Question

In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of $2\Omega$, the balancing becomes 120 cm. The internal resistance of the cell is:

• A. $1\Omega$
• B. $0.5\Omega$
• C. $4\Omega$
• D. $2\Omega$

Answer 1

Answer: (D)

As, $k{l}_1=E-ri=E-r(0)=E$

$\Rightarrow k{l}_1=E\dots (i)$
and $k{l}_2=E-r{i}_1=R{i}_1$
$\Rightarrow i_1=\frac{E}{R+r}$
$\Rightarrow k{l}_2=\frac{RE}{R+r}\dots (ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\frac{l_1}{l_2}=\frac{R+r}{R}=1+\frac{r}{R}$
$\Rightarrow r=R\left(\frac{l_1}{l_2}-1\right)\dots (iii)$
where, $l_1=$ balancing length of potentiometer wire $=240cm$
$l_2=$ balancing length after shunting $=120cm$ and $R=$ shunting resistance $=2\Omega$
Putting all the values in Eq. (iii), we get
$r=\left(\frac{240}{120}-1\right)\times 2=(2-1)\times 2=2\Omega$