In a moving coil galvanometer, the current sensitivity increases by `30%` .
When the coil resistance is doubled, calculate the percentage change in its voltage sensitivity.

Answer

To solve the problem, we need to analyze the relationship between current sensitivity (I_s), voltage sensitivity (V_s), and resistance (R) in a moving coil galvanometer.

### Step-by-step Solution:

1. **Understand the Definitions:**
- Current sensitivity (I_s) is defined as the deflection per unit current.
- Voltage sensitivity (V_s) is defined as the deflection per unit voltage.

2. **Relate Current and Voltage Sensitivity:**
- The relationship between current sensitivity and voltage sensitivity can be expressed as:
\[ V_s = \frac{I_s}{R} \]
where R is the resistance of the coil.

3. **Initial Conditions:**
- Let the initial current sensitivity be \( I_s \).
- The initial voltage sensitivity is given by:
\[ V_s = \frac{I_s}{R} \]

4. **Change in Current Sensitivity:**
- The problem states that the current sensitivity increases by 30%. Therefore, the new current sensitivity \( I_s' \) is:
\[ I_s' = 1.3 I_s \]

5. **Change in Resistance:**
- The resistance of the coil is doubled, so the new resistance \( R' \) is:
\[ R' = 2R \]

6. **Calculate New Voltage Sensitivity:**
- The new voltage sensitivity \( V_s' \) can be calculated using the new current sensitivity and the new resistance:
\[ V_s' = \frac{I_s'}{R'} = \frac{1.3 I_s}{2R} = \frac{1.3}{2} \cdot \frac{I_s}{R} = 0.65 V_s \]

7. **Determine the Percentage Change in Voltage Sensitivity:**
- The percentage change in voltage sensitivity can be calculated as:
\[ \text{Percentage Change} = \frac{V_s' - V_s}{V_s} \times 100\% \]
- Substituting \( V_s' \):
\[ \text{Percentage Change} = \frac{0.65 V_s - V_s}{V_s} \times 100\% = \frac{-0.35 V_s}{V_s} \times 100\% = -35\% \]

8. **Conclusion:**
- The voltage sensitivity decreases by 35%.

### Final Answer:
The percentage change in voltage sensitivity is a decrease of **35%**.

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