Question

A galvanometer, having a resistance of 50 Ω gives a full scale deflection for a current of 0.05 A. the length in meter of a resistance wire of area of cross-section 2.97 × 10^–2cm² that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is : (Specific resistance of the wire = 5 × 10^–7Ωm)

• A. 9
• B. 6
• C. 3
• D. 1.5

Answer 1

Answer: (C)

3

Answer 2

Given G = 50 Ω, i_g = 0.05 Amp., i = 5A, A = 2.97 × 10^–2 cm² and ρ = 5 × 10^–7Ω-m

By using $\frac{i}{i_{g}} = 1 + \frac{G}{S}$

⇒ $S = \frac{G.i_{g}}{(i - i_{g})} \Rightarrow \frac{\rho l}{A} = \frac{Gi_{g}}{(i - i_{g})} \Rightarrow l = \frac{Gi_{g}A}{(i - i_{g})\rho}$ on putting values l = 3 m