Search any question & find its solution
Question:
Answered & Verified by Expert
A galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \mathrm{~V}$ along with a resistance of $2950 \Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
Options:
Solution:
2308 Upvotes
Verified Answer
The correct answer is:
$4450 \Omega$
Current through the galvanometer
\(I=\frac{3}{(50+2950)}=10^{-3} \mathrm{~A}\)
Current for 30 divisions \(=10^{-3} \mathrm{~A}\)
Current for 20 divisions \(=\frac{10^{-3}}{30} \times 20\)
\(=\frac{2}{3} \times 10^{-3} \mathrm{~A}\)
For the same deflection to obtain for 20 divisions, let resistance added be R
\(\begin{aligned}
& \therefore \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 \mathrm{R})} \\
& \text {or } \mathrm{R}=4450 \Omega
\end{aligned}\)
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.