Question

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

• A. 4450 Ω
• B. 5050 Ω
• C. 5550 Ω
• D. 6050 Ω

Answer 1

Answer: (A)

4450 Ω

Answer 2

Current in the galvanometer

i_g = $\frac { 3 } { 2950 + 50 }$= 10^–3 A

i_g' = $\frac { 20 } { 30 }$× 10^–3 = $\frac { 2 } { 3 }$× 10^–3A

= $\frac { 2 } { 3 }$× 10^–3

∴ R = 4450 Ω