JEE Main 2025 · 23 January · evening · Single correct · +4 / −1
A galvanometer having a coil of resistance 30 Ω need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be 30/X Ω, where X is
- (a) 447
- (b) 298
- (c) 149
- (d) 596
Answer and solution
Answer (c) 149
Solution
Physics sir · 4 steps
The shunt carries everything the galvanometer cannot: 3 A minus 20 mA, at the same voltage as the coil.
- Split the current. At full scale the galvanometer takes I_g = 20 mA = 0.02 A. The rest, I - I_g = 3 - 0.02 = 2.98 A, must go through the shunt.
- Same voltage. The shunt is in parallel with the coil, so both have the same potential difference across them: I_g G = (I - I_g) S.
S = (I_g G)/(I - I_g) = (0.02 × 30)/2.98 = 0.6/2.98 Ω
- Match the form. Multiply the top and bottom by 50 to get 30 on top: 0.6/2.98 = 30/149 Ω.
- So. The shunt is 30/149 Ω, so X = 149, option (c).
The idea to keep. For an ammeter, the shunt takes I - I_g, not the full I. Equate the voltages across coil and shunt, I_g G = (I - I_g)S, and you have the shunt.
Answer from the NTA final answer key. Solution by PrepWiser.