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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 A. `R gt r//2`B. `R lt r//2`C. `R gt r//3`D. `R lt r//3` |
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Answer» Correct Answer - A 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` |
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