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NCERT Class XII Chapter
Current Electricity
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Question : 24 of 24
Marks: +1, -0
Figure 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.
Solution:  
In the open circuit, the balance point is obtained for the emf of 1.5 V.
E = kl1kl_1 ⇒ 1.5 = k × 76.3 ... (i)
When the external circuit is connected, a current is drawn from the cell of 1.5 V in external resistance of 9.5 Ω. Now the balance point is obtained for terminal potential
V = kl2kl_2 ⇒ V = k × 64.8 ... (ii)
Now we know the relation
r = (l1l2−1)\left(\frac{l_1}{l_2} - 1\right) R or r = (76.364.8−1)\left(\frac{76.3}{64.8} - 1\right) 9.5 = 1.69 Ω
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