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\,\muF. 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\,\muF. What is it good for? Any time you want to let a small range of frequencies through, but nothing else. Radio comes to mind.
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