Question

Figure 3.35 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.

Answer 1

Internal resistance of the cell = r

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

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

New balance point of the circuit, l₂ = 64.8 cm 

Current flowing through the circuit = I The relation connecting resistance and emf is,

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