A galvanometer coil has a resistance of `15 Omega` and the metre shows full scale deflection for a current of 4 mA. To convert the meter into a voltmeter of range 0 to 18 V, the required resistance is

Options

A. `5885 Omega "in series"`
B. `4485 Omega "in series"`
C. `5885 Omega "in series"`
D. `4485 Omega "in series"`

Explanation

To solve the problem of converting a galvanometer into a voltmeter, we need to calculate the required resistance that must be connected in series with the galvanometer. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of galvanometer, \( R_g = 15 \, \Omega \)
- Full-scale deflection current, \( I_g = 4 \, \text{mA} = 4 \times 10^{-3} \, \text{A} \)
- Desired voltage range of the voltmeter, \( V = 18 \, \text{V} \)

2. **Understand the Formula:**
To convert the galvanometer into a voltmeter, we need to add a resistance \( R \) in series with the galvanometer. The formula to calculate the required resistance \( R \) is given by:
\[ R = \frac{V}{I_g} - R_g \]
where:
- \( V \) is the maximum voltage range of the voltmeter,
- \( I_g \) is the full-scale deflection current of the galvanometer,
- \( R_g \) is the resistance of the galvanometer.

3. **Substitute the Values:**
Plug in the values into the formula:
\[ R = \frac{18 \, \text{V}}{4 \times 10^{-3} \, \text{A}} - 15 \, \Omega \]

4. **Calculate the First Term:**
Calculate \( \frac{18}{4 \times 10^{-3}} \):
\[ \frac{18}{4 \times 10^{-3}} = \frac{18}{0.004} = 4500 \, \Omega \]

5. **Subtract the Galvanometer Resistance:**
Now, subtract the galvanometer resistance from this value:
\[ R = 4500 \, \Omega - 15 \, \Omega = 4485 \, \Omega \]

6. **Final Answer:**
The required resistance to be connected in series with the galvanometer to convert it into a voltmeter of range 0 to 18 V is:
\[ R = 4485 \, \Omega \]