Talk:RL circuit

Complex Impedance Methods

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

$Z_{L}\ =\ j\omega L$

the complex impedance is

$Z_{L}\ =\ Ls$

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)

If you look slightly further down the page, you'll see that Laplace domain stuff is included there. -Splash^talk 12:52, 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)

The entire section 'Complex impedance' has no reference to RL circuits. Yes, it's probably needed to understand the frequency domain methods used in the rest of the article, but if each of the hundreds of articles that use these methods, RC circuit, LC circuit, RLC circuit, amplifier, Colpitts oscillator, etc. has an introductory section on complex impedance and eigenvalues, we'll get tremendous article bloat. This section should be removed and replaced with a warning that readers will have to know complex impedance and Laplace transform to understand what follows. --Chetvorno^TALK 13:02, 9 October 2008 (UTC)

Well, as it happens, I created Template:Complex Z, call as {{Complex Z}}, for just such occassions in the articles I write. I also think that introducing the s-domain in this article is of limited use to the readers, many of whom are not familiar with the method. Yes, it is a more powerful concept, but in something this simple jω-domain or direct solution of the diff equations will do for most cases. Save it for articles that can really make use of itElliptical filter for instance. SpinningSpark 17:20, 9 October 2008 (UTC)

RL RC similiarites

I was looking down the RL and RC articles, and I noticed they are not in the same layout. I believe since there are many similarities between RL and RC circuits it would be beneficial to someone reading the two if the layouts were the same. The one that stood out the most to me was the Natural Response. On the RC page it is at the top and called 'natural response.' On the RL page it is farther down and called 'zero input response' and has an aka. 'natural response' —Preceding unsigned comment added by 209.142.189.102 (talk) 19:39, 3 February 2011 (UTC)

SVG Conversion

Sorry about the SVG conversion missing the italics. I didn't notice that in preview. The SVG is actually correct, with italics, but the thumbnail is being generated incorrectly. I will pursue the correction of the thumbnail generation before applying the SVG change again. — Preceding unsigned comment added by 24.19.214.109 (talk) 17:35, 23 October 2011 (UTC)

Energy

I have removed the following from the article;

Delivered power

The energy delivered to the resistor during any interval of time after the switch has been opened for a circuit.

$\omega ={\frac {1}{2}}LI_{0}^{2}(1-e^{-2{\frac {R}{L}}t})$

While t ≥ 0.

This is very unclear for a number of reasons. The title says power but the expression is in units of energy. It is not explained what switch is opening. The meaning of I₀ is not given. I am also not convinced it is correct. If we assume the switch in question is intitially shorting the circuit and when opened connects a supply and I₀ is the final current in the circuit. The boundary condition at t=∞ will have a steady power of V²/R dissipated in the resistor. The energy expression should therefore have a term in t as the energy will be steadily increasing. The boundary power will not depend on L as is implied by the given expression. I also make it that there is a another exponential term in the energy equation.

This needs more context on what the source is discussing. It would also benefit from not using omega as a symbol for energy (I think the source was probably using w). It is probably also more appropriate for the time domain considerations section. SpinningSpark 22:11, 5 November 2013 (UTC)

Fixed the problem. I improperly created the header.—CKY2250 ^ταικ 00:33, 6 November 2013 (UTC)

Energy delivered

The energy delivered to the resistor during any interval of time after the switch has been opened for a circuit.

$\omega ={\frac {1}{2}}LI_{0}^{2}(1-e^{-2{\frac {R}{L}}t})$

While t ≥ 0. After the switch has been opened. As t approaches infinite the energy dissipated in the resistor approaches the initial energy stored in the inductor.

Edited from original.00:33, 6 November 2013 (UTC)

---

I have restored the version of the deleted passage above. Please don't alter it, my comments about it will make no sense if you do. In answer to your response, one would have to close a switch putting L and R in parallel in order to get the result stated. Opening a switch makes no sense. I really think that you need to provide a diagram to go with this before it can be understood by readers. At the very least an explanation of what is being switched from and to is needed. SpinningSpark 01:43, 6 November 2013 (UTC)

Do you know of any good diagram maker or would multisim be the best.—CKY2250 ^ταικ 02:26, 6 November 2013 (UTC)

I use Inkscape for diagrams, but Multisim will do just fine. SpinningSpark 10:10, 6 November 2013 (UTC)

Ok just got some time into my day here is the diagram that goes with the equation.—CKY2250 ^ταικ 15:49, 15 November 2013 (UTC)

That's a parallel circuit, not a series circuit as claimed. And you can't use ω for energy, it is used for something else in the article, as it is across about all science and engineering disciplines. Your source seems to be using w (for work), not ω. SpinningSpark 16:26, 15 November 2013 (UTC)

There is a switch. It charges the inductor to full capacity then it opens so it is series. At t=0 then there is only an inductor and a resistor. I know the equation is correct for the diagram. —CKY2250 ^ταικ 16:40, 15 November 2013 (UTC)

I'm not disputing the equation is correct for the diagram. In the context of this article series circuit means that the L and the R are in series so there is no way this should be placed under the series circuit heading. It could, however, be placed under parallel circuit. Oh, and please be consistent with using lower case t for time (not T=0) and explain the meaning of I₀ in the article. SpinningSpark 19:01, 15 November 2013 (UTC)

Just to clarify things the current stored in the inductor is equal to the source current. L and R are in series after T=0 which is a closed series circuit. Before t=0 the inductor is charging for a really long time and becomes the same value as the source current. The equation is only relevant when t >=0. Hence why this is a series problem, so I have no idea what you're thinking and I'm thinking.—CKY2250 ^ταικ

I will fix the T, but I₀ is different from I_s. I₀ is the current of the inductor when it has been charged by the source for a very long time, so once it switches it is then the initial current.—CKY2250 ^ταικ 19:19, 15 November 2013 (UTC)

I know how the circuit works and you don't need to explain to me what I₀ is. The point is that you should not introduce symbols into an article without defining them. It is completely perverse to describe a circuit that was a parallel circuit before the source was disconnected and has not changed after the source has been disconnected as suddenly becoming a series circuit. SpinningSpark 21:27, 15 November 2013 (UTC)

With or without the R_o it is still the same current. This is a fundamental circuit architecture. It is defined how it is to show that the inductor at full capacity. You are just seeing something for the first time are not understanding it, you do know that when a inductor is in contact with a source current for a very long time that it is short right, and then the current will skip the resistor there since the current flowing through the resistor would be negligible. I still don't see how you are classifying this as a parallel circuit when it is about t≥0 so it is only one inductor with full capacity and one resistor. You could think of it this way. Only circuit you see is an inductor with full charge and a resistor, however that would be impossible in real life, since there is no source added before the start.—CKY2250 ^ταικ 22:37, 15 November 2013 (UTC)

100% same idea just different way of showing it. Here is an open source I found [1], showing my book would not be allowed.—CKY2250 ^ταικ 23:24, 15 November 2013 (UTC)

References

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