A bulb is connected to a battery of emf 10 V so that the resulting current is 10 mA. When the bulb is connected to 220 V mains , the current is 50mA. Choose the correct alternative (s)

Options

A. In the first case , the resistance of the bulb is `1KOmega` and in second case ,it is `4.4kOmega`
B. It is not possible since ohm's law is not followed
C. the increase in resistance is due to heating of the filament of the bulb when it is connected to 220 V mains
D. None of these

Answer

To solve the problem, we need to determine the resistance of the bulb in two different scenarios and analyze the implications of the results.

### Step-by-Step Solution:

1. **Identify Given Values:**
- For the first scenario:
- Voltage (V1) = 10 V
- Current (I1) = 10 mA = 10 × 10^-3 A
- For the second scenario:
- Voltage (V2) = 220 V
- Current (I2) = 50 mA = 50 × 10^-3 A

2. **Calculate Resistance in the First Scenario:**
- Using Ohm's Law: \( R = \frac{V}{I} \)
- For the first case:
\[ R_1 = \frac{V_1}{I_1} = \frac{10 \text{ V}}{10 \times 10^{-3} \text{ A}} = \frac{10}{0.01} = 1000 \, \Omega = 1 \, k\Omega \]

3. **Calculate Resistance in the Second Scenario:**
- For the second case:
\[ R_2 = \frac{V_2}{I_2} = \frac{220 \text{ V}}{50 \times 10^{-3} \text{ A}} = \frac{220}{0.05} = 4400 \, \Omega = 4.4 \, k\Omega \]

4. **Analyze the Results:**
- The resistance of the bulb increased from 1 kΩ to 4.4 kΩ when the voltage was increased from 10 V to 220 V.
- This indicates that the resistance of the bulb is not constant and changes with temperature.

5. **Conclusion:**
- The increase in resistance is due to the heating of the filament of the bulb when connected to the higher voltage mains. This follows the behavior of conductors, where resistance increases with temperature.