Physics quiz question: voltmeter reading along "top" of series circuit

Physics 205B Quiz 5, spring semester 2015
Cuesta College, San Luis Obispo, CA

An ideal 4.5 V emf source is connected to an ideal voltmeter, a light bulb, and two resistors, as shown at right. The voltmeter reading is:
(A) 0.26 V.
(B) 0.39 V.
(C) 2.1 V.
(D) 2.4 V.

Correct answer (highlight to unhide): (D)

Since the ideal voltmeter has an infinite resistance, no current flows through it, and only along the lower loop of the circuit (through the ideal emf source, the top left resistor, the light bulb, and then through the lower right resistor). Since the resistors and light bulb are wired in series, the equivalent resistance of the circuit is just the arithmetic sum of their individual resistances:

R_eq = 8.0 Ω + 1.5 Ω + 8.0 Ω = 17.5 Ω.

From Ohm's law, the current flowing through this simplified (one ideal emf and one equivalent resistor) is:

I = ε/R_eq = (4.5 V)/(17.5 Ω) = 0.2571428571 A.

Starting from the contact on the left, the difference in voltage detected by the voltmeter is the drop due to the top left resistor plus the drop due to the light bulb:

∆V = (–I·R_resistor) + (–I·R_{light bulb}),

∆V = –I·(R_resistor + R_{light bulb}),

∆V = –(0.2571428571 A)·(8.0 Ω + 1.5 Ω) = –2.4428571429 V,

or to two significant figures, the voltmeter reading is 2.4 V.

(Response (A) is the current I = ε/R_eq flowing through the circuit; response (B) is the voltage drop ∆V = I·R_{light bulb} of just the 1.5 Ω light bulb; response (C) is the voltage drop ∆V = I·R_resistor of just the 8.0 Ω resistor.)

Sections 30882, 30883
Exam code: quiz05aL7y
(A) : 16 students
(B) : 6 students
(C) : 9 students
(D) : 16 students

Success level: 34%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.62