Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time and the…
Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively $360 \mathrm{~cm}$ and $420 \mathrm{~cm}$. If emf of $A$ is $2.4 \mathrm{~V}$, the emf of the second cell $B$ is
$2.8 \mathrm{~V}$
$3.2 \mathrm{~V}$
$3.0 \mathrm{~V}$
$2.6 \mathrm{~V}$
Solution
If $E_1$ and $E_2$ are emf's of cells with $l_1$ and $l_2$ balance lengths in a potentiometer experiment, then, $\frac{E_1}{E_2}=\frac{l_1}{l_2}$
or $\quad E_2=E_1 \cdot \frac{l_2}{l_1}$
Here, $E_1=24 \mathrm{~V}, l_1=360 \mathrm{~cm}$ and $l_2=420 \mathrm{~cm}$ Substituting values in Eq. (i), we get
$
E_2=24 \times \frac{420}{360}=28 \mathrm{~V}
$
So, emf of cell $B$ is $2.8 \mathrm{~V}$