Question

Potentiometer wire length is $10\text{m}$, having a total resistance of $10\Omega$. If a battery of emf $2\text{V}$ (of negligible internal resistance) and a rheostat are connected to it then the potential gradient is $2{\text{mV m}}^{-1}$. Find the resistance imparted through the rheostat:

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

Answer 1

The correct option is B $990\Omega$

Given,

$L_{AB}=10\text{m}{R}_{AB}=10\Omega E=2\text{V}\text{potential gradient}\varphi =20\text{mV/m}$


Let's say resistance imparted by the rheostat is $R$.

Current in the circuit is,

$i=\frac{V}{10+R}=\frac{2}{(10+R)}$

So,$\Delta V_{AB}=i\times R_{AB}=\frac{2}{10+R}\times 10=\left(\frac{20}{10+R}\right)$

Potential gradient,

$\varphi =\frac{\Delta V_{AB}}{L_{AB}}=\frac{20}{(10+R)10}=\frac{2}{10+R}$

$\frac{2}{10+R}=2\times {10}^{-3}{\text{Vm}}^{-1}$

$\frac{2}{10+R}=2\times {10}^{-3}\Rightarrow 1={10}^{-2}+R\times {10}^{-3}$

$R\times {10}^{-3}=1-0.01$

$R=990\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.