JEE Main 2024 Physics question
A galvanometer of resistance 100 Ω when connected in series with 400 Ω measures a voltage of upto 10 V. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is x × 10 Ω. The value of x is:
Answer: (C)
Solution
= 10/500 = 0.02 A. Shunt S = = 0.02 × 100/9.98 ≈ 0.2 Ω = 20 × 10⁻² Ω.