A moving coil galvanometer of resistance `50 Omega` gives a full scale deflection , when a current of `0.5` m A is passed through it. To convert it to a voltmeter of range , 10 V , the resistance required to be placed in series is

Options

A. `1995 Omega`
B. `2000 Omega`
C. `10000 Omega`
D. `19950 Omega`

Explanation

To solve the problem of converting a moving coil galvanometer into a voltmeter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer (R_g) = 50 Ω
- Full scale deflection current (I_g) = 0.5 mA = 0.5 × 10^(-3) A
- Desired voltage range (V) = 10 V

2. **Calculate the Total Current Required:**
- The current that will flow through the galvanometer at full scale deflection is given as 0.5 mA.

3. **Use Ohm's Law to Find the Required Resistance:**
- The voltage across the galvanometer when the full scale deflection current flows can be calculated using Ohm's Law:
\[ V_g = I_g \times R_g \]
- Substituting the values:
\[ V_g = (0.5 \times 10^{-3}) \times 50 = 0.025 \text{ V} = 25 \text{ mV} \]

4. **Calculate the Total Resistance Needed for the Voltmeter:**
- The total voltage (V) that we want to measure is 10 V. Therefore, the total resistance (R_total) needed to achieve this voltage with the given current is:
\[ R_{total} = \frac{V}{I_g} = \frac{10}{0.5 \times 10^{-3}} = 20000 \, \Omega \]

5. **Calculate the Required Series Resistance (R):**
- The total resistance required (R_total) is the sum of the galvanometer resistance (R_g) and the series resistance (R):
\[ R_{total} = R_g + R \]
- Rearranging gives:
\[ R = R_{total} - R_g = 20000 - 50 = 19950 \, \Omega \]

### Final Answer:
The resistance required to be placed in series with the galvanometer to convert it into a voltmeter of range 10 V is **19950 Ω**.