Q. 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
A) 4450 Ω
B) 5050 Ω
C) 5550 Ω
D) 6050 Ω
Explanation
[Solution from source] Full-scale current = 3/(50 + 2950) = 1 mA for 30 divisions. For 20 divisions I = (2/3) mA, so total resistance = 3/(2/3 × 10⁻³) = 4500 Ω and the series resistance = 4500 − 50 = 4450 Ω. 4500 Ω would forget to subtract the galvanometer's own resistance. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.