Cuesta College, San Luis Obispo, CA
A light bulb and a resistor are connected to two ideal emf sources. The __________ has a greater potential difference. (A) light bulb.
(B) resistor.
(C) (There is a tie.)
(D) (Not enough information is given.)
Correct answer: (C)
(The two emf sources are in series with each other, such that the equivalent emf in this circuit is 9.0 V + 3.0 V = 12.0 V.)
The resistor and the light bulb are in parallel with each other, and can be analyzed separately using Kirchhoff's loop rule, where the potential rises (voltages supplied by the emfs) equal the potential drops (voltages used by the resistors). There are two possible round-trip loops we can make, one of which is the loop starting from lower left-hand corner, through the 3.0 V emf source, through the 15 Ω resistor, and through the 9.0 V emf source before returning to the starting point in the lower left-hand corner
(3.0 V) + (9.0 V) = ∆VR.
Similarly, applying the loop rule in the counterclockwise direction, starting from lower left-hand corner, through the 3.0 V emf source, through the 1.3 Ω light bulb, and through the 9.0 V emf source before returning to the starting point in the lower left-hand corner results in:
(3.0 V) + (9.0 V) = ∆Vr.
From inspection, the potential drop across the resistor VR is 12.0 V, and the potential drop across the light bulb ∆Vr is also 12.0 V, meaning they have the same drop in potential (but different amounts of currents), as is typical for any two resistive circuit elements in parallel with each other.
Sections 30882, 30883
Exam code: quiz04Br7w
(A) : 4 students
(B) : 16 students
(C) : 3 students
(D) : 0 students
Success level: 13%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25
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