Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 18.63
An ideal 16.0 V emf source is connected to ideal light bulbs and an ideal resistor, as shown at right. Find the amounts of power used by each light bulb. Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and electric power. Solution and grading rubric:
- p:
Correct. Determines power of 8.0 Ω light bulb by direct application of P = (∆V)2/R, or first solves for I through light bulb, and then uses P = I·∆V or I2·R. For the 5.0 Ω light bulb, must first solve for the current passing through the lower loop, then uses power = I2·R, or uses voltage drop across the 5.0 Ω light bulb to use P = I·∆V or (∆V)2/R. - r:
Nearly correct, but includes minor math errors. Typically solves correctly for the power used by the 8.0 Ω light bulb, but has conceptual errors in applying Kirchhoff's rules and Ohm's law to determine current passing through and/or voltage drop across the 5.0 Ω light bulb. - t:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. - 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.
Grading distribution:
Section 70856
p: 7 students
r: 4 students
t: 0 students
v: 0 students
x: 0 students
y: 0 students
z: 0 students
A sample "p" response (from student 1025). Note that the numerical results here are correct, but there are several errors in setting up calculations:
- the equivalent resistance is set up incorrectly--it should be written Req = ((5.0 Ω + 4.0 Ω)–1 + (8.0 Ω)–1)–1, but the resulting value is correct.
- the power dissipated by the 5.0 Ω light bulb is set up incorrectly--it should be written P = I2⋅R = (1.778 A)2·(5.0 Ω), but the resulting value is correct.
A sample "r" response (from student 1010):
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