If the current current sensitivity of a moving coil galvanometer is increased by `20%`, its resistance also increases by `1*5` times. How will the voltage sensitivity of the galvanometer be affected?

Explanation

To solve the problem, we will analyze the effect of changes in current sensitivity and resistance on voltage sensitivity step by step.

### Step 1: Understand the Definitions
- **Current Sensitivity (Is)**: It is defined as the deflection (θ) per unit current (I) passing through the galvanometer. Mathematically, it is given by:
\[ I_s = \frac{\theta}{I R} \]
- **Voltage Sensitivity (Vs)**: It is defined as the deflection (θ) per unit voltage (V) across the galvanometer. Mathematically, it is given by:
\[ V_s = \frac{\theta}{V} = \frac{\theta}{I R} \]
Since \( V = I \cdot R \), we can express voltage sensitivity in terms of current sensitivity:
\[ V_s = \frac{I_s}{R} \]

### Step 2: Calculate the New Current Sensitivity
The current sensitivity is increased by 20%. If the original current sensitivity is \( I_s \), the new current sensitivity \( I_s' \) will be:
\[ I_s' = I_s + 0.2 I_s = 1.2 I_s = \frac{6}{5} I_s \]

### Step 3: Calculate the New Resistance
The resistance of the galvanometer is increased by 1.5 times. If the original resistance is \( R \), the new resistance \( R' \) will be:
\[ R' = 1.5 R \]

### Step 4: Calculate the New Voltage Sensitivity
Using the new current sensitivity and new resistance, we can find the new voltage sensitivity \( V_s' \):
\[ V_s' = \frac{I_s'}{R'} = \frac{\frac{6}{5} I_s}{1.5 R} \]
Substituting the value of \( R' \):
\[ V_s' = \frac{6}{5} \cdot \frac{I_s}{1.5 R} = \frac{6}{5} \cdot \frac{1}{1.5} V_s \]
Calculating \( \frac{6}{5} \cdot \frac{1}{1.5} \):
\[ \frac{6}{5} \cdot \frac{1}{1.5} = \frac{6}{5} \cdot \frac{2}{3} = \frac{12}{15} = \frac{4}{5} \]
Thus, we have:
\[ V_s' = \frac{4}{5} V_s \]

### Step 5: Determine the Percentage Change in Voltage Sensitivity
To find the percentage decrease in voltage sensitivity, we calculate:
\[ \text{Percentage Change} = \frac{V_s - V_s'}{V_s} \times 100 = \frac{V_s - \frac{4}{5} V_s}{V_s} \times 100 = \frac{1 - \frac{4}{5}}{1} \times 100 = \frac{1}{5} \times 100 = 20\% \]

### Conclusion
The voltage sensitivity of the galvanometer decreases by 20%.

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