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I am currently going through the MIT 6.002 lecture videos to help prepare me for my upcoming electronics final, and I came across something that I didn't quite understand

https://youtu.be/WT-qzgaKeGI?list=PL9F74AFA03AA06A11&t=2402

url point to the video and the point in time I am confused about

More specifically:

So far I have been understanding the general idea of how the low/high/band pass (and band stop) filters work, and how they can be created. I understood where he got the part of the graph at low frequencies because the inductor becomes a short, the voltage goes across that and very little drops across the capacitor where Vout is being read. What confused me is the next part, where he says at high frequencies the capacitor is a short, which i understood, but to me that meant the voltage drop happens across the capacitor now, not the inductor, and so Vout gets the voltage drop and you have a high pass filter.

I know from other videos, other classes, and the end of the video that the filter is a band pass, but I am having trouble building that intuition, so I was wondering if anyone here can help explain to me what happens to cause that band pass behavior

I could look at the transfer function, which Im sure will get me the answer, but Im really hoping to build up my intuition about these so I can go on and understand more complicated circuits. Thank you for any help!

As a side note: The way the MIT lectures have been teaching me to look at the circuits, and the way I have been from other classes, is that you look at what happens at low frequencies (capacitors go to open, inductors go to short) and high frequencies (the opposite) and see what happens at your Vout (where you are measuring). So far, I have gotten all of the series examples right, but this one is throwing me off. Am I missing something? Also, I keep thinking that voltage across parallel components is the same, which makes me think the 0 at high frequencies makes sense, yet since the voltage is always across at least one path, I expect it to be a band stop with purely that logic... I know Im missing some key logic here!

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  • \$\begingroup\$ A 'short' usually means 0 ohms. You can't get a voltage drop across 0 ohms. \$\endgroup\$ – brhans Apr 27 '17 at 22:01
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    \$\begingroup\$ Oh wow. That makes perfect sense... The capacitor becomes a short, but no voltage is across it because voltage is a potential difference right? so in this case, when either is a short, the current travels down that path but there is no voltage. In the case of a series rlc, depending upon the placements, a short still lets current travel to the resistor and thus there is voltage across the resistor... is this correct? \$\endgroup\$ – ChrisM Apr 27 '17 at 22:31
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What confused me is the next part, where he says at high frequencies the capacitor is a short, which i understood, but to me that meant the voltage drop happens across the capacitor now, not the inductor, and so Vout gets the voltage drop and you have a high pass filter.

Like brhans says in comments, at high frequencies a capacitor behaves like a short circuit. This means that high frequency currents go through it easily, and very little voltage can be sustained across it.

Also, I keep thinking that voltage across parallel components is the same, which makes me think the 0 at high frequencies makes sense, yet since the voltage is always across at least one path,

With a parallel connection, the same voltage will be found across all paths. That's what "parallel" means.

I expect it to be a band stop with purely that logic... I know Im missing some key logic here!

Whether a RLC parallel circuit acts as a band pass or band stop depends how it's connected:

schematic

simulate this circuit – Schematic created using CircuitLab

In the band pass version, at low frequencies, L1 shunts the signal to ground. At high frequencies, C1 shunts the signal to ground. Only near resonance, the admittance of those two parts cancels and (assuming R2 is relatively high-valued), the signal passes through to the load.

In the band-stop version, at low frequencies the signal passes through L2. At high frequencies the signal passes through C2. Only near resonance, the admittances of these two elements cancel, and (again assuming R3 has high enough value) the signal is blocked from reaching the load.

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  • \$\begingroup\$ Thank you for your detailed answer! Those two circuits helped me a lot. So in general, will a simple circuit with an inductor and a capacitor always act as some sort of band filter, whether pass or stop depending upon the orientation? \$\endgroup\$ – ChrisM Apr 29 '17 at 1:41

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