Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 18.73
An ideal voltmeter is connected to a switch in a circuit with an ideal 12.0 V emf source and two light bulbs, as shown at right. While the switch remains open, the voltmeter reading is:(A) 0 V.
(B) 4.5 V.
(C) 7.5 V.
(D) 12.0 V.
(E) ∞.
Correct answer (highlight to unhide): (C)
No current will flow through the ideal voltmeter, only through the lower part of the circuit (indicated in green). Since the 3.0 Ω and the 5.0 Ω light bulbs are in series, the equivalent resistance of the circuit is their arithmetic sum, Req = 8.0 Ω. This means that the current in this circuit is:Icircuit = εeq/Req = (12.0 V)/(8.0 Ω) = 1.5 A.
The voltmeter will "feel" the total difference in potential of the emf and the 3.0 Ω light bulb together (where its connection to the circuit is shown in blue). The emf and the 3.0 Ω light bulb contribute a voltage rise of ε and a voltage drop of I·R, respectively:
∆V = +ε – I·R = +(12.0 V) – (1.5 A)·(3.0 Ω) = +12.0 V – 4.5 V = +7.5 V.
(Alternatively, as wired the voltmeter also "feels" just the voltage drop of the 5.0 Ω light bulb, which would be ∆V = – I·R = –(1.5 A)·(5.0 Ω) = 7.5 Ω. This would be expected from applying Kirchhoff's loop rule for circuits. Note that as in the laboratory component of this course, the affect of the polarity (and thus the ± sign of voltmeter readings) of wiring the voltmeter to a circuit is neglected.)
Sections 30882, 30883
Exam code: quiz05b4L8
(A) : 18 students
(B) : 1 student
(C) : 6 students
(D) : 10 students
(E) : 3 students
Success level: 16%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33
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