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?

Total initial resistance

= G+ R = 50Ω + 2950Ω = 3000Ω

Current,I= 3v/3000Ω = 1 × 10−3A=1mA

current for 30 divisions = I = 10−3A

⇒ current for 1 division = I/30 = 1/30 × 10−3 A

⇒ current for 20 divisions = 20I/30 = 20/30 × 10−3 A = 2 × 10−3/3 A

now, equivalent resistance will be , Req = 3v/(2 × 10−3/3) = 9000/2 = 4500 Ω

as it is given, resistance of galvanometer is 50 Ω

so, resistance in series should be , R = Req – RG

= 4500 – 50 = 4450 Ω

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