Question

An electric bulb is rated 220V, 11W. The resistance of its filament when it glows with a power supply of 220V is:

• A. 4400 \( \Omega \)
• B. 440 \( \Omega \)
• C. 400 \( \Omega \)
• D. 20 \( \Omega \)

Hint:

For an electrical bulb, you can calculate the resistance from its power and voltage using \( R = \frac{V^2}{P} \).

Answer 1

The Correct Option is A

The electric bulb is rated at 220V and 11W. We need to find the resistance of its filament when powered by 220V. We use the power formula: \[P = \frac{V^2}{R}\] where: \( P \) is the power (11W), \( V \) is the voltage (220V), \( R \) is the resistance. Solving for \( R \): \[R = \frac{V^2}{P}\] Substitute the values: \[R = \frac{220^2}{11}\] Calculate \( 220^2 \): \[220^2 = 48400\] Then, divide by the power: \[R = \frac{48400}{11} = 4400 \, \Omega\] The filament's resistance is \( 4400 \, \Omega \). The correct answer is option (a).