A galvanometer has a coil of resistance 200 Ω with a full scale deflection at 20 μ A. The value of resistance to be added to use it as an ammeter of range (0-20) mA is :

Solution

Physics sir · 4 steps

Most of the 20 mA must bypass the coil, so put a small shunt in parallel and share the voltage.

1. Shunt in parallel. To make an ammeter, a small resistance S goes in parallel with the coil. Only I_g = 20 μA may pass through the coil. The remaining I - I_g flows through the shunt.
2. Equal voltages. Parallel branches have the same potential difference across them. So I_g G = (I - I_g) S.

   S = (I_g G)/(I - I_g)

3. Put in numbers. I_g G = 20 × 10⁻⁶ × 200 = 4 × 10⁻³ V. The shunt current is I - I_g = 20 mA - 0.02 mA = 19.98 × 10⁻³ A. So S = (4 × 10⁻³)/(19.98 × 10⁻³) ≈ 0.2002 Ω.
4. So. The shunt needed is about 0.20 Ω, which is option (c).

The idea to keep. Ammeter means a small shunt in parallel, and voltmeter means a large resistance in series. For a shunt, equate the voltages: I_g G = (I - I_g)S.

Answer from the NTA final answer key. Solution by PrepWiser.