A galvanometer of resistance 50 is connected to a battery of 3 V along with a resistance of 2950 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
- 6050
- 4450
- 5050
- 5550
Answer
(2) 4450
Solution
Initial current I1 = 350 + 2950 = 1 10-3 A (30 divisions)
For 20 divisions, I2 = 2030 10-3 = 23 10-3 A
50 + R = 3(2/3) 10-3 = 4500
R = 4450
Answer: (2)