Physics 205B Quiz 4, spring semester 2017
Cuesta College, San Luis Obispo, CA
Two different types of light bulbs, a resistor, and an ideal emf source are connected together. There is 0.32 A of current flowing through the emf. The resistance of the resistor is:
(A) 0.8 Ω.
(B) 2.1 Ω.
(C) 4.0 Ω.
(D) 4.7 Ω.
Correct answer: (C)
The resistor and the two light bulbs are in series with each other, such that their equivalent resistance is given by:
Req = R1 + R2 + R3,
Req = R1 + 0.50 Ω + 0.20 Ω.
The current flowing through this equivalent one emf, one resistor circuit is then given by Ohm's law:
I = ε/Req,
0.32 A = (1.5 V)/(R1 + 0.50 Ω + 0.20 Ω),
which can be solved for the unknown resistance of the resistor:
R1 + 0.70 Ω = (1.5 V)/(0.32 A),
R1 = (1.5 V)/(0.32 A) – 0.70 Ω = 4.7 Ω – 0.70 Ω,
or to two significant figures, the resistance of the resistor is 4.0 Ω.
(Response (A) is 1.5 V – 0.70 Ω; response (B) is (1.5 V)/(0.70 Ω); response (D) is Req.)
Sections 30882, 30883
Exam code: quiz04Br7w
(A) : 2 students
(B) : 3 students
(C) : 11 students
(D) : 7 students
Success level: 48%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.34
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