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:
- \( 12 \, \Omega \)
- \( 96 \, \Omega \)
- \( 48 \, \Omega \)
- \( 100 \, \Omega \)
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to find the resistance of a moving coil galvanometer using the concept of shunt resistance. A shunt is applied to decrease the deflection, and we can use the following formula:
The deflection of a galvanometer is proportional to the current passing through it. Therefore, the deflection is inversely proportional to the shunt resistance applied.
Given:
- Initial deflection, \( D_1 = 25 \) divisions
- Final deflection, \( D_2 = 5 \) divisions
- Shunt resistance, \( S = 24 \, \Omega \)
Using the formula based on proportionality of deflections:
\(\frac{D_1}{D_2} = 1 + \frac{G}{S}\)
Where \( G \) is the resistance of the galvanometer coil.
Substituting the given values into the formula, we have:
\(\frac{25}{5} = 1 + \frac{G}{24}\)
Simplifying the equation:
\(\frac{25}{5} = 5\)
Thus,
\(5 = 1 + \frac{G}{24}\)
Rearranging the terms gives:
\(5 - 1 = \frac{G}{24}\) \(4 = \frac{G}{24}\)
Finally, solving for \( G \) (the resistance of the galvanometer):
\(G = 4 \times 24 = 96 \, \Omega\)
Thus, the resistance of the galvanometer coil, \( G \), is \( 96 \, \Omega \).
The correct answer is, therefore, \( 96 \, \Omega \).
Approach Solution -2
Step 1: Define Variables: - Let x represent the current per division. - Initially, the full-scale deflection current Ig for 25 divisions is:
Ig = 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:
Ig = 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 - Ig = 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:
Ig ⋅ G = (I - Ig) ⋅ S
- Substituting Ig = 5x, I - Ig = 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
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