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)

$32 \,\Omega$

Answer 2

Required potential gradient $=1\, mV\, cm^{-1}$ $=\frac{1}{10} V\, m^{-1}$
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
$\, \, \, 0.4=\frac{2}{8+R} \times 8$
$\frac{4}{10}=\frac{16}{8+R},8+R =40$
$\therefore \, \, \, R = 32 \,\Omega$