In an ammeter, 5 % of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be :

Solution

Physics sir · 4 steps

The voltage across G is the voltage across the whole ammeter, so divide it by the full current.

1. Shunt in parallel. An ammeter is a galvanometer with a small shunt resistance S in parallel. Parallel parts share the same voltage, so the voltage across G equals the voltage across the whole ammeter.
2. Voltage across G. The galvanometer carries I_g = 0.05 I, so the voltage across it is V = 0.05 I G. The rest, 0.95 I, goes through the shunt, which gives S = (0.05 G)/0.95 = G/19.

   Rₐₘₘₑₜₑᵣ = V/I = (0.05 I G)/I = G/20

3. Check. Combine G and G/19 in parallel: (G × G/19)/(G + G/19) = (G²/19)/(20G/19) = G/20. Both methods agree.
4. So. The resistance of the ammeter is G/20, which is option (c).

The idea to keep. Adding a shunt in parallel always lowers the resistance below G. If a fraction f of the current goes through G, the ammeter resistance is fG, because both share one voltage.

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