The 60 Ω galvanometer of the previous setup is converted to an ammeter with a shunt rₛ = 0.02 Ω, then placed in the 3 V, 3 Ω circuit. The current it now reads is:

1. Shunt in parallel: effective resistance $= \dfrac{R_G r_s}{R_G + r_s} = \dfrac{60\times0.02}{60.02} \approx 0.02\ \Omega$. 2. Total circuit resistance: $0.02 + 3 = 3.02\ \Omega$. 3. Current: $I = \dfrac{3}{3.02}$. 4. $I = 0.99$ A — close to the ideal value, unlike the 0.048 A of the bare galvanometer. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.161_