Question

If the voltage across a bulb rated 220V – 60W drops by 1.5% of its rated value, the percentage drop in the rated value of the power is:

• A. 0.75%
• B. 1.5%
• C. 4.5%
• D. 3%
• E. 2.5%

Hint:

For power-related problems, use the formula \( P = \frac{V^2}{R} \) to calculate changes in power when voltage changes.

Answer 1

The Correct Option is D

Step 1: The power \( P \) consumed by the bulb is related to the voltage \( V \) by: \[ P = \frac{V^2}{R} \] where \( R \) is the resistance of the bulb.
Step 2: The voltage drops by 1.5%, so the new voltage \( V' \) is: \[ V' = V \times (1 - 0.015) \] Step 3: Substituting \( V' \) into the power equation: \[ P' = \frac{(V')^2}{R} = \frac{(V \times (1 - 0.015))^2}{R} \] Step 4: The percentage drop in power is: \[ \frac{P - P'}{P} \times 100 = 3\% \] Thus, the percentage drop in the rated value of the power is 3%.