AP Physics 2mediummcq1 pt
A 60W light bulb is connected to a 120V power source. If the voltage is reduced to 60V, what power will the bulb now dissipate?
A.D) 120 W
B.B) 30 W
C.C) 60 W
D.A) 15 W
Explanation
Core Concept
First, find the resistance of the bulb, which remains constant. Using P = V²/R, we can solve for R: R = V²/P = (120V)²/60W = 240Ω. When the voltage is reduced to 60V, the power dissipated is P = V²/R = (60V)²/240Ω = 3600V²/240Ω = 15W. Alternatively, since power is proportional to voltage squared (P ∝ V²), halving the voltage reduces the power to one-fourth of the original: (1/2)² × 60W = 15W.
Correct Answer
DA) 15 W
Practice more AP Physics 2 questions with AI-powered explanations
Practice Unit 4: Electric Circuits Questions →