# Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell.

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.

Solution:

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, l= 64.8 cm

$r=\left(\frac{l_{1}-l_{2}}{l_{2}}\right) R$

$=\frac{76.3-64.8}{64.8} \times 9.5=1.68 \Omega$

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

Current flowing through the circuit = I

The relation connecting resistance and emf is,