The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is doubled calculate, by what factor does the voltage sensitivity change?

Options

A. Becomes `3/5` times
B. Becomes `2/5` times
C. No change
D. Decreases by a factor of `7/5`

Answer

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

### Step-by-Step Solution:

1. **Understanding Current Sensitivity (Is):**
The current sensitivity (Is) of a galvanometer is defined as the deflection (θ) per unit current (I). Mathematically, it can be expressed as:
\[ I_s = \frac{\theta}{I} \]

2. **Understanding Voltage Sensitivity (Vs):**
The voltage sensitivity (Vs) is defined as the deflection per unit voltage (V). Since voltage (V) can be expressed as \( V = I \cdot R \), we can rewrite the voltage sensitivity as:
\[ V_s = \frac{\theta}{V} = \frac{\theta}{I \cdot R} = \frac{I_s}{R} \]

3. **Given Information:**
- The current sensitivity increases by 20% when the resistance is doubled.
- Therefore, the new current sensitivity (Is') is:
\[ I_s' = 1.2 \cdot I_s \]
- The new resistance (R') is:
\[ R' = 2R \]

4. **Calculating New Voltage Sensitivity (Vs'):**
Using the new current sensitivity and the new resistance, we can find the new voltage sensitivity:
\[ V_s' = \frac{I_s'}{R'} = \frac{1.2 \cdot I_s}{2R} \]

5. **Substituting for Vs:**
We know that:
\[ V_s = \frac{I_s}{R} \]
Therefore, we can express Vs' in terms of Vs:
\[ V_s' = \frac{1.2 \cdot I_s}{2R} = \frac{1.2}{2} \cdot \frac{I_s}{R} = \frac{1.2}{2} \cdot V_s = 0.6 \cdot V_s \]

6. **Finding the Factor of Change:**
The voltage sensitivity changes by a factor of:
\[ \text{Factor} = \frac{V_s'}{V_s} = 0.6 \]

### Final Answer:
The voltage sensitivity changes by a factor of **0.6**.