A galvanometer of resistance 100 Ω when connected in series with 400 Ω measures a voltage of upto 10 V. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is x × 10⁻² Ω. The value of x is :

Solution

Physics sir · 3 steps

Use the voltmeter data to find the full-scale current first; the shunt follows from that one number.

1. Full-scale current. As a voltmeter, the total resistance is 100 + 400 = 500 Ω and full scale is 10 V. So the galvanometer gives full deflection at I_g = 10/500 = 0.02 A.
2. Shunt in parallel. To make an ammeter, a small shunt S goes in parallel. At 10 A, only I_g passes through the galvanometer and the rest goes through S. Both share the same voltage.

   I_g G = (I - I_g) S ⇒ S = (0.02 × 100)/(10 - 0.02) = 2/9.98 ≈ 0.2 Ω

3. So. S ≈ 0.2 Ω = 20 × 10⁻² Ω, so x = 20, option (b).

The idea to keep. Any galvanometer conversion starts from its full-scale current I_g. A voltmeter adds series resistance; an ammeter adds a parallel shunt carrying I - I_g at the same voltage.

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