Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 18.21
An ideal emf source is connected to a light bulb and a resistor, as shown at right. The __________ has a greater potential difference. (A) light bulb.
(B) resistor.
(C) (There is a tie.)
(D) (Not enough information is given.)
Correct answer (highlight to unhide): (B)
From applying Kirchhoff's loop rule in the clockwise direction around this circuit, the voltage rise due to the emf (ε = 4.5 V) must equal the sum of the voltage drop due to the 0.8 Ω light bulb (Ibulb·Rbulb) and the voltage drop due to the 3.0 Ω resistor (Iresistor·Rresistor), such that:
ε = Ibulb·Rbulb + Iresistor·Rresistor.
Since from Kirchhoff's junction rule (where there are no junctions in this simple circuit), the current flowing through all parts of this circuit must be the same:
Ibulb = Iresistor.
Because the resistance of the resistor is greater than the resistance of the light bulb, then the voltage drop of the resistor must be greater than the voltage drop of the light bulb:
Ibulb·Rbulb < Iresistor·Rresistor.
Sections 30882, 30883
Exam code: quiz04mCnC
(A) : 9 students
(B) : 25 students
(C) : 2 students
(D) : 0 students
Success level: 69%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.28
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