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 2V, 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

The correct option is C $32\Omega$

the total resistance after adding resistance R in series with potentiometer wire
=$8\Omega +R\Omega$
the current in the circuit will be
$i=V/R_{eq}=\frac{2V}{R+8}$
the potential drop across potentiometer wire will be $i{R}_{potentiometer}=\frac{2V}{R+8}\times 8$
the potential gradient along length 4 m is given as
$=V_{p}otetiometer=\frac{\frac{2V}{R+8}\times 8}{4}$
the gradient is given as $1mV/cm=.1V/m$
$\frac{\frac{2V}{R+8}\times 8}{4}=.1$
$R=32\Omega$