Question

Primary circuit of a potentiometer is shown in the diagram. Determine the maximum potential difference, which we can measure by using this potentiometer.

• A. $\frac{20}{31}\text{V}$
  
• B. $\frac{2}{31}\text{V}$
  
• C. $\frac{5}{9}\text{V}$
  
• D. $\frac{5}{8}\text{V}$
  

Answer 1

The correct option is A $\frac{20}{31}\text{V}$


Current flowing through the primary cell is,
$I=\frac{E}{r+R_1+R}=\frac{2}{1+20+10}$

$\Rightarrow I=\frac{2}{31}\text{A}$

$V_{AB}=IR=\frac{2}{31}\times 10=\frac{20}{31}\text{V}$

The maximum potential difference, which we can measure by a potentiometer, is equal to the potential difference across the entire potentiometer wire.

Hence, $\left(A\right)$ is the correct answer.

$$\begin{array}{c}\text{why this Question ?}\\ \text{Any potential difference which is less than the potential difference maintained across the potentiometer wire can be measured using potentiometer.}\end{array}$$