Talk:RC circuit/Archive 1

Parallel RC Discussion
How about the same discussion of the parallel RC circuit?


 * Actually, the parallel RC circuit as shown in the article is incorrect. You would never place a voltage source across a parallel combination of a resistor and a capacitor.  The input source should be a current source, in which case the circuit would function as the exact dual of the series circuit. -- Rdrosson 03:27, 6 November 2005 (UTC)


 * If you think about it, the series circuit is really just a Thevenin equivalent power source driving a capacitor. So the dual circuit would have to be a Norton equivalent power source driving an inductor.  But that, of course, would be an RL circuit.  So the parallel RC circuit, with a current source, is actually the dual of the series RL circuit. Likewise, the series RC circuit is the dual of the parallel RL circuit.  -- Rdrosson 03:39, 6 November 2005 (UTC)

if input signal not pure sinusoid
What if the input signal is not a pure sinusoid? Why is the article restricting the frequency domain analysis only to pure imagnary frequencies? There is a much more general form involving Laplace transforms where, instead of using


 * $$ Z_C = { 1 \over j \omega C } $$

the complex impedance is


 * $$ Z_C = { 1 \over Cs } $$

where s is a complex number


 * $$ s \ = \ \sigma + j \omega $$

Sinusoidal steady state is then a special case where


 * $$ \sigma \ = \ 0 $$

and


 * $$ s \ = \ j \omega $$

This approach then enables you to use some interesting and powerful techniques:


 * Solution of the differential equations using polynomial functions of s
 * Laplace transfomations of inputs and outputs to derive complex valued functions in terms of s
 * Analysis using complex valued transfer functions, also in terms of s, that are simple ratios of the complex valued input and output functions
 * Identification of poles and zeros of the transfer functions, and plotting the poles and zeros in the complex s-plane
 * Calculation of gain as the magnitude of the transfer function and phase angle as the argument of the transfer function.
 * Frequency domain analysis involving not only pure sinusoids but also damped sinusoids
 * Fourier decomposition and analysis of arbitrary (non-sinusoidal) signal inputs and outputs.


 * -- Rdrosson 12:45, 4 November 2005 (UTC)

Laplace domain stuff

 * If you look slightly further down the page, you'll see that Laplace domain stuff is included there. -Splash talk 12:51, 4 November 2005 (UTC)
 * Yes, I see what you are saying. But actually, the article barely scratches the surface of these ideas, and everything prior to the mention of Laplace Transforms can be vastly simplified by a much more general and elegant set of techniques.  If you read the the list of bullet points I created (above), I don't see any of these concepts other than one brief mention of Laplace Transforms in the article.  Furthermore, even the discussion of Laplace misses the key point:  you don't need to restrict the input signals to sinusoids -- you can represent and analyze the behavior for virtually any input signal .   -- Rdrosson 20:09, 4 November 2005 (UTC)
 * I don't honestly see the utility of the section of the complex impedance of a capacitor, since it just duplicates that in capacitor. We don't need to repeat basic information. I've removed it. The bit about the pole/zero is also probably covered better elsewhere and doesn't actually have any context here at all. It would be better discussed elsewhere is context, with just a reference from here. This article, is after all, about RC circuits, not poles and zeros of transfer functions. The generalisation of the Laplace stuff is good, though, except that there are some undefined symobls that need fixing. I have limited internet seconds now, so can't doit myself.-Splash talk 19:04, 7 November 2005 (UTC)
 * I also dislike the fact that $$j\omega$$ stuff is not in the analysis at all anymore. It's the perspective from which everybody studies it first, and it should be the way we present it first. Laplace may be prettier, but it's less instructive from a fundamentals of circuitiry perspective. -Splash talk 19:07, 7 November 2005 (UTC)

j &omega;
Are you looking at the same article that I am? I see j ω all over the place. Of course, now that you unilaterally removed the section on Complex Impedance, the fact that s = j ω is no longer in the article, so it's pretty difficult for people to make the connection. Why don't you simply revert the article back to the way it was before I started making any changes at all?

If you notice, I didn't really remove any information from the article. All I did was to add new information that was not already there.

Why would you want to have Pole-Zero diagrams in an article about RC circuits? Do you mean, besides the fact that they provide a phenomenally easy way to understand what's going on? Of course it was out of context -- Wikipedia is a work in progress -- you cannot expect someone to do it all in one sitting. So I guess it is better to do nothing than to at least get started moving down a path, even if it is not completed on Day 1. Well, good for you, thanks for undoing all of the hard work that I have done over the last few days. Oh, and congratulations on your open mind and willingness to consider someone else's point of view before trashing their ideas. -- Rdrosson 21:24, 7 November 2005 (UTC)
 * If all your good work had been in those few lines of mathematics, the article would be much the poorer. On the other hand, I left almost all of your edits in. There's really no need to get all angry about it &mdash; after all, you can revert me quite easily. The mathematics is not introduced from the j omega perspective, and my personal feeling is that it should be. My personal feeling is nothing to get annoyed with, surely? I didn't revert all your changes, because nearly all of them were good. Take a few deep breaths, and re-read my messages, and you'll find them much less offensive than you currently think they are. And as for unilateralism, just about every Wikipedia edit is unilateral: yours were too. -Splash talk 16:33, 8 November 2005 (UTC)

Step response
Is the step response the same thing as the impulse response? Because if it is it should says so, and if its not - the step response should be added. Fresheneesz 23:29, 9 December 2005 (UTC)
 * No, it's not. The step response is in the Time domain considerations section. Unfortunately, the recent rearrangement of the page was a little haphazard and needs fixing. Jump in... -Splash talk 23:31, 9 December 2005 (UTC)

Voltage source?
I don't think it has to be driven by a voltage source... why would it? Fresheneesz 06:05, 14 April 2006 (UTC)

Integrator Circuit
Should the statement: "...Consider the output across the capacitor at high frequency..." instead read "...at low frequency..."? I'm not certain about this, but doesn't the voltage across the capacitor go to 0 at high frequencies, so the integrating circuit, the one whose output voltage waveform closely matches the area of that of the input voltage, should only be observed at low frequencies? 134.226.1.229 20:19, 9 January 2007 (UTC)

Definition seems wrong, what about multiple caps and resistors?
The definition says that a RC circuit has only one cap and one resistor. An RC circuit should be any circuit comprised of only resistors and caps, of any complexity. In that case I'd say the article was too focused on the simple canonical case. Also I don't think its necessary to give the transfer function for the case where the output is across the resistor and when its across the cap. I'm gonna try making the series example a bit more concise if no one objects. Roger 01:12, 12 May 2007 (UTC)


 * Any circuit with multiple resistors, multiple sources and one capacitor can be put into the form of a circuit consisting of one source, one resistor, and one capacitor. While I agree with you that multiple capacitors would still make an 'RC' circuit in the broadest sense (plural Rs and Cs); in the stricter sense that is most common, RC refers to a resistor and a capacitor (the singular sense) such that we have a circuit with a single time constant.
 * I certainly don't object to making this article more concise. In fact, there is much repetition of this material in RL circuit so have a go at that one too.  The fact is, I do believe that the majority of the 'filter' theory here is out of place.  Go for it.  Alfred Centauri 04:23, 12 May 2007 (UTC)
 * Well the definition says "It consists of a resistor and a capacitor, either in series or in parallel, driven by a voltage or current source", that excludes multiple resistors as well. I don't think the convention of calling a circuit a "RC" (Resistor/Capacitor) circuit is meant to literally mean one resistor and one capacitor since there's alot of network theory that applies to circuits having only resistors and capacitors (including multiples). I'll work on shortening the rest of the article, but I think the definition needs changing. Roger 04:38, 12 May 2007 (UTC)


 * Done. Alfred Centauri 13:43, 12 May 2007 (UTC)

Time Domain Plots
I've increased the line weight on my plots by 600%, it should make them much easier to see (you're right! the small thumbs were impossible to see [I set a large thumb size]). Rather than get in a revert war over it, if you're okay with them, give me permission to replace them or do it yourself. Here's how they look now.--Ktims 00:36, 1 April 2006 (UTC)

This formula $$\,\!V_C(t) = V\left(1 - e^{-t/RC}\right)$$ $$\,\!V_R(t) = Ve^{-t/RC}$$ give the same result as 0.999^1000 = 0.367695425 $$\tau \ = \ RC $$, The discharge time is      744,760591 $$\tau \ $$, example 0.999^744 760.590 = 4.94065646 × 10-324, 0.999^744 760.591 = 0 on google calc. 1-result tau = charge tau. Pawem1 --213.199.225.33 12:15, 17 September 2007 (UTC)

English, Layman Version
This is great for eggheads who already know the subject up and down, but how about something for the layman, who, say, wants to build an RC circuit as 1/10th of a second a timer? More practical and less theoretical would be helpful. --68.97.208.232 14:13, 8 May 2006 (UTC)
 * Seconded. I agree that the physics and math involved are important, but I would also like to see a more practical approach to this article. As it is now, it seems straight from a physics textbook, and the raw information isn't really all that useful without a great deal of educational context. At the very least, I think this article could use an explanation of why such a circuit is useful, and why it is useful for those applications. --Ktims 07:52, 9 May 2006 (UTC)

Another request for a section which describes RC for dabblers. My question is: when a make an RC circuit and attach to my microcontroller, how exactly are the electrons flowing? What makes the electrons start and stop so that a series of peaks are created and can be timed? Thanks for what is here so far. 59.183.11.174 14:02, 29 November 2006 (UTC)John

Yeah I agree, this article should have a simple section for students starting out in understanding electronics. The coverage is way too complex for beginners. —Preceding unsigned comment added by 207.191.134.162 (talk) 05:22, 23 September 2008 (UTC)

I came to this page looking for two things, an explanation of why an RC circuit does what it does, and what it is commonly used for. I really don't see answers to either one of those questions here at all. The entire article reads like a math proof. I understand that the writer(s) of the article are worried about what experts will think of the accuracy of their page. But most people coming to this page are here because they are seeking an explanation of what these circuits are FOR in some intermediate to beginners electronics project. It's like reading a biochemical description of what aspirin does at the molecular level when what one is looking for is an answer to the question "what does aspirin do". --Dean


 * I added a small paragraph in the lead to help. I think it still needs an illustration in the lead, maybe something more elaborate than first-order.  Dicklyon (talk) 15:19, 24 June 2009 (UTC)

The simplest analog IIR filter ?
Anyone please comment on the above characterisation of the RC circuit found at Recurrence relation. Cuddlyable3 (talk) 13:30, 8 April 2009 (UTC)


 * It's nonsense; the terms IIR and FIR and recurrence relation are not applicable to continuous-time filters. Dicklyon (talk) 21:12, 8 April 2009 (UTC)


 * Agreed. I see it's corrected now. Cuddlyable3 (talk) 20:06, 24 June 2009 (UTC)

simplify
Sorry but this is not a good article. The statement "The angular frequency s is, in general, a complex number, "  is wrong. w is the angular frequency. I think the article would be improved if it used jw in place of s as s is used for laplace transforms. Arydberg (talk) 16:21, 14 July 2009 (UTC)
 * s describes the more general case, jω applies only to steady-state sinusoids and the article correctly moves from the general to the more specific. However you are quite right, the angular frequency, ω is always real, s should be described as the complex frequency.  I have made the necessary change to the article.  Sp in ni  ng  Spark  19:42, 24 July 2009 (UTC)

I wrote the following. Any comments?

RC Circuits
Consider the following circuit



When the switch is moved to the up position the capacitor will begin to charge. The time required for this is determined by the product of the capacitance and the resistor. for the elements shown we have R = 1K or 1 x 103 ohms and C = 1000uF ( 1000 x 10-6  or RC = 1 .  This results in a time constant of 1 second or it takes or it takes 1 second for the capacitor to charge to 63 % of it’s final value.

The charge of the capacitor is shown in the following graph.



Now lets move the switch to the down position. The capacitor will begin to discharge through the resistor. The discharge is shown in the following plot.



If we want to see the math behind all this consider the top schematic with the switch in the down position. Then by Kirchhoff's law the sum of all the voltages around the circuit must equal zero. The resistor voltage is = i x R,   The capacitor voltage is  = $$\frac{1} {C} $$ $$\int i dt$$.

or     i x R  = $$\frac{-1} {C} $$ $$\int i dt$$

If we take the derivative of this

we get:     R $$\frac{di} {dt} $$  =  $$\frac{-1} {C}  i$$

and rearrange terms $$\frac{di} {i} $$ =  $$\frac{-dt} {RC} $$

Then integrate both sides we get :       $$ln ( i) = \frac{-t} {RC} $$

or      $$ i = e$$-t/RC

Which is the equation for the 2nd curve above. —Preceding unsigned comment added by Arydberg (talk • contribs) 13:27, 24 July 2009 (UTC)


 * Similar equations and diagrams are already in the article. The existing diagrams are much better resolution svg format.  Sp in ni  ng  Spark  19:28, 24 July 2009 (UTC)
 * Sorry about moving your note. It seems to be fixed.   You are correct about svg being a better format but i cannot use svg.   IMHO what is needed is a introduction to rc filters and that is what i am attempting to do.    It seems that much of the traffic is from people who want to learn about electronics.
 * Arydberg (talk) 02:11, 13 August 2009 (UTC)

Voltage source

If we move on to a RC series network supplied by a AC sinusoidal voltage source and develop the impedance of the network we should start by defining the voltage source. Rather than the common notation of cycles per second we will use radians per second represented by $$\omega$$. Note that there are $$ 2 \pi$$ radians in 360 degrees  so 60 cycles per second is equal to  $$ 60 x 2 \pi $$  radians per second or  377 radians per second. For simplicity we will let the peak amplitude of the voltage = 1 volt. Also we will use Z to represent impedance. Impedance is similar to resistance but it may be a complex number and can represent capacitors, inductors or entire complex networks. It can even represent a resistor but usually the convention is to use R for resistance.

A simplistic form of the voltage is $$ V = sin( \omega  t)$$ but there is a more precise way of defining voltage. This is to let $$V = e$$-j$\omega$t where j = square root of -1,    $$\omega $$  is in radians per second  and t = time in seconds.

As it happens  ej$\omega$t = cos $$\omega$$t +j  sin $$\omega$$t.

from V = ej $\omega$ t

then dv/dt = j $$\omega$$ ej w t

or dv/dt = j $$\omega$$ V

This is not a trivial result. It is termed an operator and makes possible a hudge simplification of the mathmatics that follows. It was one of many contributations of Oliver Heaviside to electrical engineering and lead to Laplace transforms.

R C Series Circuit

OK now that the voltage is defined we need to look at the two elements R and C.

For the resistor from i = V /R  or

$$R = \frac{V}{i}$$

The above is simple if we do the same thing for the capacitor we we can find the impedance of a capacitor ( Zc):

from i = C dV/dt  and dv/dt = j $$\omega$$t  V  so    i = C j*W*V  or  Zc = V/i =  1/C * j*W

$$Zc=\frac{ 1} {jC\omega}$$

So now we have a circuit with two impedance elements in series. it looks like this.



Just as we add the resistance of two resistors by adding R1 to R2 we can add these two elements by:

$$Z = R + \frac{ 1}{j \omega C}$$

If we want to plot the impedance of this network and we let R = 1 ohm and C = 1 farad we will get the plot shown below. We can see that as W ( or $$\omega$$) becomes large the impedance approches R and as $$\omega$$ becomes very small or approaches DC the impedance becomes very large.



Arydberg (talk) 19:36, 12 August 2009 (UTC) —contribs) 13:12, 12 August 2009 (UTC)

Linear RC circuit with voltage gain
This is more of a curiosity than a comment on the article. I recently came across an RC circuit with voltage gain. The author of the book says that the circuit has a maximum voltage gain of 1.15. I found this difficult to believe, so I simulated it in PSpice and analysed the circuit algebraically. It does indeed behave as advertised, with a Vo/Vs of about 1.15 V/V at about 1 kHz. I'm not trying to plug the book, and to prove it I have redrawn the circuit here so that you don't have to follow the link above.

I'm not claiming anything supernatural, since the circuit obviously doesn't have power gain, but I just can't understand intuitively how it works. I understand how LC circuits can have voltage gain, but I've never seen it happen in a linear RC circuit before. Is there any way to explain it, other than by saying "do the sums"? --Heron 10:15, 30 May 2007 (UTC)


 * Unless I've made an error, the transfer function is:


 * $$T(s) = \frac{1 + s[R_2 C_2 + (R_1 + R_2) C_1]}{s^2 R_1 R_2 C_1 C_2 + s[R_2 C_2 + (R_1 + R_2) C_1] + 1}$$


 * Comparing this to the standard form:


 * $$T(s) = \frac{1 + \frac{1}{Q} (\frac{s}{s_0})}{(\frac{s}{s_0})^2 + \frac{1}{Q}(\frac{s}{s_0}) + 1}$$


 * yields:


 * $$s_0 = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}$$


 * $$Q = \frac{1}{\sqrt{\frac{R_2 C_2}{R_1 C_1}} + \sqrt{\frac{R_1 C_1}{R_2 C_2}} + \sqrt{\frac{R_2 C_1}{R_1 C_2}}}$$


 * Using the values specified yields:


 * $$f_0 = 1592 Hz \, $$


 * $$Q = \frac{1}{2.0001} $$


 * So, we have a sum of a 2nd order LPF and a 1st order BPF where both filters have the same Q (about 1/2) and the same cut-off / center frequency. The sum of these two low Q filters gives a magnitude greater than unity somewhat below the cut-off frequency.


 * Looking at this circuit in the time domain with a 'hand waving' argument, observe that C1 'sees' R1 in series with and impedance while C2 'sees' R2 in parallel with some impedance. Thus, the phase of the current through C2 lags the phase of the current through C1.  This implies that, at points in time, there is a current 'down' through C2 and a current 'up' through C1.  But, this is what we need for Vo to exceed Vs.  The current 'up' through C1 causes a voltage across R1 that adds to Vs.  Alfred Centauri 15:40, 30 May 2007 (UTC)


 * Thanks for that insight. I agree with your transfer function, but I couldn't have turned that into a filter type without your help.  The hand-waving explanation is also ingenious, but one thing troubles me.  You say that the current up C1 causes a voltage across R1 that adds to Vs, which I understand, but isn't that an instantaneous current?  It might only be true at some points in the cycle, as it would be true of any two out-of-phase sinusoids.  How do you generalise from that to saying that Vo(RMS) > Vs(RMS)? --Heron 21:10, 30 May 2007 (UTC)


 * [[Image:RC-with-gain-simulation.png|right|200px]]
 * Yes, it's instantaneous current. Take a look at the transient analysis screenshot.  See that with the appropriate phase relationship, the two sinusoids add constructively.  To answer your question about generalizing to the the AC case, as long as there is a current 'up' through C1 when Vs is at max positive peak, we are assured that Vo(RMS) > Vs(RMS).  This seemed like the case to me as long as we are below the cutoff frequency.  Alfred Centauri 22:29, 30 May 2007 (UTC)


 * Thanks. I've just played around with my version of the transient simulation, so I can see now how V(R1) adds to Vs below Fcutoff (right down to DC).  Last time I only looked at the AC simulation, which didn't help me much.  --Heron 12:59, 31 May 2007 (UTC)


 * Thanks for bringing this circuit to our attention. I didn't suspect that a passive RC filter could give a voltage gain.  BTW, I just thought of a better 'hand waving' argument.  First, see that the branch containing R1 and C1 has a much larger impedance than the branch with R2 and C2 so the R2 C2 branch behaves more or less like a standard 1st order LPF.  At low frequencies, the voltage across R2 leads Vs by close to 90 deg.  But, the voltage across R2 'drives' the R1 C1 branch.  Thus, at low frequencies, the voltage across R1 leads the voltage across R2 by close to 90 deg.  But this means that the voltage across R1 leads Vs by close to 180 deg which is what we need to add constructively.  Alfred Centauri 13:52, 31 May 2007 (UTC)


 * That explanation is easier to grasp. I consider the mystery solved. --Heron 19:25, 31 May 2007 (UTC)


 * The book "Fast Analytical Techniques for Electrical and Electronic Circuits" by Vatche Vorperian discusses this circuit, and more complicated versions, quite extensively (making use of the extra element theorem). The reason why you get a >1 voltage gain is because you can design the zero to occur before the two poles. Roger 20:08, 31 May 2007 (UTC)


 * Roger, that is a good observation but I have to point out that having the zero come before the poles doesn't necessarily give insight as to why this passive RC filter has a gain exceeding unity. Up until Heron brought this circuit to my attention, I assumed that any passive RC filter with a zero before any poles would necessarily have a DC gain less than unity.  Thanks for the reference, I'm going to take a look at it.  Alfred Centauri 23:17, 31 May 2007 (UTC)


 * The circuit approaches its maximum gain of sqrt(4/3) (approximately 1.1547 or 1.25 decibels) when the impedance of the second stage is high compared to the first stage, the products RC of the two resistor-capacitor pairs are equal, and the angular frequency is sqrt(2)/RC. —Preceding unsigned comment added by 208.53.195.38 (talk) 18:00, 4 March 2011 (UTC)

For me, the key to understanding intuitively how this circuit works is to realise that the sum of the magnitudes of the voltages across the resistor and the capacitor in an RC circuit is greater than unity (relative to the input voltage). When the impedances of the two elements are equal, |VR2| = |VC2|, the sum is at its maximum of &radic;2. They are, of course, in quadrature and sum algebraically to unity. However, if VR2 is passed through a phase shift circuit such that that the output is brought into phase with VC2 then the output of the entire circuit could indeed be &radic;2. Phase shifters can be built entirely out of passive components and this is exactly what the second RC circuit can be understood to be doing. It cannot, however, simultaneously output both a 90° phase shift and maximum voltage. With C1 very large compared with R1 the phase shift approaches 90° but VC2 approaches zero. With C1 very small the output is maximum but the phase shift approaches zero. As a compromise, and for simplicity, choose R1 and C1 such that their impedances are equal in magnitude and large enough not to significantly drop the voltage across R2 (they can be made arbitrarily large since the whole circuit has no load impedance specified). VC1 is then 1/&radic;2 of VR2, or 1/2 of the input VS and 45° out of phase with VC2 (see phasor diagram). 1/&radic;2 + 1/2&ang;45° = &radic;5/2 &asymp; 1.118&ang;18.4°. As pointed out in the post above this is not the maximum output but is easily visualised in phasor diagrams.  Spinning Spark  16:30, 1 July 2011 (UTC)


 * Here are a few thoughts that can help the intuitive understanding of this odd circuit:
 * Structure. R2-C2 form an RC circuit with two outputs: a differentiating (VR2) and an integrating (VC2) one. The differentiating output drives the RC circuit R1-C1 that has only an integrating output (VC1). So, the upper RC circuit depends on the lower RC circuit. The two integrating outputs are connected in series so that their voltages add thus forming the total output voltage VOUT = VC1 + VC2.
 * Operation. In the beginning, the input voltage VIN begins increasing. As VIN > VC1 + VC2, the input voltage source passes currents through R1 and R2 to C1 and C2 and they begin charging. The voltages across them begin increasing: VC1 rapidly increases in the beginning as its input voltage VR2 is maximal; then, it slows down as VR2 decreases. VC2 continuosly increases and "lifts" the output voltage through C1 (a charged capacitor shifts voltage variations). As a result, the output voltage (the sum) increases thus following and even exceeding the input voltage in the moment when it reaches its maximum.
 * The input voltage has reached its maximum and begins decreasing. VIN drops below VOUT and VC1 begins decreasing since C1 begins discharging. But (interesting!) this process is slowed down since VC2 continues increasing and "lifts" the output voltage through C1. So, the output voltage is a sum of two voltages: the decreasing VC1 and the increasing VC2. Obviously, the result is a voltage following (little exceeding) the input one. Circuit dreamer (talk, contribs, email) 14:29, 5 July 2011 (UTC)

What is Cs?
This question deleted by the poster. This term apparently means "C times s". 3dimen (talk) 04:20, 8 August 2011 (UTC)


 * Yes. Maybe that confusion is why it's often written with the s before the C as in 1/sC.  Should we change it that way more generally?  Dicklyon (talk) 07:13, 8 August 2011 (UTC)

Link to RC calculator
This ends up at a site that has taken to automaticlly banning any IP address that connects to it so it is no longer useful. 202.0.86.162 (talk) 01:50, 3 February 2012 (UTC)

What's so natural?
I see nothing in the section titled "natural response" that gives the nonexpert any hint about how that title relates to the material in the section. Said differently, how and to whom is that title useful? The people who already know this stuff don't need the title, and the people who don't know the stuff don't understand the title; so the title is useful to nobody.—PaulTanenbaum (talk) 02:44, 3 February 2012 (UTC)


 * I agree this should be better explained. The natural response of this, or any, circuit is its behaviour when not being driven by an external source of energy.  I also think that the sentence "[w]hen a circuit consists of only a charged capacitor and a resistor, the capacitor will discharge its stored energy through the resistor" is grossly misleading.  A capacitor and resistor connected in series and to nothing else will not result in a transfer of charge from anywhere to anywhere.  Implicitly, the calculation is assuming that the the circuit is connected to zero volts, in other words a short-circuit, effectively putting the two elements in parallel rather than series.  Spinning  Spark  09:31, 3 February 2012 (UTC)

Make this page useful to someone who isn't a PhD or EE
Good teaching and education starts with tailoring information to audience. The disclaimer at the top of this article - "This article relies on knowledge of the complex impedance representation of capacitors and on knowledge of the frequency domain representation of signals " is comical. If one knew all those things, well, one probably wouldn't be looking for this page.

RC filters are one of the first things that someone exploring electronic theory runs into. They can be conceptually explained in a few short paragraphs before falling into all the obtuse and complex math, which perhaps some might come to a Wiki page for, but more likely would be the interest of the minority. —Preceding unsigned comment added by 38.119.114.42 (talk) 05:54, 17 March 2008 (UTC)


 * Since you were able to type this comment, I assume your fingers aren't broken and thus I wonder why you haven't made the changes yourself? The editors here are unpaid volunteers and don't really give a rat's ass for armchair editors.  If you have the chops, fix it.  If not, then I suggest you come down down off of your high horse and show some respect to those that have actually taken the time to edit Wikipedia.   Alfred Centauri (talk) 02:25, 19 March 2008 (UTC)

You both have good points; by converting all caps heading to normal, I hope to have put OP's comment into a form less offensive, and nicely invite him to follow up on your suggestions. Dicklyon (talk) 04:49, 19 March 2008 (UTC)

This is classic Wiki nonsense, no insight into applications. Most people hit the page looking applied electronics shortcuts.... and what you get is text book extractions is that same ole wiki font... utterly useless. — Preceding unsigned comment added by 24.222.194.29 (talk) 12:35, 22 February 2012 (UTC)
 * Rather than insult editors on this page, you could add something to the article yourself. After all, the article was created by volunteers just like you, not by someone you paid for the service.  By the way, how do you know what most people are looking for, do you have access to some data that we don't?  Spinning  Spark  13:52, 22 February 2012 (UTC)

Sigma (as in s = sigma + jw)
Sigma (as in s = sigma + jw) hs the dimension of Nepers/s NOT radians/s. Just saying. — Preceding unsigned comment added by 24.101.25.254 (talk) 15:45, 29 April 2016 (UTC)

Universal dielectric response
I just removed the following addition from the article lead,

It may be that UDR can explain properties of dielectrics in terms of RC networks – I don't know anything about that – but it is certainly not true that RC networks have the general property of a frequency power law. Easily disproved by contradiction. The simplest example is a series RC network whose admittance is given by,


 * $$Y= \frac {sC}{1 + sCR}$$

This is a rational function of frequency, but it is not a power law. Further, rational functions are not peculiar to RC circuits. All finite, linear, lumped element circuits can be described by rational functions. SpinningSpark 16:30, 4 September 2018 (UTC)


 * Percolation through RC networks under AC conditions is a topic worth discussing in this article. In particular the role of the fraction of C or R elements and the scaling of admittance with frequency...
 * Known as the UDR, this is an emergent phenomenon of increasing interest. In particular with random RC networks. This robust power law response has been detailed in numerous publications. The frequency range in which emergent power law behaviour is seen is determined by the network components. This RC network model and the emergent UDR is very useful not just in electronic engineering but also for gaining information about heterogeneous systems from impedance spectroscopy measurements. — Preceding unsigned comment added by Brotter121 (talk • contribs) 08:55, 5 September 2018 (UTC)


 * I've removed the heading of your post. The convention when replying on talk pages is to indent your post, not start a new thread.  Percolation theory, as it relates to electrical properties, is concerned with the analysis of essentially infinite networks of distributed elements.  As I stated above, this article is about finite, lumped networks.  It is intended as an article on simple, basic networks.  This material does not really belong on this page.  It certainly does not belong in the lead.  The lead section is meant to be a summary of the article body and this is introducing something completely new.  It is also unsourced, which is not really an acceptable thing to do for an advanced concept like this.  And as I said above, it is demonstrably incorrect as stated. SpinningSpark 18:18, 5 September 2018 (UTC)


 * A more suitable place might be one of our articles on percolation, such as percolation theory, percolation threshold, or percolation critical exponents. Or perhaps an entirely new article on percolation as it relates to electrical theory. SpinningSpark 18:18, 5 September 2018 (UTC)