Question

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 $\Omega$)

• A. 50
• B. 100
• C. 115
• D. 80

Answer 1

Answer: (C)

$\frac{2.5}{(30+5+R)}A$ ...(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 $\Omega$