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  • A galvanometer of resistance  is connected to a battery of <img alt="3V" src="https://entra

A galvanometer of resistance 50 \Omega is connected to a battery of 3V 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

Option: 1

5050 \Omega


Option: 2

5550 \Omega


Option: 3

6050 \Omega


Option: 4

4450 \Omega


Answers (1)

best_answer

Total initial resistance =50 \Omega+2950 \Omega=3000 \Omega
EMF of the cell, E=3 \mathrm{~V}
Current, \frac{E}{R}=\frac{3 \mathrm{~V}}{3000 \Omega}=1 \times 10^{-3} \mathrm{~A}=1 \mathrm{~mA}
\therefore The current for full-scale deflection of 30 divisions is 1 mA
\therefore Current for a deflection of 20 divisions,
\mathrm{I}=\frac{1 \mathrm{~mA}}{30} \times 20=\frac{2}{3} \mathrm{~mA}
Let x be the resistance of the circuit,
x=\frac{E}{I}=\frac{3 V}{\left(\frac{2}{3} \times 10^{-3} P\right)}=4500 \Omega
But the resistance of the galvanometer is 500 \Omega
\therefore Resistance to be added =(4500-50)=4450 \Omega

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