Question

The deflection in a moving coil galvanometer falls from 25 divisions to 5 divisions when a shunt of \( 24 \, \Omega \) is applied. The resistance of the galvanometer coil will be:

• A. \( 12 \, \Omega \)
• B. \( 96 \, \Omega \)
• C. \( 48 \, \Omega \)
• D. \( 100 \, \Omega \)

Answer 1

The Correct Option is B

Step 1: Define Variables: - Let x represent the current per division. - Initially, the full-scale deflection current I_g for 25 divisions is:

I_g = 25x

Step 2: After Applying the Shunt: - With the shunt applied, the deflection drops to 5 divisions, so the new current through the galvanometer becomes:

I_g = 5x

- The remaining current bypasses through the shunt, giving the total current I:

I = 25x

- The total current I is divided between the galvanometer and the shunt. So, the current through the shunt is:

I - I_g = 25x - 5x = 20x

Step 3: Using the Shunt Resistance: - The shunt resistance S = 24Ω. - Since the potential difference across the galvanometer and the shunt must be equal, we have:

I_g ⋅ G = (I - I_g) ⋅ S

- Substituting I_g = 5x, I - I_g = 20x, and S = 24Ω:

5x ⋅ G = 20x ⋅ 24

Step 4: Solving for G: - Cancel x from both sides:

5G = 20 × 24

- Simplify:

G = \( \frac{20 \times 24}{5} \) = 4 × 24 = 96Ω

So, the correct answer is: 96Ω