A moving coil galvanometer 'A' has 200 turns and a resistance of `100Omega`. Another meter 'B' has 100 turns resistance 400 . All the other quantities are same in both the cases. The current sensitivity of A is

Options

A. double that of B
B. 2.5 times of B
C. 5 times of B
D. `1//5^(th)` of B

Answer

To find the current sensitivity of galvanometer A, we can use the formula for current sensitivity of a moving coil galvanometer, which is given by:

\[ S = \frac{N \cdot B \cdot A}{C} \]

Where:
- \( S \) is the current sensitivity,
- \( N \) is the number of turns,
- \( B \) is the magnetic induction,
- \( A \) is the area of the coil,
- \( C \) is the restoring torque.

Since the problem states that all other quantities are the same for both galvanometers A and B, we can focus on the number of turns and the resistance to find the ratio of their current sensitivities.

### Step-by-Step Solution:

1. **Identify the parameters for both galvanometers**:
- For galvanometer A:
- Number of turns, \( N_A = 200 \)
- Resistance, \( R_A = 100 \, \Omega \)
- For galvanometer B:
- Number of turns, \( N_B = 100 \)
- Resistance, \( R_B = 400 \, \Omega \)

2. **Write the formula for current sensitivity**:
The current sensitivity for galvanometer A is:
\[ S_A = \frac{N_A \cdot B \cdot A}{C} \]
And for galvanometer B:
\[ S_B = \frac{N_B \cdot B \cdot A}{C} \]

3. **Find the ratio of current sensitivities**:
The ratio of the current sensitivities of A and B can be expressed as:
\[ \frac{S_A}{S_B} = \frac{N_A}{N_B} \]

4. **Substitute the values**:
Substituting the values of \( N_A \) and \( N_B \):
\[ \frac{S_A}{S_B} = \frac{200}{100} = 2 \]

5. **Conclusion**:
This means that the current sensitivity of galvanometer A is 2 times that of galvanometer B:
\[ S_A = 2 \cdot S_B \]

### Final Answer:
The current sensitivity of A is double that of B.