Question

In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length $\ell$ is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance $R$. If $R$ is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be

• A. $\ell$
• B. $2\ell$
• C. $\ell /2$
• D. $\ell /4$

Answer 1

Answer: (C)

Balancing length $I$ will give emf of cell
$\therefore E=Kl$
Here K is potential gradient.
If the cell is short circuited by resistance 'R'
Let balancing length obtained be l' then $V=k{l}^{\prime }$
$r=\left(\frac{E-V}{V}\right)R$
$\Rightarrow V=E-V$ [$\because -r=R$ given]
$\Rightarrow 2V=E$
or,$2K{l}^{\prime }=Kl$