A galvanometer has a coil of resistance `100 Omega` and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 V range, the resistance required to be added will be.

Options

A. ` 1000 Omega`
B. ` 900 Omega`
C. `1800 Omega`
D. `500 Omega`

Answer

To solve the problem of converting a galvanometer into a voltmeter, we need to follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer (G) = 100 Ω
- Full scale deflection current (I_max) = 30 mA = 30 × 10^(-3) A
- Desired maximum voltage (V_max) = 30 V

2. **Understand the Configuration:**
- To convert the galvanometer into a voltmeter, we need to add a resistance (R) in series with the galvanometer. The total resistance in the circuit will then be G + R.

3. **Apply Ohm's Law:**
- The relationship between voltage, current, and resistance is given by Ohm's Law:
\[ V = I \times R \]
- For the voltmeter, we can express the maximum voltage as:
\[ V_{max} = I_{max} \times (G + R) \]

4. **Rearranging the Formula:**
- Rearranging the formula to find R:
\[ G + R = \frac{V_{max}}{I_{max}} \]
- Substitute the known values:
\[ G + R = \frac{30 \, \text{V}}{30 \times 10^{-3} \, \text{A}} = 1000 \, \Omega \]

5. **Calculate the Required Resistance (R):**
- Now, we can find R:
\[ R = 1000 \, \Omega - G = 1000 \, \Omega - 100 \, \Omega = 900 \, \Omega \]

6. **Final Answer:**
- The resistance required to be added to the galvanometer to convert it into a voltmeter with a range of 30 V is **900 Ω**.