Capacitive reactance  understanding in practical manner - Page 1
Forum section: electronics / general

author: kgavionics
author_rank: Regular Contributor
timestamp: July 12, 2012, 02:37:09 am
content: hii understand the math formulas of capacitive reactance but i want to know why the capacitor behave like a short circuits in high frequencies and as an open circuit in low frequencies.i'm not looking for mathematical explanations( i'm good in math) but i'm looking for practical explanations.thank you in advance

author: Psi
author_rank: Super Contributor
timestamp: July 12, 2012, 02:58:06 am
content: Ok, basic real world explanation..It doesn't turn into an actual/real short circuit. (it's an effective short/open circuit to the signal)With high frequency, large capacitor size and limited current into the capacitor (RC filter) it takes time to charge/discharge the cap.If the waveform is changing very fast the capacitor doesn't get the time needed to reflect these changes, so the output signal is almost non-existent, it's like it's shorted out.- If you slow down the frequency the capacitor now has long enough to start charge/discharging, some of the signal now appears on the capacitor.- If you decrease the resistance (increase current flow) more energy can flow into the capacitor in a given time, so more of the signal appears on the capacitor.- If you decrease the size of the capacitor it needs less energy to charge/discharge so same effect.Note:If the frequency is really high and the capacitor is large then the current that must flow into the capacitor to change its voltage in that short time is HUGE.So there doesn't have to be an physical resistor component in an RC filter because, at this current, the wires of the circuit and source impedance act as a sufficient limit to the current into the capacitor.So, to sum up...input signal ---> (cap to ground) --> output signalThe output can be ~100% of the input all the way down to ~0% of the input. Which is the two extremes we label as open circuit or short circuit for a capacitor with respect to an AC signal.If the cap is in series, likeInput signal ------||------ output signalThen it's the same thing, it's just that the reference is to the other side of the capacitor instead of a common ground, so any DC is preserved.

author: kgavionics
author_rank: Regular Contributor
timestamp: July 12, 2012, 03:43:39 am
content: that's the explantion that i was looking for100000 thanks you answer my question Psi

author: IanB
author_rank: Super Contributor
timestamp: July 12, 2012, 03:49:46 am
content: Quote from: kgavionics on July 12, 2012, 02:37:09 amThe practical explanations are the same as the mathematical explanations. Mathematics is the glue between engineering and the real world. Rather than saying "i'm good in math", you should accept your limitations and say "my mathematics is not yet good enough to understand this, can you give me some other ways to think about the problem?"

author: kgavionics
author_rank: Regular Contributor
timestamp: July 12, 2012, 04:00:05 am
content: Quote from: IanB on July 12, 2012, 03:49:46 amno my friend, understanding equations only doesn't mean that you understand the practical aspect of the phenomena.i accept my limitations , because i not perfect like you Sir.we are here in the forum to learn.Fortunately there's some gentelman like Psi in this forum that are humble enough to give a good answers.

author: Psi
author_rank: Super Contributor
timestamp: July 12, 2012, 04:04:43 am
content: I find the maths for a problem becomes obvious and simple once you have real world knowledge of something. It's really hard (for me) to learn the math first, not because i'm bad at math but because i have no frame or reference about the problem and how it relates to the real world.

author: kgavionics
author_rank: Regular Contributor
timestamp: July 12, 2012, 04:07:30 am
content: Quote from: Psi on July 12, 2012, 04:04:43 amyou are like me Psi, now the equation Xc=1/wc is more obvious for me.Thank you again Psi

author: Psi
author_rank: Super Contributor
timestamp: July 12, 2012, 04:10:06 am
content: you're welcome

author: IanB
author_rank: Super Contributor
timestamp: July 12, 2012, 04:10:50 am
content: Quote from: kgavionics on July 12, 2012, 04:00:05 amOf course it does!The capacitor equations tell you exactly the same story that Psi told you. But if you don't understand the equations you can't read the story...By the way, the equation you need is this one:I = C dV/dtThat is what tells you how a capacitor works. The Xc = 1/wc equation is just a bit of fakery to simplify things.

author: kgavionics
author_rank: Regular Contributor
timestamp: July 12, 2012, 04:13:47 am
content: thank you for you advice Sir Enstein...loooooooooool.

author: Psi
author_rank: Super Contributor
timestamp: July 12, 2012, 04:32:24 am
content: There is no right or wrong way to look at it. It's just how your brain works and everyone is different.Some people need theory to understand practical operation while others need practical understand to "get" the theory.Might be to do with left / right brain dominance, but i dunno.

author: bradleytron
author_rank: Regular Contributor
timestamp: July 12, 2012, 07:17:04 am
content: Quote from: Psi on July 12, 2012, 04:04:43 amRichard Feynman said this very same thing in regard to the subject of physics and in how best to approach it, you make an excellent point.