First slide

potentiometer

Question

In an experiment to measure the internal resistance of a cell, by a potentiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5 resistance and is at a length of 3 m when the cell is shunted by a 10 $Ω$ resistance. The internal resistance of the cell is then

Moderate

A

1.5 $Ω$
B

10 $Ω$
C

15 $Ω$
D

1 $Ω$

Solution

In case of internal resistance measurement by potentiometer,
$\frac{V_{1}}{V_{2}} = \frac{l_{1}}{l_{2}} = \frac{\{{ER}_{1} / (R_{1} + r)\}}{\{{ER}_{2} / (R_{2} + r)\}} = \frac{R_{1} (R_{2} + r)}{R_{2} (R_{1} + r)\}}$
Here $l_{1} = 2 m, l_{2} = 3 m, R_{1} = 5 Ω and R_{2} = 10 Ω$
$∴ \frac{2}{3} = \frac{5 (10 + r)}{10 (5 + r)}$
or $20 + 4 r = 30 + 3 r$
or $r = 10 Ω$

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