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

Question

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

TestHub · Accepted Answer

Current through the galvanometer I = ( 50 + 2950 ) 3 ​ Current for 30 divisions Current for 20 divisions ​ = 1 0 − 3 A = 1 0 − 3 A = 30 1 0 − 3 ​ × 20 = 3 2 ​ × 1 0 − 3 A ​ For the same deflection to obtain for 20 divisions, let resistance added be R ∴ 3 2 ​ × 1 0 − 3 = ( 50 + 1 R ) 3 ​ or R = 4450Ω ​

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