If the current in an electric bulb increases by 1%, what will be the change in the power of the bulb? (Assume that the resistance of the filament of the bulb remains constant.)

Question

Accepted Answer

(3) increases by 2%. Given: the current rises by 1% and the filament resistance R stays the same. Idea: with R fixed, write the power in terms of the current alone, P=I²R. For a small change, (Δ P)/P=2 (Δ I)/I. (Δ P)/P=2×1%=2%. Exactly: (1.01)²=1.0201, a rise of 2.01%, which is 2% to the accuracy of the question. So the power increases by 2%.

If the current in an electric bulb increases by 1%, what will be the change in the power of the bulb? (Assume that the resistance of the filament of the bulb remains constant.)

Step-by-step solution

Why the other options are wrong