Question

A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is :

• A. 48 Ω
• B. 32 Ω
• C. 40 Ω
• D. 44 Ω

Answer 1

The correct option is B

32 Ω

Figure alongside shows a potentiometer wire of length $L=4m$ and resistance $RAB=8\Omega$. Resistance connected in series is R. When an accumulator of emf $ϵ=2V$ is used, we have current I given by,
$I=\frac{ϵ}{R+R_{AB}}=\frac{2}{8+R}$
The resistance per unit length of the potentiometer wire is given by
$\frac{R_{AB}}{L}=\frac{8}{4}=2\Omega /m$
The potential gradient is given by
$\frac{I{R}_{AB}}{L}=\frac{2}{8+R}\times \frac{R_{AB}}{L}=\frac{2\times 2}{8+R}$
For a potential gradient 1mV per cm $=\frac{1\times {10}^{-3}}{{10}^{-2}}=0.1V/m$
We have $\frac{4}{8+R}=0.1$
$\therefore 8+R=40\therefore R=32\Omega$