A moving-coil galvanometer of resistance 100 Ω is converted into an ammeter using a shunt of 0.1 Ω. Its maximum-deflection galvanometer current is 100 μA. The full-scale ammeter current equals (10ⁿ + 0.1) mA. Find n.
Physics · Mixed Chapter
JEE Mains

Question

A moving-coil galvanometer of resistance 100 Ω is converted into an ammeter using a shunt of 0.1 Ω. Its maximum-deflection galvanometer current is 100 μA. The full-scale ammeter current equals (10ⁿ + 0.1) mA. Find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Galvanometer and shunt are in parallel: The potential drops across galvanometer and shunt are equal: IₛRₛ = I𝓰R𝓰. Step 2 — Current through shunt: I𝓰 = 100 μA = 0.1 mA, R𝓰 = 100 Ω, Rₛ = 0.1 Ω. Iₛ = I𝓰(R𝓰/Rₛ) = 0.1×(100/0.1) = 100 mA. Step 3 — Total ammeter current: I = I𝓰 + Iₛ = 0.1 + 100 = 100.1 mA. Step 4 — Compare with requested expression: I = (10ⁿ + 0.1) mA = (10² + 0.1) mA. Answer: n = 2.