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A galvanometer of resistance 50Ω is connected to a battery of 3V along with a resistance of 2950Ω 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

6050Ω

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B

4450Ω

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C

505Ω

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D

5550Ω

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Solution

The correct option is B

4450Ω


Step 1. Given data

A galvanometer of resistance 50Ω is connected to a battery of 3V along with a resistance of 2950Ω 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 3V connection across it corresponds to 20 divisions.

Step 2. Formula to be used.

According to Kirchhoff's law,

If V1,V2,.....Vn shows the individual potential difference across them, then by Kirchhoff’s law of voltage we get,
V=V1+V2+.....+Vn
For the same is the effective resistance in series is R, then current I in the circuit is given by ohms law i.e.

V=IR

Step 3. Calculate current and total initial resistance.

Total initial resistance is,

RG+R1=50+2950

=3000Ω

ε=3V

Therefore,

Current =3V30000Ω

=1×10-3A

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 i=1mA×23 as full deflection scale for 1mA which is equal to 30 divisions.

3V=3000Ω×1mA

=xΩ×23mA

x=3000×1×32

=4500Ω

But the galvanometer resistance is 50Ω.

So,

The resistance to be added =4500-50Ω

=4450Ω

Hence, option B is correct answer.


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