A light bulb is rated at 100 W for a 220 V ac supply . The resistance of the bulb is

Options

A. `284 Omega`
B. `384 Omega`
C. `484 Omega`
D. `584 Omega`

Explanation

To find the resistance of the light bulb rated at 100 W for a 220 V AC supply, we can use the formula for electrical power:

1. **Identify the formula for power**:
The power (P) consumed by a resistor can be expressed as:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the resistor and \( R \) is the resistance.

2. **Rearrange the formula to solve for resistance (R)**:
We can rearrange the formula to find the resistance:
\[ R = \frac{V^2}{P} \]

3. **Substitute the given values**:
We know the power \( P = 100 \) W and the voltage \( V = 220 \) V. Substituting these values into the formula gives:
\[ R = \frac{(220)^2}{100} \]

4. **Calculate \( V^2 \)**:
First, calculate \( 220^2 \):
\[ 220^2 = 48400 \]

5. **Calculate the resistance**:
Now substitute \( 48400 \) back into the equation:
\[ R = \frac{48400}{100} = 484 \, \Omega \]

6. **Conclusion**:
The resistance of the light bulb is \( 484 \, \Omega \).