A galvanometer having coil of resistance `100 Omega` shows full-scale deflection when 1 mA current is passed through it. What resistance is needed to be connected in series with the coil so that it may be used as a voltmeter which may measure a potential difference of 5 volts?

Options

A. `49 Omega`
B. `490 Omega`
C. `4,900 Omega`
D. `49.000 Omega`

Explanation

To solve the problem, we need to determine the resistance that must be connected in series with the galvanometer so that it can function as a voltmeter capable of measuring a potential difference of 5 volts.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer (G) = 100 Ω
- Full-scale deflection current (I_G) = 1 mA = 1 × 10^(-3) A
- Desired potential difference (V) = 5 V

2. **Use the Formula for Voltmeter:**
The total resistance (R_total) needed to measure the voltage can be expressed as:
\[ R_{total} = R_G + R_S \]
where \( R_S \) is the series resistance we need to find.

3. **Apply Ohm's Law:**
The current through the galvanometer at full-scale deflection can also be expressed using Ohm's Law:
\[ I_G = \frac{V}{R_{total}} \]
Substituting for \( R_{total} \):
\[ I_G = \frac{V}{R_G + R_S} \]

4. **Rearranging the Equation:**
We can rearrange this equation to find \( R_S \):
\[ R_G + R_S = \frac{V}{I_G} \]
Therefore,
\[ R_S = \frac{V}{I_G} - R_G \]

5. **Substituting the Known Values:**
Now substitute the known values into the equation:
\[ R_S = \frac{5 \text{ V}}{1 \times 10^{-3} \text{ A}} - 100 \text{ Ω} \]

6. **Calculating \( R_S \):**
\[ R_S = 5000 \text{ Ω} - 100 \text{ Ω} = 4900 \text{ Ω} \]

7. **Final Answer:**
The resistance needed to be connected in series with the galvanometer is:
\[ R_S = 4900 \text{ Ω} \]