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Question -

Figure shows a 2.0 V potentiometer used forthe determination of internal resistance of a 1.5 V cell. The balance point ofthe cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in theexternal circuit of the cell, the balance point shifts to 64.8 cm length of thepotentiometer wire. Determine the internal resistance of the cell.



Answer -

In the open circuit, the balance point isobtained for the emf of 1.5 V.

When the external circuit is connected, acurrent is drawn from the cell of 1.5 V in external resistance of 9.5 E1. Nowthe balance point is obtained for terminal potential

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