Question

A battery is connected to a potentiometer and the balance point is obtained at $84\text{cm}$ along the wire. When its terminals are connected by a $5\Omega$ resistor, the balance point changes to $70\text{cm}$.

$\left(ii\right)$ Find the new position of the balance point when $5\Omega$ resistor is changed by $4\Omega$ resistor.

​​​​​​​$[$1 Mark$]$

• A. $26.5\text{cm}$
• B. $52\text{cm}$
• C. $67.2\text{cm}$
• D. $83.3\text{cm}$

Answer 1

The correct option is C $67.2\text{cm}$

The formula for internal resistance $r$ of cell using potentiometer is given by
$r=\left(\frac{l_1-l_2}{l_2}\right)R$
Where
$l_1$ - Balancing length when only the battery is connected.
$l_2$ - Balancing length when the battery is shunted by a resistance
$r=\left(\frac{84-70}{70}\right)\times 5=\frac{14}{70}\times 5$
$r=1\Omega$
Internal resistance of cell, $r=1\Omega$
Given: $R^{\prime }=4\Omega$
$l_1=84\text{cm}$
Let the new balancing length when $4\Omega$ resistor is connected be $l_3$
$\therefore r=\left(\frac{l_1-l_3}{l_3}\right)R^{\prime }$
$1=(\frac{l_1}{l_3}-1)4$
$\frac{1}{4}=\frac{l_1}{l_3}-1$
$\frac{l_1}{l_3}=\frac{5}{4}$
$l_3=\frac{4}{5}{l}_1=\frac{4}{5}\times 84$
$l_3=67.2\text{cm}$
$\therefore$ option c is correct.