Question

A potentiometer wire has length $4m$ and resistance $8\Omega$. 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 $1mV$ per cm on the wire is

• A. $44\Omega$
• B. $48\Omega$
• C. $32\Omega$
• D. $40\Omega$

Answer 1

Answer: (C)

Required potential gradient $=1mVc{m}^{-1}=\frac{1}{10}V{m}^{-1}$
Length of potentiometer wire, $l=4m$

So potential difference across potentiometer wire $=\frac{1}{10}\times 4=0.4V$...(i)
In the circuit, potential difference across $8\Omega =1\times 8=\frac{2}{8+R}\times 8$...(ii)
Using equation (i) and (ii), we get
$0.4=\frac{2}{8+R}\times 8$
$\frac{4}{10}=\frac{16}{8+R},8+R=40$
$\therefore R=32\Omega$