The current senstivity `(S_i) `of a moving coil galvanometer of resistance 200 Omega is I division per mA. Its voltage senstivity is

Options

A. `5 xx 10^(-2)`div/mV
B. `5 xx 1-^(-4)`div/mV
C. `5 xx 10^(-3)`div/mV
D. `5 xx 10^(-1)`div/mV

Explanation

To find the voltage sensitivity of a moving coil galvanometer, we can use the relationship between current sensitivity and voltage sensitivity. The current sensitivity \( S_i \) is given as \( I \) divisions per mA, and the resistance \( R \) of the galvanometer is 200 Ω.

### Step 1: Understand the relationship between current sensitivity and voltage sensitivity
The voltage sensitivity \( S_v \) of a galvanometer can be expressed in terms of its current sensitivity \( S_i \) and its resistance \( R \) using the formula:

\[ S_v = S_i \times R \]

### Step 2: Convert current sensitivity to a suitable unit
Since \( S_i \) is given in divisions per mA, we need to convert it to divisions per A for consistency with the resistance in ohms.

If \( S_i = k \) divisions per mA, then:

\[ S_i = k \times 1000 \text{ divisions per A} \]

### Step 3: Substitute values into the formula
Now, substituting \( S_i \) and \( R \) into the formula for voltage sensitivity:

\[ S_v = (k \times 1000) \times 200 \]

### Step 4: Simplify the expression
This simplifies to:

\[ S_v = 200k \times 1000 = 200000k \text{ divisions per V} \]

### Final Result
Thus, the voltage sensitivity \( S_v \) of the moving coil galvanometer is:

\[ S_v = 200000k \text{ divisions per V} \]