Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 18.112
An ideal 12 V emf source is connected to three resistors, as shown at right. The __________ resistor has the least amount of current flowing through it. (A) 2.0 Ω.
(B) 4.0 Ω.
(C) 8.0 Ω.
(D) (There is a two-way tie.)
(E) (There is a three-way tie.)
(F) (Not enough information is given.)
Correct answer: (B)
From Kirchhoff's junction rule, the current passing through the 8.0 Ω resistor is equal to the sum of the currents passing through the 2.0 Ω resistor and the 4.0 Ω resistor:
I8 = I2 + I4,
thus the 8.0 Ω resistor has the most current passing through it, as it must be greater than either I2 or I4.
This I8 current remains constant as it flows through the 8.0 Ω resistor and the 12 V emf, but then it splits the junction at the upper right part of the circuit, with some of that I8 current going through the 2.0 Ω resistor, and the rest separately going through the 4.0 Ω resistor. As a hunch, more current would go through the past of least resistance, so we would expect that more current would go through the 2.0 Ω resistor than through the 4.0 Ω resistor. Then in order of greatest to least amount of currents:
I8 > I2 > I4.
Let's set out to formally prove the above assertion that more current passes through the 2.0 Ω resistor than through the 4.0 Ω resistor.
Kirchhoff's loop rule can be applied to any round-trip loop in a circuit, where the voltage supplied ("potential rises" from the emf) is equal to the voltages used ("potential drops" from the resistors). Applying this to the counterclockwise loop starting from the lower right-hand corner of the circuit, which then passes through the 12 V emf source, the 2.0 Ω resistor, the 8.0 Ω resistor, and back to the lower right-hand corner:
12 V = ∆V2 + ∆V8.
Similarly for the counterclockwise loop that starts from the lower right-hand corner of the circuit, which then passes through the 12 V emf source, the 4.0 Ω resistor, the 8.0 Ω resistor, and back to the lower right-hand corner:
12 V = ∆V4 + ∆V8.
Rearranging these two results from separate Kirchhoff's loop rule loops:
12 V – ∆V8 = ∆V2, and
12 V – ∆V8 = ∆V4.
As the left-hand side of both equations above are the same, then the right-hand side of both equations above are the same as well, so we set them equal to each other, and apply Ohm's law to both sides:
12 V – ∆V8 = 12 V – ∆V8,
∆V2 = ∆V4,
I2·(2.0 Ω) = I4·(4.0 Ω).
Thus more current flows through the 2.0 Ω resistor, and less current flows through the 4.0 Ω resistor. This demonstrates the general rule that for any set of resistors in parallel, the path of least resistance will carry the most current, which in this case is the 2.0 Ω resistor.
Section 30882
Exam code: quiz04eQu7
(A) : 1 student
(B) : 3 students
(C) : 6 students
(D) : 17 students
(E) : 3 students
(F) : 0 students
Success level: 9%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25
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