Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2.0\, V$ and a negligible internal resistance. The potentiometer wire itself is $4\, m$ long. When the resistance $R$, connected across the given cell, has values of (i) infinity (ii) $9.5\, \Omega$ the balancing lengths on the potentiometer wire are found to be $3 \,m$ and $2.85\, m$, respectively. The value of internal resistance of the cell is

• A. $0.25\, \Omega$
• B. $0.95\, \Omega$
• C. $0.5\, \Omega$
• D. $0.75\, \Omega$

Answer 1

Answer: (C)

$0.5\, \Omega$

Answer 2

The internal resistance of the cell is
$r = \bigg(\frac{l_1}{l_2}-1\bigg) R$
Here, $l_1=3\, m,t_2=2.85\, m, R=9.5\, \Omega$
$\therefore \, \, \, \, r = \bigg(\frac{3}{2.85}-1\bigg) (9.5\, \Omega)= \frac{0.15}{2.85} \times 9.5\, \Omega=0.5\, \Omega$