JEE Main 2024 Physics question
A galvanometer has a resistance of 50 Ω and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance:
Answer: (C)
Solution
R = V/I − G = 100/0.005 − 50 = 19950 Ω.