A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to red

Question

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

TestHub · Accepted Answer

Current through the galvanometer I = 3 ( 50 + 2950 ) = 10 - 3 A Current for 30 divisions = 10 - 3 A Current for 20 divisions = 10 - 3 30 × 20 = 2 3 × 10 - 3 A For the same deflection to obtain for 20 divisions, let resistance added be R ∴ 2 3 × 10 - 3 = 3 50 + R o r R = 4450 Ω

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