# In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 $\Omega$ is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Given,
The balance point of cell in open circuit l1 = 350 cm
The balance point  of the cell with resistance  l2 = 300 cm
External Resistance  $R= 9\Omega$
The internal resistance of the cell is , $r= \left ( \frac{l_{1}}{l_{2}}-1 \right )R$
$r=\left ( \frac{l_{1}-l_{2}}{l_{2}} \right )R$
or ; $r= \left ( \frac{350-300}{300} \right )\times 9= 9\times \frac{50}{300}= 1\cdot 5\Omega$
Therefore, the internal resistance of the cell is $1\cdot 5\Omega$

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