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 a resistance $R$, connected across the given cell, has values of,
(i) infinity
(ii) $10\, \Omega$
The balancing lengths on the potentiometer wire are found to be $60\, cm$ and $50\, cm$ respectively. The value of internal resistance of the cell is:-

• A. $0.25\, \Omega$
• B. $0.95\, \Omega$
• C. $0.5\, \Omega$
• D. $2\, \Omega$

Answer 1

Answer: (D)

$\varepsilon=\phi \times \ell_{1}$ ... (i)
$\frac{\varepsilon R}{ R + r }=\phi \times \ell_{2}$ ... (ii)
$r =\left(\frac{\ell_{1}-\ell_{2}}{\ell_{2}}\right) R =\left(\frac{60-50}{50}\right) \times 10$
$\Rightarrow r =2\, \Omega$