## 20130414

### Physics quiz question: resistor with least current

Physics 205B Quiz 4, spring semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 18.112

An ideal emf source is connected to three resistors. 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.)

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.

Kirchhoff's loop rule is then applied separately to the path that passes through the 12 V emf source, the 2.0 Ω resistor, and the 8.0 Ω resistor, back to just before the 12 V emf source:

0 = +(12 V) – I2·(2.0 Ω) – I8·(8.0 Ω),

or to the current that instead passes through the 12 V emf source, the 4.0 Ω resistor, and the 8.0 Ω resistor, and again back to just before the 12 V emf source:

0 = +(12 V) – I4·(4.0 Ω) – I8·(8.0 Ω).

Eliminating common terms in the above two equations shows that current that passes through the 2.0 Ω resistor cannot be equal to the current that passes through the 4.0 Ω resistor:

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.

So in order of greatest to least amount of currents:

I8 > I2 > I4.

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