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)

The internal resistance of the cell is
$r=\left(\frac{l_{1}}{l_{2}}-1\right) R$
Here, $l_{1}=3 \,m , l_{2}=2.85\, m , R=9.5 \,\Omega$
$\therefore r=\left(\frac{3}{2.85}-1\right)(9.5 \Omega)$
$=\frac{0.15}{2.85} \times 9.5 \,\Omega=0.5 \,\Omega$