AP Physics C - Resistors and Resistance
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- เผยแพร่เมื่อ 30 ก.ค. 2024
- A brief introduction to resistors, resistance, and resistivity for calculus-based physics students such as those taking AP Physics C. For more information, please visit APlusPhysics.com.
You deserve way more views! I think that your videos are honestly one of the best resources for reviewing AP Physics.
Thanks Benjamin! Thrilled to hear this was helpful for you!
Him: *Says anything*
Me: “Write that down, Write that down!
Have been self studying for the upcoming AP exams, your videos have been a godsend
Have you checked out the resources and help forums at the APlusPhysics site?
hi Mr Dan , i really like your lessons and they help me a lot
im having hard time studying this chapter ( electric current and Ohm's law ) so can you help me out with this please ?
Hello Sir i have a little project and need help i bought nichrome 0.32 and i need 1 meter of the wire to heat up to 300° what resistor do i need and will 7.4v lipo be sufficient enough? Thank you
Hey Mr. Fullerton,
I was wondering where I could find practise questions for AP physics C mechanics and electricity/magnestism
+shloak patio The College Board website and learnapphysics.com are great places to start. :-)
At 9:12, the video said that V=molar mass/density. If you cancel out the units for molar mass and density (g/mol and g/cm cubed, respectively), you get cm cubed/mol which is different from the volume unit of centimeters cubed. When you use molar mass, do you ignore the moles part of the unit, or just assume that you are working with one mole?
I think we assume we're working with one mole, that way you get mass units?
Hi Mr. Fullerton. Thx so much for these videos! I'm a little bit confused by the silver wire II problem. why N=Na/V? shouldn't Na, mole mass be chemistry stuff? are we required to know these for AP physics C exam?
+黄泓皓 Yes, you should know these for the AP-C exam.
Mr. Fullerton, thanks for another great video as usual!
Question: you postulate that the charge carrier density N should be avogadros number divided by the volume. Could you please explain why this is the case?
Thanks, and make it a great day! :-)
Also, why do you not take the total volume of the wire, multiply it by the mass density, convert to atoms and find the number of charge carriers that way? In this case,t hat would be N * A, right?
Hi Nathan. Because we're talking about silver, we assume one elementary charge per atom. The charge density, then, is Avogadro's # (# of atoms per mole, or in this case, # of elementary charges per mole), divided by the volume. When we then rearrange using molar density, we can use the given information to determine the charge carrier density (in elementary charges). If you wanted the actual charge density, you just multiple that by the charge of an elementary charge, in Coulombs.
The reasoning is sound, not sure it would be N*A as you haven't specifically defined N and A for this approach.
Thank you Mr. Fullerton for you kind reply.
This helped me a lot!!
7:42 isnt it .005?
Marc Manalo Nope. .5mm is .0005 m.