When a resistance of 5 Ω is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA, however when 1050 Ω resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ Ω.

Solution

Physics sir · 4 steps

The same full-scale current I_g flows through the coil in both circuits, so write two equations and divide.

1. Shunt circuit. At full scale, I_g goes through the coil and 0.25 - I_g goes through the 5 Ω shunt. Both are in parallel, so I_g G = (0.25 - I_g) × 5. This rearranges to I_g (G + 5) = 1.25.
2. Series circuit. Now the same I_g flows through the coil and the 1050 Ω resistor together, across 25 V. So I_g (G + 1050) = 25.

   (G + 1050)/(G + 5) = 25/1.25 = 20

3. Solve for G. G + 1050 = 20G + 100, so 19G = 950 and G = 50 Ω. Check: I_g = 1.25/55 ≈ 22.7 mA, and 22.7 mA × 1100 Ω ≈ 25 V.
4. So. The resistance of the galvanometer is 50 Ω.

The idea to keep. The full-scale current I_g of a galvanometer does not change. When you have two conversions of the same meter, write one equation for each and divide to cancel I_g.

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