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, $kl_1 = E - ri = E - r(0) = E$

$\Rightarrow \, k l_{1}=E\, \dots(i) $
and $k l_{2}=E-r i_{1}=R i_{1}$
$\Rightarrow i_{1}=\frac{E}{R+r}$
$\Rightarrow k l_{2}=\frac{R E}{R+r}\, \dots(ii)$
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 $=240\,cm$
$l_{2}=$ balancing length after shunting $=120\, cm$ and $R=$ shunting resistance $=2 \Omega$
Putting all the values in Eq. (iii), we get
$r=\left(\frac{240}{120}-1\right) \times 2=(2-1) \times 2=2 \Omega$