Physics quiz question: comparing potential differences in serial circuit

Physics 205B Quiz 4, spring semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 18.48(b)

Two ideal emf sources are connected to two identical light bulbs, R1 = R2, as shown at right. The __________ light bulb has the greater potential difference.
(A) R1.
(B) R2.
(C) (There is a tie.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

From Kirchhoff's junction rule (in this simple circuit, there being no junctions), the same current I must pass through all elements in this circuit. Applying Kirchhoff's loop rule to the current that passes through the 12 V emf source, R1 light bulb, the 6.0 V emf source, and the R2 light bulb, back to just before the 12 emf source results in:

0 = +(12 V) – I·R1 +(6.0 V) – I·R2,

I·R1 + I·R2 = +(18 V),

where the two left-hand terms in the above equation respectively correspond to the potential differences across the R1 light bulb and the R2 light bulb, which must be equal to each other, as the same current I passes through them, and they have the same resistances R1 = R2.

Section 30882
Exam code: quiz04eQu7
(A) : 14 students
(B) : 6 students
(C) : 13 students
(D) : 0 students

Success level: 39%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45

No comments: