Figure `6.13` shows a `2.0 V` potentiometer used for the determination of internal resistance of a `1.5 V` cell. The balance point of the cell in open circuit is `76.3 cm`. Whan a resistor of `9.5 Omega` is used in the external circuit of the cell, the balance point shifts to `64.8 cm`, length of the potentiometer. Dentermine the internal resistance of the cell.

Figure `6.13` shows a `2.0 V` potentiometer used for the determination

1.	Figure `6.13` shows a `2.0 V` potentiometer used for the determination of internal resistance of a `1.5 V` cell. The balance point of the cell in open circuit is `76.3 cm`. Whan a resistor of `9.5 Omega` is used in the external circuit of the cell, the balance point shifts to `64.8 cm`, length of the potentiometer. Dentermine the internal resistance of the cell.
Answer» Here, `l_(1) =76.3cm , l_(2) = 64.8cm,` `r=?, R=9.5Omega` (br) Now, `r=((l_(2)-l_(2))/l_(2))R = ((76.3-64.8)/(64.8))xx9.5 =1.68Omega`

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