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

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, 5050 Ω, 5550 Ω, 6050 Ω, 4450 Ω

MARKS App · 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

Search any question & find its solution

Looking for more such questions to practice?