Physics 205B Quiz 5, spring semester 2013
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Questions 18.21-18.23
An ideal 9.0 V emf source is connected to an ideal ammeter, three light bulbs, and an open switch, as shown at right. When the switch is closed, the brightness in the 10.0 Ω bulb will:
(A) decrease to zero.
(B) decrease slightly, but remain lit.
(C) remain the same.
(D) increase.
(E) (Not enough information is given.)
Correct answer (highlight to unhide): (C)
The upper 10.0 Ω bulb portion of this circuit is in parallel to the switched portion of this current, and Kirchhoff's circuit loop rule can be applied independently to these two loops. For the upper loop, whether the switch in the lower loop is open or closed, the 9.0 V rise from the ideal emf must be equal to the ∆V drop across the 10.0 Ω resistor:
∆Vrise = ∆Vdrop,
9.0 V = I10·(10.0 Ω),
such that the current I10 passing through the 10.0 Ω bulb = (9.0 V)/(10.0 Ω) = 0.90 A, regardless of whether the switch in the lower loop is open or closed. This means that the brightness of (the power dissipated by) the 10.0 Ω bulb will be:
P = I102·R10 = (0.90 A)2·(10.0 Ω) = 8.1 watts,
and remain constant regardless of whether the switch in the lower loop is open or closed. (Note that this would not be true if the ideal emf source were replaced by a real battery with a non-zero internal resistance.)
Section 30882
Exam code: quiz05c4Rb
(A) : 0 students
(B) : 21 students
(C) : 8 students
(D) : 3 students
(E) : 0 students
Success level: 25%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.23
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