The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is dubled 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 understand the relationship between current sensitivity (Is), voltage sensitivity (Vs), and resistance (R) in a moving coil galvanometer.

### Step-by-Step Solution:

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

2. **Understand Voltage Sensitivity (Vs)**:
- Voltage sensitivity (Vs) is defined as the deflection (θ) per unit voltage (V). It can be expressed as:
\[ V_s = \frac{\theta}{V} \]
- Since \( V = IR \), we can also express voltage sensitivity in terms of current sensitivity and resistance:
\[ V_s = \frac{I_s}{R} \]

3. **Initial Conditions**:
- Let the initial current sensitivity be \( I_s \) and the initial resistance be \( R \).
- Thus, the initial voltage sensitivity can be expressed as:
\[ V_s = \frac{I_s}{R} \]

4. **Changes in Sensitivity**:
- According to the problem, the current sensitivity increases by 20% when the resistance is doubled.
- Therefore, the new current sensitivity \( I_s' \) becomes:
\[ I_s' = 1.2 I_s \]
- The new resistance \( R' \) is:
\[ R' = 2R \]

5. **Calculate New Voltage Sensitivity (Vs')**:
- Using the new values, the new voltage sensitivity \( V_s' \) can be calculated as:
\[ V_s' = \frac{I_s'}{R'} = \frac{1.2 I_s}{2R} \]
- Simplifying this expression gives:
\[ V_s' = \frac{1.2}{2} \cdot \frac{I_s}{R} = 0.6 \cdot V_s \]

6. **Conclusion**:
- The new voltage sensitivity \( V_s' \) is 0.6 times the initial voltage sensitivity \( V_s \). Thus, the factor by which the voltage sensitivity changes is:
\[ \text{Factor} = \frac{V_s'}{V_s} = 0.6 \]

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

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