Q. A potentiometer wire of length 10 m and resistance $30\,\Omega$ is connected in series with a battery of emf 2.5 V and internal resistance of $5\,\Omega$ and an external resistance R. If the fall of potential along the potentiometer wire is 50 V/mm, the value of R is (in $\Omega$)
Solution:
$\frac{2.5}{(30 + 5 + R)} A $ ...(i)
Also I = $\frac{50 \times 10^{-6}}{10^{-3} \times \frac{30}{10}}= \frac{50}{3} \times 10^{-3} A$ .....(ii)
Equating (i) and (ii) $\frac{50}{3} \times 10^{-3} = \frac{2.5}{30 + 5 + R}$
i.e. R = 115 $\Omega$
