Question

An ideal battery of emf $2 \, V$ and a series resistance $R$ are connected in the primary circuit of a potentiometer of length $1 \, m$ and resistance $5 \, \Omega$ . The value of $R$ , to give a potential difference of $5 \, mV$ across $10 \, cm$ of potentiometer wire is

• A. $1 8 0 \, Ω$
• B. $1 9 0 \, Ω$
• C. $1 9 5 \, \Omega $
• D. $2 0 0 \, Ω$

Answer 1

Answer: (C)

$\textit{I}=\frac{\text{E}}{\left(\text{R} + \text{r}\right)}=\frac{2}{\left(\text{R} + 5\right)} \, \text{A}$
Therefore, the potential difference across the potentiometer wire of lengths L = 100 cm is
$\text{V}=I\text{r}=\frac{2}{\left(\text{R} + 5\right)}\times 5=\frac{10}{\left(\text{R} + 5\right)} \, \text{V}$
Given, $V=5 \, mV=5\times 10^{- 3} \, V$ for 10 cm of wire. Hence, we have
$5\times \left(10\right)^{- 3}=\frac{1}{\left(\text{R} + 5\right)}\Rightarrow \text{R}=195 \, \Omega$