Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell.
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.
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
$r=\left(\frac{l_{1}-l_{2}}{l_{2}}\right) R$
$=\frac{76.3-64.8}{64.8} \times 9.5=1.68 \Omega$
Therefore, the internal resistance of the cell is 1.68Ω.
Current flowing through the circuit = I
The relation connecting resistance and emf is,
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.