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

We are given that the electric bulb is rated at 220V and 11W. We are tasked with finding the resistance of its filament when it glows with a power supply of 220V. To solve this, we use the formula for electrical power: \[ P = \frac{V^2}{R} \] where:
\( P \) is the power consumed by the bulb (11W),
\( V \) is the voltage across the bulb (220V),
\( R \) is the resistance of the bulb.
Rearranging the formula to solve for \( R \), we get: \[ R = \frac{V^2}{P} \] Substitute the known values for \( V \) and \( P \): \[ R = \frac{220^2}{11} \] First, square the voltage value: \[ 220^2 = 48400 \] Now, divide by the power: \[ R = \frac{48400}{11} = 4400 \, \Omega \] Thus, the resistance of the filament is \( 4400 \, \Omega \). Therefore, the correct answer is option (a).