## 20120507

### Physics midterm problem: unknown resistor with voltmeter, ammeter

Physics 205B Midterm 2, spring semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 18.75

An ideal 12.0 V emf source is connected to two ideal light bulbs, an ideal resistor, an ideal voltmeter and an ideal ammeter, as shown at right. The voltmeter has a reading of 4.2 V, and the ammeter has a reading of 0.50 A. The two light bulbs have the same resistance r. Determine the resistance R of the resistor. Show your work and explain your reasoning.

Solution and grading rubric:
• p:
Correct. Applies Kirchhoff's loop rule and Ohm's law to find the voltage drop across the resistor R = 12.0 V - 4.2 V = 7.8 V, such that its resistance R = ΔV/I = (7.8 V)/(0.50 A) = 16 Ω.
• r:
Nearly correct, but includes minor math errors. May have confused voltmeter reading as absolute voltage value, rather than voltage difference.
• t:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. May have included all three resistor voltage drops in same loop rule equation, but at least demonstrates a methodical application of Kirchhoff's rules, Ohm's law, and equivalent resistance.
• v:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
• x:
Implementation of ideas, but credit given for effort rather than merit.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.

Section 30882
Exam code: midterm02B1rD
p: 9 students
r: 7 students
t: 10 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 4027), methodically applying Kirchhoff's junction and loop rules:
Another sample "p" response (from student 0514), explicitly applying Ohm's law to the resistor R, given the voltmeter and ammeter readings: