Question

A potentiometer wire has length $4 \,m$ and resistance $8 \,\Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\, V$, so as to get a potential gradient $1 \,mV$ 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 $=1 \,m \,V \, cm ^{-1}=\frac{1}{10} \,V \,m^{-1}$
Length of potentiometer wire, $l=4 \,m$

So potential difference across potentiometer wire $=\frac{1}{10} \times 4=0.4 \, V$...(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 $
$\therefore R=32 \, \Omega$