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.

 

 

 

 
 
 
 
 

Answers (1)
S safeer

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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