Question

Primary circuit of the potentiometer is shown in figure. Determine the potential gradient (means potential drop per unit length of potentiometer wire).

• A. $\frac{20}{21}\text{V/m}$
• B. $\frac{2}{21}\text{V/m}$
• C. $\frac{5}{9}\text{V/m}$
• D. $\frac{5}{8}\text{V/m}$

Answer 1

The correct option is B $\frac{2}{21}\text{V/m}$

Given:
$E=2\text{V}$
$r=1\Omega ;R=10\Omega ;R_1=10\Omega$

So, current $i$ passes through circuit will be
$i=\frac{E}{r+R_1+R}=\frac{2}{1+10+10}$

$\therefore i=\frac{2}{21}\text{A}$

So, potential drop across wire $\text{AB}$ will be

$V_{AB}=iR=\frac{2}{21}\times 10=\frac{20}{21}\text{V}$

Therefore, potential gradient $K$ is

$K=\frac{V_{AB}}{L}=\frac{20/21}{10}=\frac{2}{21}\text{V/m}$

Hence, option (b) is the correct answer.