A potentiometer arrangement is shows in Fig. `6.62`. The driver cell has emf `e` and internal resistance `r`. The resistance of potentiometer wire `AB` is `R.F` is the cell of emf `e//3` and internal resistance `r//3`. Balance point `(J)` can be obtained for all finite value of

Options

A. `R gt r//2`
B. `R lt r//2`
C. `R gt r//3`
D. `R lt r//3`

Answer

A

Explanation

Current in `AB = I = (e)/(R + r)`
potential difference across `AB = IR = (eR)/(R + r)`
To obtain balanced point, `(eR)/(R + r) gt (e)/(3)` or `R gt r//2`