A galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \mathrm{~V}$ along with a…

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
  1. $5050 \Omega$
  2. $5550 \Omega$
  3. $6050 \Omega$
  4. $4450 \Omega$

Solution


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
$\begin{aligned} & \therefore \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 R)} \\ & \text{or } R=4450 \Omega \end{aligned}$

Asked in: NEET 2008 (Screening)

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