Physics quiz question: power dissipated by light bulb in series circuit

Physics 205B Quiz 5, spring semester 2016
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to two light bulbs (each with a 1.2 Ω resistance), and a 0.50 Ω resistor. The electrical power used by the top light bulb is:
(A) 2.1 W.
(B) 5.1 W.
(C) 12 W.
(D) 30 W.

Correct answer (highlight to unhide): (B)

The equivalent resistance of this series circuit is the arithmetic sum of their resistances:

R_eq = 1.2 Ω + 1.2 Ω + 0.50 Ω = 2.9 Ω.

From Kirchhoff's junction rule, the amount of current flowing through the emf source and all three resistive circuit elements must be the same, as they are in series with each other, and the current flowing this circuit is given by applying Ohm's law to the entire equivalent circuit:

I_eq = ε/R_eq = (6.0 V)/(2.9 Ω).

Since we now know the resistance value of the top light bulb, and the amount of current flowing through it, the power dissipated by the top light bulb is then:

P_{top light bulb} = I_eq·ΔV_{top light bulb} = I_eq·(I_eq·R_{top light bulb}) = I_eq²·R_{top light bulb},

P_{top light bulb} = ((6.0 V)/(2.9 Ω))²·(1.2 Ω) = 5.1367419737 W,

or to two significant figures, 5.1 W.

(Response (A) is the numerical value for the current flowing through the circuit I_eq = ε/R_eq = (6.0 V)/(2.9 Ω); response (C) is ((6.0 V)/(2.9 Ω))²·(2.9 Ω), which is the amount of power supplied by the emf source to rest of the circuit; response (D) is (6.0 V)²/(1.2 Ω), which would be the power dissipated by the top light bulb if it were to use up all of the voltage supplied by the emf (which would violate Kirchhoff's loop rule!).)

Sections 30882, 30883
Exam code: quiz05Tt1p
(A) : 4 students
(B) : 16 students
(C) : 13 students
(D) : 5 students

Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.79