Physics 205B Midterm 2, spring semester 2016
Cuesta College, San Luis Obispo, CA

A "AA" alkaline battery with an emf of 1.5 V and an internal resistance of
r = 0.90 Ω is attached to an ideal voltmeter, with a
R = 2.0 Ω light bulb that is wired in parallel with an open switch. Discuss why the voltmeter will have a lower reading after the switch is closed. Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and properties of voltmeters.
Solution and grading rubric:
- p:
Correct. Recognizes that when the switch is open, the voltmeter will
have a non-zero reading, and have a lower (zero) reading when the switch is
closed, using one of two similar arguments:- when the switch is open, there is a non-zero ΔV = +1.5 V − I⋅r reading,
and when the switch is closed, from Kirchhoff's loop rule the voltage
rise of +1.5 V from the emf must now exactly equal the −I⋅r voltage
drop of the internal resistance of the battery, such that the voltmeter
reading is now zero; or
- when the switch is open, there is a non-zero ΔV = − I⋅R reading,
and when the switch is closed, since the light bulb R is bypassed by a
zero resistance switch, making ΔV = 0.
- r:
Nearly correct, but includes minor math errors. Does not
sufficiently show numerically or qualitatively how voltmeter reading when switch
is open is higher versus when the switch is closed.
- t:
Nearly correct, but approach has conceptual errors, and/or
major/compounded math errors. At least has a conceptual understanding of how a
voltmeter measures a potential difference, and how the switch changes the
current flow when it is open versus when it is closed.
- v:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at applying Kirchhoff's rules, Ohm's law, and
equivalent resistance.
- x:
Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Kirchhoff's rules, Ohm's law, and
properties of voltmeters.
- y:
Irrelevant discussion/effectively blank.
- z:
Blank.
Grading distribution:
Sections 30882, 30883
Exam code: midterm02Mc4s
p: 7 students
r: 17 students
t: 4 students
v: 12 students
x: 2 students
y: 0 students
z: 0 students
A sample "p" response (from student 3158):
Another sample "p" response (from student 5433):