ r)`
`therefore l_(1)=l_(2)((R+r)/(R ))`