An electric bulb is rated 60 W , 220 V . The resistance of its filament is

Options

A. `708 Omega`
B. `870 Omega`
C. `807 Omega`
D. `780 Omega`

Answer

To find the resistance of the filament of an electric bulb rated at 60 W and 220 V, we can use the formula relating power (P), voltage (V), and resistance (R):

1. **Identify the given values**:
- Power (P) = 60 W
- Voltage (V) = 220 V

2. **Use the formula for power**:
The formula for electrical power is given by:
\[ P = \frac{V^2}{R} \]
Rearranging this formula to solve for resistance (R), we get:
\[ R = \frac{V^2}{P} \]

3. **Substitute the known values into the formula**:
\[ R = \frac{(220)^2}{60} \]

4. **Calculate \(220^2\)**:
\[ 220^2 = 48400 \]

5. **Now substitute this value back into the equation**:
\[ R = \frac{48400}{60} \]

6. **Perform the division**:
\[ R = 806.67 \, \Omega \]

7. **Round the answer**:
The resistance of the filament is approximately:
\[ R \approx 807 \, \Omega \]

Thus, the resistance of the filament is **807 ohms**.