In an attempt to increases the current sensitivity of a moving coil galvanometer, it is found that its resistance becomes double while the current sensitivity increases by 10%. The voltage sensitivity of the galvanometer changes by

Options

A. `40%`
B. `-45%`
C. `55%`
D. `-55%`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Initial Current Sensitivity
Let the initial current sensitivity of the galvanometer be denoted as \( I_s \).

### Step 2: Calculate Final Current Sensitivity
The problem states that the current sensitivity increases by 10%. Therefore, the final current sensitivity \( I_s' \) can be calculated as:
\[ I_s' = I_s + 0.1 I_s = 1.1 I_s \]

### Step 3: Define Initial and Final Resistance
Let the initial resistance of the galvanometer be \( R \). According to the problem, the resistance doubles, so the final resistance \( R' \) is:
\[ R' = 2R \]

### Step 4: Calculate Initial Voltage Sensitivity
The voltage sensitivity \( V_s \) is defined as the ratio of current sensitivity to resistance:
\[ V_s = \frac{I_s}{R} \]

### Step 5: Calculate Final Voltage Sensitivity
Using the final current sensitivity and final resistance, the final voltage sensitivity \( V_s' \) can be expressed as:
\[ V_s' = \frac{I_s'}{R'} = \frac{1.1 I_s}{2R} \]
This can be rewritten as:
\[ V_s' = \frac{1.1}{2} \cdot \frac{I_s}{R} = 0.55 V_s \]

### Step 6: Calculate Change in Voltage Sensitivity
To find the change in voltage sensitivity, we can use the formula:
\[ \Delta V_s = \frac{V_s' - V_s}{V_s} \times 100\% \]
Substituting the values we found:
\[ \Delta V_s = \frac{0.55 V_s - V_s}{V_s} \times 100\% = \frac{-0.45 V_s}{V_s} \times 100\% = -45\% \]

### Final Answer
The change in voltage sensitivity is:
\[ \Delta V_s = -45\% \]

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