20190504

Physics quiz question: power dissipated by a resistor in series to parallel light bulbs

Physics 205B Quiz 5, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 9.0 V emf source is connected to a resistor, and two light bulbs that are each connected to ideal ammeters. The electrical power used by the 3.0 Ω resistor is:
(A) 2.6 W.
(B) 3.4 W.
(C) 21 W.
(D) 27 W.

Correct answer (highlight to unhide): (C)

The 0.5 Ω and the 2.5 Ω light bulbs are in parallel with each other, and are together in series with the 3.0 Ω resistor, such that their equivalent resistance is given by:

Req = (3.0 Ω) + ((0.5 Ω)–1 + (2.5 Ω)–1)–1,

Req = (3.0 Ω) + (2.4 Ω–1)–1 = 3.4166666667 Ω.

Then Ohm's law is applied to the entire circuit, to find the current flowing through the entire circuit:

Icircuit = ε/Req,

Icircuit = (9.0 V)/(3.4166666667 Ω) = 2.6341463415 A.

So the power used by the 3.0 Ω resistor is then given by:

P = I2·R,

P = (2.6341463415 A)2·(3.0 Ω) = 20.8161808453 W,

or to two significant figures, the power used by the 3.0 Ω resistor is 21 W.

(Response (A) is the current flowing through the 3.0 Ω resistor; response (B) is the equivalent resistance of the circuit; response (D) is (9.0 V)2/(3.0 Ω), which would be the power used by the 3.0 Ω resistor if it were somehow able to use up all of the 9.0 V from the emf source that is supplied to the circuit as a whole, leaving nothing for the light bulbs.)

Sections 30882, 30883
Exam code: quiz05eXpL
(A) : 1 student
(B) : 5 students
(C) : 4 students
(D) : 29 students

Success level: 11% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33

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