A galvanometer of resistance 50 is connected to a battery of 4 V along with a resistance of 3950 in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 10 divisions, the resistance in series should be equal to:

Step-by-step solution

The current I flowing through the galvanometer for a full-scale deflection of 30 divisions is given by Ohm's law: I = V R_ g + R_ s = 4 50 + 3950 = 4 4000 = 10⁻³ A The current per division is 10⁻³ 30 A/division . To obtain a deflection of 10 divisions, the new current I' required is: I' = 10 10⁻³ 30 = 1 3 10⁻³ A Let the new series resistance be R'_ s . The circuit equation is: I' = V R_ g + R'_ s 1 3 10⁻³ = 4 50 + R'_ s 50 + R'_ s = 4 3 10³ = 12000 R'_ s = 12000 - 50 = 11950 Answer: 11950 ohm