Question

A galvanometer of resistance $50\mathrm{\Omega }$ is connected to a battery of $3\mathrm{V}$ along with a resistance of $2950\mathrm{\Omega }$ 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

• A. $6050\mathrm{\Omega }$
• B. $4450\mathrm{\Omega }$
• C. $505\mathrm{\Omega }$
• D. $5550\mathrm{\Omega }$

Answer 1

The correct option is B

$4450\mathrm{\Omega }$

Step 1. Given data

A galvanometer of resistance $50\mathrm{\Omega }$ is connected to a battery of $3\mathrm{V}$ along with a resistance of $2950\mathrm{\Omega }$ in series which is shown in above fiugre. A full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions.

We have to find the resistance in series that is to be connected to the above set up such that the deflection for $3\mathrm{V}$ connection across it corresponds to $20$ divisions.

Step 2. Formula to be used.

According to Kirchhoff's law,

If ${\mathrm{V}}_1,{\mathrm{V}}_2,.....{\mathrm{V}}_{\mathrm{n}}$ shows the individual potential difference across them, then by Kirchhoff’s law of voltage we get,
$\mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2+.....+{\mathrm{V}}_{\mathrm{n}}$
For the same is the effective resistance in series is $\mathrm{R}$, then current $\mathrm{I}$ in the circuit is given by ohms law i.e.

$\mathrm{V}=\mathrm{IR}$

Step 3. Calculate current and total initial resistance.

Total initial resistance is,

${\mathrm{R}}_{\mathrm{G}}+{\mathrm{R}}_1=50+2950$

$=3000\mathrm{\Omega }$

$\mathrm{\epsilon }=3\mathrm{V}$

Therefore,

Current $=\frac{3\mathrm{V}}{30000\mathrm{\Omega }}$

$=1\times {10}^{-3}\mathrm{A}$

Step 4. The value of current the resistance R to be connected in series.

If the deflection has to be reduced to $20$ divisions, current $\mathrm{i}=1\mathrm{mA}\times \frac{2}{3}$ as full deflection scale for $1\mathrm{mA}$ which is equal to $30$ divisions.

$3\mathrm{V}=3000\mathrm{\Omega }\times 1\mathrm{mA}$

$=x\mathrm{\Omega }\times \frac{2}{3}\mathrm{mA}$

$\Rightarrow$ $\mathrm{x}=3000\times 1\times \frac{3}{2}$

$=4500\mathrm{\Omega }$

But the galvanometer resistance is $50\mathrm{\Omega }$.

So,

The resistance to be added $=\left(4500-50\right)\mathrm{\Omega }$

$=4450\mathrm{\Omega }$

Hence, option $\mathrm{B}$ is correct answer.