In full scale deflection current in galvanometer of `100 ohm` resistance is 1 mA. Resistance required in series to convert it into voltmeter of range 10 V.

Options

A. `0.99 K ohm`
B. `9.9 K ohm`
C. `9.8 K ohm`
D. `10 K ohm`

Explanation

To solve the problem of converting a galvanometer into a voltmeter, we need to determine the resistance required in series with the galvanometer. Here’s a step-by-step solution:

### Step 1: Understand the given values
- The resistance of the galvanometer (Rg) = 100 ohms
- The full-scale deflection current (Imax) = 1 mA = 0.001 A
- The desired maximum voltage (Vmax) = 10 V

### Step 2: Use Ohm's Law
The relationship between voltage (V), current (I), and resistance (R) is given by Ohm's Law:
\[ V = I \times R \]

### Step 3: Calculate the total resistance needed for the voltmeter
To find the total resistance (R_total) required to measure up to 10 V at a maximum current of 1 mA, we can rearrange Ohm's Law:
\[ R_{total} = \frac{V_{max}}{I_{max}} \]
Substituting the values:
\[ R_{total} = \frac{10 \, \text{V}}{0.001 \, \text{A}} = 10,000 \, \text{ohms} \]

### Step 4: Determine the series resistance (Ra)
Since the galvanometer has its own resistance (Rg), the resistance required in series (Ra) can be calculated as:
\[ R_{total} = R_g + R_a \]
Rearranging gives:
\[ R_a = R_{total} - R_g \]
Substituting the known values:
\[ R_a = 10,000 \, \text{ohms} - 100 \, \text{ohms} = 9,900 \, \text{ohms} \]

### Step 5: Final answer
The resistance required in series to convert the galvanometer into a voltmeter of range 10 V is:
\[ R_a = 9,900 \, \text{ohms} \]