In a moving coil galvanometer, the current sensitivity \( S_I \) is directly proportional to the number of turns \( N \) of the coil. The voltage sensitivity \( S_V \) is proportional to the number of turns squared. Therefore, when the number of turns \( N \) of the coil is doubled, the current sensitivity \( S_I \) becomes double, while the voltage sensitivity \( S_V \) remains unchanged. Hence, the correct answer is that the current sensitivity is doubled, but voltage sensitivity remains unchanged.
The correct option is (D) : the current sensitivity is doubled but voltage sensitivity remains unchanged
In a moving coil galvanometer, the current sensitivity is defined as the deflection per unit current and is given by:
\( \text{Current Sensitivity} \propto NAB \) where:
If the number of turns \( N \) is doubled, the current sensitivity also doubles.
Voltage sensitivity is defined as the deflection per unit voltage:
\( \text{Voltage Sensitivity} = \frac{\text{Current Sensitivity}}{R} \), where \( R \) is the resistance of the coil.
When the number of turns is doubled, the length of the wire increases, so the resistance \( R \) also approximately doubles. Hence:
- Current sensitivity becomes \( 2 \times \text{original} \) - Resistance becomes \( 2 \times R \) - Voltage sensitivity becomes:
\( \frac{2 \times \text{original current sensitivity}}{2 \times R} = \text{original voltage sensitivity} \)
Final Answer: \( \boxed{\text{the current sensitivity is doubled but voltage sensitivity remains unchanged}} \)