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

Given,
Emf of the potentiometer wire, E = 2 V
Potential of cell, V= 1.5 V
Balance point obtained, when the cell is in open circuit, l₁= 76.3 cm
New balance point obtained, when a resistor is used in the external circuit, l₂= 64.8 cm
External resistor, R= 9.5 Ω

Now,

r = \frac{R (E - V)}{V} = \frac{R (l_{1} - l_{2})}{l_{2}} [\frac{E}{V} = \frac{l_{1}}{l_{2}}] = \frac{9.5 (76.3 - 64.8)}{64.8} ohm

= \frac{9.5 \times 11.5}{64.8} ohm = 1.686 ohm ~ 1.7 Ω

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