Question

Potentiometer wire of length $10\text{m}$, having a total resistance of $10\Omega$. A battery of emf $2\text{V}$ of negligible internal resistance and a rheostat are connected to it. If the potential gradient is $20\text{mV/m}$, find the resistance of the rheostat.

• A. $90\Omega$
• B. $990\Omega$
• C. $40\Omega$
• D. $190\Omega$

Answer 1

The correct option is A $90\Omega$


Let the resistance of Rheostat is $R$.

So,

$i=\frac{V}{R_{eq}}=\frac{2}{10+R}$

Now,

$\Delta V_{AB}=i\times R_{AB}=\frac{2}{10+R}\times 10=\frac{20}{10+R}$

Further, potential gradient $=\frac{\Delta V_{AB}}{L}$

$=\frac{20}{(10+R)\times 10}=\frac{2}{10+R}$

$\Rightarrow \frac{2}{10+R}=20\times {10}^{-3}$

$\Rightarrow 1={10}^{-1}+R\times {10}^{-2}$

$\Rightarrow R\times {10}^{-2}=1-0.1$

$\Rightarrow R=90\Omega$

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