Question

Primary circuit of potentiometer is shown in figure. Determine the maximum potential which we can measure by the given 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 passing throught the circuit,

$i=\frac{\epsilon }{r+R_1+R}$

Substituting the values, we have

$i=\frac{2}{1+20+10}$

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

Potential difference across wire $\text{AB}$,

$V_{AB}=iR=\frac{2}{31}\times 10$

$\therefore V_{AB}=\frac{20}{31}\text{volts}$

Maximum potential which we can measure by it is equal to the potential drop across wire $\text{AB}$ = $\frac{20}{31}\text{volts}$

Hence, option $\left(\text{A}\right)$ is the correct answer.