Search any question & find its solution
Question:
Answered & Verified by Expert
To the potentiometer wire of length $L$ and $10 \Omega$ resistance, a battery of emf $2.5 \mathrm{~V}$ and a resistance $R$ are connected in series. If a potential difference of $1 \mathrm{~V}$ is balanced across $L$ / 2 length, the value of $R$ in $\Omega$ will be
Options:
Solution:
2866 Upvotes
Verified Answer
The correct answer is:
$2.5$
The given situation is shown below.
Resistance of potentioneter wire, $R_{p}=10 \Omega$
BmI, $E=25 \mathrm{~V}$
Total resistance, $R^{\prime}=R_{p}+R=(10+R) \Omega$
Current through potentiometer wire,
$$
l=\frac{E}{R_{p}}=\frac{25}{10+R}
$$
Since, potential difference across length $A L=\frac{L}{\mathrm{~g}}$,
$$
\begin{aligned}
& V &=l\left(\frac{R_{g}}{2}\right) \\
\Rightarrow & & 1 &=\frac{25}{10+R} \times \frac{10}{2} \\
\Rightarrow \quad & 10+n &=125 \\
\Rightarrow & & R &=25 \Omega
\end{aligned}
$$
Resistance of potentioneter wire, $R_{p}=10 \Omega$
BmI, $E=25 \mathrm{~V}$
Total resistance, $R^{\prime}=R_{p}+R=(10+R) \Omega$
Current through potentiometer wire,
$$
l=\frac{E}{R_{p}}=\frac{25}{10+R}
$$
Since, potential difference across length $A L=\frac{L}{\mathrm{~g}}$,
$$
\begin{aligned}
& V &=l\left(\frac{R_{g}}{2}\right) \\
\Rightarrow & & 1 &=\frac{25}{10+R} \times \frac{10}{2} \\
\Rightarrow \quad & 10+n &=125 \\
\Rightarrow & & R &=25 \Omega
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.