20170415

Physics quiz question: resistive circuit element with most current

Physics 205B Quiz 5, spring semester 2017
Cuesta College, San Luis Obispo, CA

An ideal 4.5 V emf source is connected to two light bulbs, a resistor, and an ideal ammeter. The __________ has the most current flowing through it.
(A) 2.0 Ω light bulb.
(B) 3.0 Ω light bulb.
(C) 5.0 Ω resistor.
(D) (There is a two-way tie.)
(E) (There is a three-way tie.)
(F) (Not enough information is given.)

Correct answer: (E)

From Kirchhoff's junction rule, the current passing through the 4.5 V emf source is equal to the sum of the current that passes through both light bulbs, and the current that separately passes through the 5.0 Ω resistor:

Iemf = Ilight bulbs + Iresistor.

Kirchhoff's loop rule is then applied to the (clockwise) path that passes through the 4.5 V emf source, the light bulbs, and back to just before the 4.5 V emf source:

0 = +(4.5 V) – Ilight bulbs·(3.0 Ω) – Ilight bulbs·(2.0 Ω),

which can then be solved for the amount of the current that passes through the light bulbs:

Ilight bulbs·(3.0 Ω) – Ilight bulbs·(2.0 Ω) = +(4.5 V),

Ilight bulbs = +(4.5 V)/(3.0 Ω + 2.0 Ω),

Ilight bulbs = 0.90 A.

Similarly, Kirchhoff's loop rule can also be applied separately to the (clockwise) path that passes through the 4.5 V emf source, the 5.0 Ω resistor, and back to just before the 4.5 V emf source:

0 = +(4.5 V) – Iresistor·(5.0 Ω),

which can then be solved for the amount of the current that passes through the resistor:

Iresistor·(5.0 Ω) = = +(4.5 V),

Iresistor = +(4.5 V)/(5.0 Ω),

Iresistor = 0.90 A.

Thus the amount of current that passes through the two light bulbs is equal to the amount of current that separately passes through the resistor, making this a three-way tie.

Conceptually this can also be seen by looking at the equivalent resistance of the two light bulb that are connected in series:

Rlight bulbs = 3.0 Ω + 2.0 Ω = 5.0 Ω.

Since the current that passes through the 4.5 emf source splits up to either go through the light bulbs (which together have a total resistance of 5.0 Ω) or instead go through the 5.0 Ω resistor, then it can be seen that equal amounts of current will pass separately either through the light bulb or through the resistor.

Sections 30882, 30883
Exam code: quiz05vLeY
(A) : 3 students
(B) : 3 students
(C) : 5 students
(D) : 5 students
(E) : 10 students
(F) : 0 students

Success level: 38%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.75

No comments:

Post a Comment