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)

Applying Kirchhoff's loop rule to the complete circuit, the voltage supplied ("potential rises") by the 6.0 V and the 12.0 V batteries ("emf sources") must be equal to the voltage used ("potential drops") by the two light bulbs R₁ and R₂. So the light bulbs together must use up a total of 18.0 V (as the emf sources point in the same clockwise direction around the circuit):

(6.0 V) + (12.0 V) = ∆V₁ + ∆V₂,

18.0 V = ∆V₁ + ∆V₂.

Since light bulbs have the same resistance (R₁ = R₂) and the same amount of current (I₁ = I₂, from Kirchhoff's junction rule), then from Ohm's law:

∆V₁ = I₁·R₁,

∆V₂ = I₂·R₂,

Therefore ∆V₁ = ∆V₂, and each light bulb uses the same amount of voltage (same amount of potential drop), 9.0 V each.

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