The ratio of voltage sensitivity `(V_S)` and current sensitivity `(I_S)` of a moving coil galvanometer is

Options

A. `1/G`
B. `1/(G^2)`
C. `G`
D. `G^2`

Answer

To solve the problem of finding the ratio of voltage sensitivity \((V_S)\) and current sensitivity \((I_S)\) of a moving coil galvanometer, we can follow these steps:

### Step 1: Understand the Definitions
- **Current Sensitivity \((I_S)\)** is defined as the deflection \((\phi)\) per unit current \((I)\).
- **Voltage Sensitivity \((V_S)\)** is defined as the deflection \((\phi)\) per unit voltage \((V)\).

### Step 2: Express Current Sensitivity
From the definition of current sensitivity:
\[ I_S = \frac{\phi}{I} \]
Where:
- \(\phi\) = deflection
- \(I\) = current through the coil

### Step 3: Express Voltage Sensitivity
From the definition of voltage sensitivity:
\[ V_S = \frac{\phi}{V} \]
Where:
- \(V\) = voltage across the galvanometer

### Step 4: Relate Voltage and Current
Using Ohm's law, we know that:
\[ V = I \cdot G \]
Where \(G\) is the resistance of the galvanometer.

### Step 5: Substitute for Voltage in Voltage Sensitivity
Substituting \(V\) in the expression for voltage sensitivity:
\[ V_S = \frac{\phi}{I \cdot G} \]

### Step 6: Relate \(V_S\) and \(I_S\)
Now, we can express \(V_S\) in terms of \(I_S\):
\[ V_S = \frac{\phi}{I \cdot G} = \frac{I_S \cdot I}{I \cdot G} = \frac{I_S}{G} \]

### Step 7: Find the Ratio of \(V_S\) to \(I_S\)
Now, we can find the ratio:
\[ \frac{V_S}{I_S} = \frac{I_S/G}{I_S} = \frac{1}{G} \]

### Final Result
Thus, the ratio of voltage sensitivity to current sensitivity of a moving coil galvanometer is:
\[ \frac{V_S}{I_S} = \frac{1}{G} \]