Question

Question 3.24: 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.

Class 12 - Physics - Current Electricity Current Electricity Page-131

Answer 1

in open circuit , E = 1.5 V, $l_1$=76.3cm
so, in case of open circuit, the balance point is obtained for the emf of 1.5V
e.g., $\bf{E=kl_1}$
1.5 = k × 76.3 ---------(1)

when external circuit is connected, a current is drawn from the cell of 1.5V in external resistance of 9.5Ω. now the balance point is obtained for the terminal potential V
$V=kl_2\implies V= k\times64.8$-----(2)

now we know the relation between, l₁, l₂ , r and R .
e.g., r = $\left(\begin{array}{c}\frac{l_1}{l_2}-1\end{array}\right)R$
here, l₁ = 76.3 cm , l₂ = 64.8 cm and R = 9.5Ω
so, r = (76.3/64.8 - 1) × 9.5Ω
= 1.69Ω

Answer 2

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