Cuesta College, San Luis Obispo, CA
An ideal 6.0 V emf source is connected to two light bulbs, a resistor, and an ideal ammeter, and an open switch. When the switch is closed, the ammeter reading will: (A) decrease.
(B) remain constant.
(C) increase.
(D) (Not enough information is given.)
Correct answer (highlight to unhide): (A)
When the switch is open, the 1.0 Ω light bulb will be dark as no current will pass through it. The current in this circuit will start at the 6.0 V emf, pass through the ammeter, go through the 7.5 Ω resistor, and then through the 2.0 Ω resistor, and back to the 6.0 V emf.
The equivalent resistance Req of this circuit is 7.5 Ω + 2.0 Ω = 9.5 Ω.
The current I through this circuit is (6.0 V)/(9.5 Ω) = 0.63 A, which is the ammeter reading.
When the switch is closed, then the 1.0 Ω light bulb is in parallel with the 7.5 Ω resistor.
The equivalent resistance Req of this circuit is 2.0 Ω + ((1/1.0 Ω) + (1/7.5 Ω))–1 = 2.88 Ω.
The current Iemf through the emf is (6.0 V)/(2.88 Ω) = 2.08 A.
Now let's figure out how much current goes through the ammeter when the switch is closed, as the 2.08 A that passes through the 6.0 V emf will split with some either going through the switch path or going through the ammeter path, as given by Kirchhoff's junction rule:
Iemf = Iswitch + Iammeter.
Let's apply Kirchhoff's loop rule for the clockwise emf-ammeter-7.5 Ω-2.0 Ω round trip path:
voltage rises = voltage drops,
6.0 V = ∆V7.5 Ω + ∆V2.0 Ω,
and then apply Ohm's law to the right-hand side terms:
6.0 V = Iammeter·(7.5 Ω) + Iemf·(2.0 Ω),
and since we already know Iemf= 2.08 A, then:
6.0 Ω = Iammeter·(7.5 Ω) + (2.08 A)·(2.0 Ω),
0.25 A = Iammeter,
which means the ammeter reading will decrease from its previous reading of 0.63 A when the switch was still open.
Sections 30882, 30883
Exam code: quiz05z0m6
(A) : 22 students
(B) : 8 students
(C) : 4 students
(D) : 0 students
Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.70
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