Physics quiz question: light bulb "between" two emfs

Physics 205B Quiz 4, spring semester 2015
Cuesta College, San Luis Obispo, CA

Two ideal emf sources are connected to a light bulb, as shown at right. The current flowing through the 6.0 V emf source is:
(A) 3.0 A.
(B) 11 A.
(C) 12 A.
(D) 21 A.

Correct answer (highlight to unhide): (D)

The equivalent emf of this circuit (the 6.0 V emf source and the 4.5 V emf source are in series with the light bulb, both ± polarities "stacked") is the sum of these emfs. Thus the current flowing through all parts of this series circuit is given by:

I = ∆Veq/R = (6.0 V + 4.5 V)/(0.50 Ω) = (10.5 V)/(0.50 Ω) = 21 A.

(This is an application of Kirchhoff's junction rule, in the case here where there is no junction in either these series circuit.)

(Response (A) is using the equivalent emf of the difference between the two emfs (the case where they would be connected in series with opposite ± polarities); response (B) is the sum of the two emf values (with the wrong units of amperes, to two significant figures), which could also erroneously be the sum of the two emf values together with the light bulb resistance value (again with the wrong units of amperes).)

Sections 30882, 30883
Exam code: quiz04sm7H
(A) : 4 students
(B) : 0 students
(C) : 9 students
(D) : 31 students

Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.62

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