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Question

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 of the potentiometer wire. Determine the internal resistance of the cell.
420308_9ddd8035ffd54d468ef75cfc33f1e4d1.png

Solution
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Internal resistance of the cell =r

Balance point of the cell in open circuit, l1=76.3cm

An external resistance (R) is connected to the circuit with R=9.5Ω

New balance point of the circuit, l2=64.8cm

Current flowing through the circuit =l

Using the relation connecting resistance and emf is,

r=(l1l2l2)R

=76.364.864.8×9.5=1.68Ω

Therefore, the internal resistance of the cell is 1.68Ω.

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