3.23 Figure 3.34 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. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

 

Answers (1)

Given, 

the balance point of cell in open circuit = l_1=76.3cm

value of external resistance added = R=9.5\Omega

new balance point = l_2=64.8cm

let the internal resistance of the cell be r.

Now as we know,  in a potentiometer,

r=\frac{l_1-l_2}{l_2}*R

r=\frac{76.3-64.8}{64.8}*9.5=1.68\Omega

Hence, the internal resistance of the cell will be 1.68 \Omega

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