Thursday, August 2, 2012
Old finals
Once I finally get home, I will start posting answers to as many of the old final questions as I can. I have many more requests than I can probably do tonight, but I'll do what I can. Start looking for them around 7pm.
A few more example problems
on quantum/atomic physics. I should be able to post some more tomorrow evening, but these are highly relevant for the exam.
Exam 2 extra credit hints
So the problem is a tough-looking one, but I swear this is an incredibly useful circuit. The circuit I gave you is a filter, and one fair tactic is to google "RLC filters" and see if you can find a match. Inelegant, but effective. I will tell you it is a "bandpass" filter, but you'll probably spend just as much time trying to understand the pages you find as you would just working the thing out with a little help. I say that because you will run across a lot of electrical engineering pages which speak a different breed of math than you are used to, and the language barrier will be tough to overcome, not because this is a particularly tough circuit.
We know reactances add like resistances. The whole circuit has L and C in parallel, and that combination in series with a resistance. The total reactance is then $X_{tot} = X_R + X_{L||C}$, where $X_{L||C}$ is the reactance of the capacitor and inductor in parallel, which you'd find by adding the $L$ and $C$ reactances like you would parallel resistors. The resistor's reactance is just $X_R=R$. The current in the circuit is then the input voltage divided by the total reactance, $I=V_{in}/X_{tot}$.
What is the output voltage? You're measuring the output voltage across the $L$ and $C$ parallel combination, so it must be $V_{out}=IX_{L||C}$. Find the equivalent reactance of $L$ and $C$ in parallel, add the resistor's reactance, and you have the total reactance. That gets you the current, and the current times the equivalent reactance of $L$ and $C$ in parallel gives you the output. This will all be a function of frequency as well as $R$, $L$, and $C$.
Qualitatively, you can ignore the $L$ and $C$ one at a time. Without the $C$, it is a high-pass. Without the $L$, it is a low-pass. The whole circuit looks like a combination of the two - hates both high and low frequencies, but will like a middle range in between.
To sketch the plot, presume some values for $R$, $L$, and $C$. I would say perhaps $R\!=\!100\,\Omega$, $L\!=\!1\,$mH, and $C\!=\!1\,\mu$F. What is it good for? Any time you want to let a small range of frequencies through, but nothing else. Radio comes to mind.
We know reactances add like resistances. The whole circuit has L and C in parallel, and that combination in series with a resistance. The total reactance is then $X_{tot} = X_R + X_{L||C}$, where $X_{L||C}$ is the reactance of the capacitor and inductor in parallel, which you'd find by adding the $L$ and $C$ reactances like you would parallel resistors. The resistor's reactance is just $X_R=R$. The current in the circuit is then the input voltage divided by the total reactance, $I=V_{in}/X_{tot}$.
What is the output voltage? You're measuring the output voltage across the $L$ and $C$ parallel combination, so it must be $V_{out}=IX_{L||C}$. Find the equivalent reactance of $L$ and $C$ in parallel, add the resistor's reactance, and you have the total reactance. That gets you the current, and the current times the equivalent reactance of $L$ and $C$ in parallel gives you the output. This will all be a function of frequency as well as $R$, $L$, and $C$.
Qualitatively, you can ignore the $L$ and $C$ one at a time. Without the $C$, it is a high-pass. Without the $L$, it is a low-pass. The whole circuit looks like a combination of the two - hates both high and low frequencies, but will like a middle range in between.
To sketch the plot, presume some values for $R$, $L$, and $C$. I would say perhaps $R\!=\!100\,\Omega$, $L\!=\!1\,$mH, and $C\!=\!1\,\mu$F. What is it good for? Any time you want to let a small range of frequencies through, but nothing else. Radio comes to mind.
HW5 solutions
can be found here. The exam problems on this material will (promise) be easier than the HW questions - the exam problems will probably be somewhere between these problems and the optics questions on your quizzes.
These HW problems were hard (mostly) intentionally - given that we had to cover the material rather quickly, I wanted you to wrestle with some harder problems in the hopes it would help the material sink in a bit better. Whether that works or not is debatable, but I think it makes sense at this point that at least reading the solutions for these harder problems will help you understand what's happening in simpler situations, even if you didn't quite solve them correctly on the HW.
These HW problems were hard (mostly) intentionally - given that we had to cover the material rather quickly, I wanted you to wrestle with some harder problems in the hopes it would help the material sink in a bit better. Whether that works or not is debatable, but I think it makes sense at this point that at least reading the solutions for these harder problems will help you understand what's happening in simpler situations, even if you didn't quite solve them correctly on the HW.
Wednesday, August 1, 2012
Possible extra credit on exam 2
can be had by solving this problem. Partial credit will be generous, and there is no way that doing or not doing the problem can hurt your grade. I will happily provide hints if you ask relatively direct and specific questions.
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