If the current sensitivity of a galvanometer is doubled, then its voltage sensitivity will be

Options

A. doubled
B. halved
C. unchanged
D. four times

Explanation

To solve the problem, we need to understand the relationship between current sensitivity and voltage sensitivity of a galvanometer.

### Step-by-Step Solution:

1. **Define Current Sensitivity**:
- Current sensitivity (S_I) of a galvanometer is defined as the deflection (θ) per unit current (I). Mathematically, it can be expressed as:
\[ S_I = \frac{\theta}{I} \]
- This can also be represented in terms of the galvanometer's parameters:
\[ S_I = \frac{NAB}{K} \]
where N is the number of turns, A is the area of the coil, B is the magnetic field, and K is the torsional constant.

2. **Define Voltage Sensitivity**:
- Voltage sensitivity (S_V) is defined as the deflection (θ) per unit voltage (V). It can be expressed as:
\[ S_V = \frac{\theta}{V} \]
- Using Ohm's law, we know that \( V = IR \), where R is the resistance. Therefore, we can express voltage sensitivity in terms of current sensitivity:
\[ S_V = \frac{\theta}{IR} = \frac{S_I}{R} \]

3. **Relationship Between Current and Voltage Sensitivity**:
- From the above equations, we can see that:
\[ S_V = \frac{S_I}{R} \]
- This indicates that voltage sensitivity is directly proportional to current sensitivity and inversely proportional to resistance.

4. **Doubling Current Sensitivity**:
- If the current sensitivity is doubled, we have:
\[ S_I' = 2S_I \]
- Now, substituting this into the voltage sensitivity equation:
\[ S_V' = \frac{S_I'}{R} = \frac{2S_I}{R} \]
- This shows that voltage sensitivity becomes:
\[ S_V' = 2 \cdot \frac{S_I}{R} = 2S_V \]
- Therefore, the voltage sensitivity also doubles when the current sensitivity is doubled, assuming resistance remains constant.

5. **Conclusion**:
- If the current sensitivity of a galvanometer is doubled, then its voltage sensitivity will also be doubled.

### Final Answer:
The voltage sensitivity will be **doubled**.