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)
P Pankaj Sanodiya

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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