Physics 5A Quiz 1, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Comprehensive Problem 1.72
[3.0 points.] The electrical power P drawn from a generator by a light bulb of resistance R is P = V^2/R, where V is the line voltage. The resistance of bulb B is 75% greater than the resistance of bulb A. What is the ratio P_B/P_A of the power drawn by bulb B to the power drawn by bulb A, if the line voltages are the same?
(A) 0.33.
(B) 0.57.
(C) 1.3.
(D) 1.8.
Correct answer: (B)
P_A = (V_A)^2/R_A, and P_B = (V_B)^2/R_B.
With V_A = V_B, and 1.75*R_A = R_B, then:
P_B /P_A = ((V_B)^2/R_B)/((V_A)^2/R_A) = (V_B/V_A)^2 = (R_A/1.75*R_A)^2 = (1/1.75)^2 = 0.57,
to two significant figures.
The incorrect response (A) is (1/1.75)^2, response (C) is (1/0.75), and (D) is (1/0.75)^2.
Student responses
Sections 4987, 4988
(A) : 8 students
(B) : 17 students
(C) : 13 students
(D) : 12 students
The answer is actually A (.33)
ReplyDeleteYou forgot to square the fraction, which is (.57)
Response (A) would be correct if the resistances were the same, but the voltage applied to bulb A were 75% greater than the voltage applied to bulb B.
ReplyDeleteHowever, the voltages are the same, and only the resistances are different. Since the resistance is not squared, then (B) is the correct answer for the problem as stated. :{o
But thanks for the comment--nice to know that someone out there is reading this blog! :{p