Cuesta College, San Luis Obispo, CA
An ideal 1.5 V emf source is connected to a switch, a light bulb, a resistor, and an ideal voltmeter. While the switch is open, the voltmeter reading is: (A) zero.
(B) finite, non-zero value below 1.5 V.
(C) exactly 1.5 V.
(D) some finite value above 1.5 V.
(E) ∞.
Correct answer: (C)
From applying Kirchhoff's loop rule to the bottom loop of the circuit, which includes the 0.90 Ω resistor, 1.5 V emf source, and the voltmeter, the voltmeter will "read" the rise in electric potential from the 1.5 V emf source, and the drop in electric potential of the 0.90 Ω resistor:
∆Vvoltmeter = – I·(0.90 Ω) + (1.5 V).
However, no current will pass through the upper part of the circuit, which includes the 0.90 Ω resistor, the 4.0 Ω light bulb, and the open switch (which has an infinite resistance). This means that there is no drop in electric potential due across the 0.90 Ω resistor, and thus the voltmeter reading will be:
∆Vvoltmeter = – (0 A)·(0.90 Ω) + (1.5 V) = 1.5 V.
Sections 30882, 30883
Exam code: quiz05vLeY
(A) : 15 students
(B) : 5 students
(C) : 6 students
(D) : 1 student
(E) : 0 students
Success level: 23%
Discrimination index (Aubrecht & Aubrecht, 1983): –0.04
Why is there a voltmeter reading if the voltmeter is inserted into the circuit? Wouldn't the nearly infinite resistance of the voltmeter stop the current from flowing through the lower loop?
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