Help with RLC circuit question - Page 1
Forum section: electronics / beginners

author: JOJOElectro29
author_rank: Contributor
timestamp: April 25, 2025, 11:12:27 am
content: Can someone explain step by step how to do this. I'm very confused with how the voltage across the capacitor is calculated. Online information says that it is negative but then how can the current be positive?

author: PGPG
author_rank: Frequent Contributor
timestamp: April 25, 2025, 12:57:59 pm
content: When you draw the arrow specifying voltage at capacitor you have two ways of doing it and in one the voltage at capacitor at t=0 moment will be +30V and in the other way it will be -30V.Equations for the circuit will be different in both cases, but with the same result for i(t) as its direction is clearly given.

author: Picuino
author_rank: Super Contributor
timestamp: April 25, 2025, 04:29:25 pm
content: Simulation:https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0lwTOG8EHZUGZIA54DYBOA1ELEAVhABYMKBTAWjDACgAnEeVRbSauELyicWAZ045wePvGzYpfPhABmAQwA2ouiwDu4WQv3z48KlBYBjI51P8ZtxdHiUGMOJDB4MVSFlxh4MHJcPEp4AmgCbHIWdWshLh5FKGgXNzgMAnI8bjxQgioaKjBMcCpIggwWADcJeU8lAwbhPh9SZMcYvQD5IR7DSF07Qx8+IUGOUcNZeukW2A8hqeapkzNBgHNrNe3slpYAe2oQPHX+In5Ocmg8YUQzKqPaU+EfC7byMCdwTlJSQ+EL1akHeMmut0QiD4VQw8loADEII53HAkQsPB4QEwQABhVQAB1UFgAlgAXVQAOws2lhx0R80xX0x2IAkhSACYAVws5KpNPkfHpSiQEGxACU6KJiaJedSWAALH4PFhAA

author: TimFox
author_rank: Super Contributor
timestamp: April 25, 2025, 05:27:22 pm
content: The circuit as drawn omitted the reference point for measuring voltages: the implication is that the vertical line at the left edge of the circuit is "zero" or "ground".

author: MrAl
author_rank: Super Contributor
timestamp: April 25, 2025, 09:48:06 pm
content: Quote from: JOJOElectro29 on April 25, 2025, 11:12:27 amHello there,Taking the picture you show here and wording verbatim, this is not really a question of if the capacitor voltage is positive or negative alone. It's an initial value problem. When know this from the problem description where the switch is in one state (closed), then switched to another state (open). The switch is used to set up the initial conditions, then provide a time when the initial conditions plus any sources still in the circuit will drive the response. The difference between this and a circuit with no initial conditions is that when there are no initial conditions you only have to consider the sources in the circuit and not any voltage or current of any capacitor and/or inductor. In this problem though you do have to consider that also.To solve this in its entirety, you should first know how to solve RLC circuits that do not have any initial conditions. You should really be proficient at that first. That's because in this circuit we will have two different topologies and if you know how to solve RLC circuits with no initial values then you will be able to transform this circuit into a circuit that has only sources and no longer has what we call initial conditions or initial values. I'll explain this method because it will probably be the easiest way to do it for you.To start, you must solve for the initial values for the inductor and for the capacitor. The inductor will have an initial current i0 and the cap an initial voltage v0. You first solve for them, then go to the next step. So step 1 is to solve for the initial values.To solve for the initial values, we first look at the circuit and try to find a way to solve the circuit when the switch is closed. That means that 30v is applied to both the capacitor and to the inductor RL part which is parallel. Since it is easier to place the ground on the far right side (the minus terminal of the cap) we will do it that way, then after we find the solution we will consider if we need to flip the sign.Going with that idea, we can see that the cap is charged immediately to +30v. The inductor and resistor RL part has a current flow due to the response of a regular RL circuit to a step input. We need the current in the inductor i0 and that will be the current when time 't' approaches infinity. You can solve for that but once you do more circuits like this you will realize that an RL series circuit subject to a step response that lasts indefinitely will have a maximum current caused only by the resistor and voltage source. In other words, the inductor becomes a short circuit at t going to infinity.So with that we can see i0 is 30/R which is 30/2 which is 15 amps and has the direction shown in the drawing, and the capacitor voltage v0 is +30 volts. Those are the initial conditions:i0=+15 and follows the arrow, andv0=+30 with positive on the left and negative on the right (which we are calling ground for now).Now that we have that information, we can proceed to step 2. We form the second topology when the switch is opened for the first time.Step 2 is to replace the initial conditions with initial condition generators. There are other ways to do this but this is probably the easiest to understand and do the calculations.For the inductor, the initial condition is +15 amps so we place a constant current source of 15 amps in parallel to the inductor with the arrow pointing in the same direction as on the drawing. For the cap, the initial condition is +30 volts so we place a constant voltage source of +30 volts in series with the cap. For this we will place it to the left of the capacitor, in series with it. The original +30v voltage source gets switched out of the circuit, so now we ignore that. We only have the constant current source and the new capacitor voltage source to consider.Step 3 is where you put your knowledge of how to solve an RLC circuit that has constant sources driving it. One way to do this is through the use of the concept of superposition, or you can do it another way if you like (such as with regular Nodal Analysis). The result of this analysis will be a time domain function that will describe the voltage across the capacitor if that is to be the end result, or the current through the inductor, or the voltage across the resistor, etc., or all of the possible solutions for voltages and currents.So to recap each step:1. Solve for the initial conditions.2. Replace the values with appropriate initial condition generators, set the current through the inductor (not the initial condition value) to zero and cap voltage to zero (not the initial condition value).3. Solve the multiple source circuit that results from step 2.One small note. Once the initial condition generators are placed into the circuit, the values for the associated components now include those generator values. For example, if at time t1 you find that the cap voltage is +2.1 volts and the generator was +30 volts, then you add the two to get 32.1 volts and that is the value of the actual capacitor voltage in the original circuit at time t1 after the switch has opened.If you choose to place the ground on the left side of the cap, that means the voltage across the cap should be reported as negative instead of positive.This might sound like a lot to take in, but it gets easy once you do a couple of these problems.There are other ways to do this but this is probably the most straightforward. There are no tricks to remember except that in the end the voltage across the original cap is not the same as in the transformed cap because the transformed cap still has the initial condition voltage source in series with it, and that must be added to the value obtained across the transformed circuit capacitor. This is also true for the original inductor and its new constant current source.

author: Benta
author_rank: Super Contributor
timestamp: April 25, 2025, 10:14:54 pm
content: Sorry, but for classic analysis in the time domain, you'll have to set up a differential/integral equation for the circuit. Superposition will not work.I'm not able to use mathematical notation here, so it's hard to show what I mean.Using Laplace analysis is easier, but presumes you're familiar with complex numbers.

author: sharow
author_rank: Contributor
timestamp: April 26, 2025, 05:40:18 am
content: \$i(t) = 15-15e^{-4t}\$ . Sound about right?My calculation say that capacitor is charged instantly. \$\frac{30}{4}\delta(t)\$ ah, switch isopenat t=0. not close. that more make sense. oops.

author: SteveThackery
author_rank: Super Contributor
timestamp: April 26, 2025, 11:21:49 am
content: So we have a simple damped RLC resonant circuit, with initial conditions of i = 15A and Vc= 30V. I'm embarrassed to admit that I have completely forgotten how to calculate i(t).Is anyone able to remind us of how the calculation proceeds? Thanks. 😊

author: PGPG
author_rank: Frequent Contributor
timestamp: April 26, 2025, 01:13:05 pm
content: Quote from: SteveThackery on April 26, 2025, 11:21:49 amI am exactly at the same state.I would probably be able to write differential equation. But how to solve it ?Compare it with that before study taking a part in physics competition I invented integrals (we didn't had at math yet) and my physics part solution took 1/2 of page and then at 3 pages I was calculating integral of sine or cosine without using integral and I got the correct result. But I don't know if I would be able to do it now.

author: ledtester
author_rank: Super Contributor
timestamp: April 26, 2025, 01:38:03 pm
content: The attachment gives the formulas for the damping factor and frequency of oscillations for a damped RLC circuit but for the case when the initial current through the inductor is 0.To account for the initial inductor current we write:$$Q(t) = e^{-\alpha t}\left[ A\cos(\omega t) + B\sin(\omega t) \right]$$where \$\alpha = R/2L\$ and \$\omega\$ is the w' in the attachment.Then solve for \$A\$ and \$B\$ using \$Q(0) = A\$ and initial current = \$Q'(0) = -\alpha\cdot B \cdot \omega\$.

author: sharow
author_rank: Contributor
timestamp: April 26, 2025, 02:28:54 pm
content: It was difficult. I may have chosen the hard way.Actually, I'm not sure this is the right answer.I would like to know if there is a model answer.

author: MrAl
author_rank: Super Contributor
timestamp: April 26, 2025, 04:27:21 pm
content: Quote from: Benta on April 25, 2025, 10:14:54 pmHi,Sorry back at ya! Just because you do not know how to use superposition with two sources does not mean that you cannot do it that way.It's true you can set up a differential equation, but you can also use Laplace Transforms and then superposition becomes easier.For example, if we are to calculate the voltage across the resistor after the switch is open, which we will call t0 for t=0, you first calculate the voltage across the resistor due to the capacitor initial condition generator voltage source, then calculate the voltage across the resistor due to the initial condition current source of the inductor, then add the two. The result is the voltage across the resistor with time vR(t).This is fairly quick and simple but you do have to know how to use Laplace Transforms, and because we are using Laplace Transforms throughout, we don't even have to use complex math. We just have to use algebra and some rules for using Laplace Transforms and Inverse Transforms.The differential equation is not too hard to set up either so we can do it that way too. I'm waiting for a reply from the original poster to see what they are more familiar with. I can show the math both ways at a later date. There's also yet another more canonical method that is similar to differential equations that is used but I'll hold off on that too.

author: MrAl
author_rank: Super Contributor
timestamp: April 26, 2025, 04:34:39 pm
content: Quote from: sharow on April 26, 2025, 02:28:54 pmHi,That plot looks correct but I'll check it better later.[LATER: sorry can't check your result you do not show the entire result part of the image is cut off, the shape looks right though]The solution should look like e^(-a*t)*(A*sin(w*t)+B*cos(w*t)) which can be simplified into e^(-a*t)*C*sin(w*t+ph) where ph is a phase shift. It's a damped sinusoidal.

author: MrAl
author_rank: Super Contributor
timestamp: April 26, 2025, 04:35:40 pm
content: Quote from: mawyatt on April 26, 2025, 02:32:04 pmAre you sure it is critically damped? I don't think it is but I'll check better a little later.

author: MrAl
author_rank: Super Contributor
timestamp: April 26, 2025, 04:53:26 pm
content: The solution using Laplace Transforms and superposition is:vR(t)=30*e^(-2*t)*(sin(2*t)+cos(2*t))which simplifies into:vR(t)=30*sqrt(2)*e^(-2*t)*sin(2*t+pi/4)or the cosine form with ph=-pi/4.We can go over this solution and others if anyone wants to see the entire solution.

author: SteveThackery
author_rank: Super Contributor
timestamp: April 26, 2025, 07:26:21 pm
content: Thanks to everyone explaining the calculations. I need to study them for a while, but I appreciate it very much.

author: Benta
author_rank: Super Contributor
timestamp: April 26, 2025, 08:35:34 pm
content: Quote from: MrAl on April 26, 2025, 04:27:21 pmTry to read what I wrote: "in the time domain". Is that ambiguous?Differential equations do not allow superposition like "adding the C charge curve to the L current curve".Moving to Laplace notation transfers the problem into thefrequencydomain, andthenyou're dealing with normal algebra.That's the whole point of Laplace notation.I did a sim just for fun. It's indeed a damped resonance as expected:

author: mawyatt
author_rank: Super Contributor
timestamp: April 27, 2025, 12:34:42 am
content: Quote from: MrAl on April 26, 2025, 04:35:40 pmYou are correct, it's not critically damped but slightly underdamped. As critical damping is when zeta = 1, not 0.707 as we incorrectly stated.We should remove our post as it's incorrect and potentially confusing.Thanks for pointing this outBest

author: IanB
author_rank: Super Contributor
timestamp: April 27, 2025, 03:03:45 am
content: Quote from: JOJOElectro29 on April 25, 2025, 11:12:27 amThere are two ways to approach a problem like this. One way is to apply standard formulas for the circuit in question (this is a series RLC circuit, which is analyzed in textbooks). Another way, which may promote better understanding, is to analyze it from first principles.If we take the first principles approach, it may look like this. (What follows below will repeat much information given above in the thread, but it may also serve as confirmation of that.)If we follow the current round the circuit when the switch is open, then we can label the voltages across each component as \$v_L\$, \$v_R\$ and \$v_C\$, where the sign of each voltage is in the same sense as the direction of the current. The current itself is already identified as \$i\$, and Kirchhoff's current law says the current must be the same everywhere around the circuit. Then, from first principles, we can write equations for the circuit as follows:Inductor:$$v_L=L\frac{di}{dt}$$Resistor:$$v_R=iR$$Capacitor:$$i=C\frac{dv_C}{dt}$$Kirchhoff's voltage law:$$v_L+v_R+v_C=0$$This gives four equations in four unknowns. The initial condition occurs while the switch is closed, and by inspection this will be as below:$$\begin{align}v_L&=0\,\mathrm{V}\\v_R&=30\,\mathrm{V}\\v_C&=-30\,\mathrm{V}\\i&=15\,\mathrm{A}\end{align}$$The four equations above plus initial conditions could be solved numerically using a suitable tool, or, by appropriate algebraic manipulation and substitutions, they can be reduced to a single second order linear differential equation, which can be solved using textbook techniques.For my own interest, I solved the four equations numerically, and got the following result:

author: MrAl
author_rank: Super Contributor
timestamp: April 27, 2025, 03:27:49 am
content: Quote from: Benta on April 26, 2025, 08:35:34 pmHi there,I am happy you were able to confirm the results of the type of the response.However, it does not matter if you work in the frequency domain or the time domain, and probably even in the quefrency domain (have not needed that yet however), as long as you know how to write the differential equations in the time domain. Superposition works in any system as long as the system is linear. We could look into this more if you like as well as look at an example.I mentioned Laplace Transforms mostly so that we could work with impedances, but that is not necessary really, just seems to make it easier (due to the pure algebra).Also, today you do not even have to understand complex numbers in order to use Laplace techniques, unless you want to do a frequency domain analysis. That would transform 's' into "j*w" of course.I'll post the solution steps in a little while.

author: MrAl
author_rank: Super Contributor
timestamp: April 27, 2025, 03:35:29 am
content: Quote from: mawyatt on April 27, 2025, 12:34:42 amHey there,No problem, and I would expect you to catch errors that I make so I can get a chance to correct them.All the best to you too

author: MrAl
author_rank: Super Contributor
timestamp: April 27, 2025, 03:38:13 am
content: Here is a step-by-step method for solving the original circuit. Follow along steps B through G.I chose to show this method because it makes it easier for someone to solve if they already know how to solve circuits with no initial conditions. Of course there are a multitude of other methods out there.I did not show the math to go with the circuits yet giving the original poster time to think about it.Any questions or corrections please feel free to post.

author: mawyatt
author_rank: Super Contributor
timestamp: April 27, 2025, 02:55:14 pm
content: Quote from: MrAl on April 27, 2025, 03:35:29 amShould have realized a*S^2 + b*S + c, b^2 must be less than 4*a*c for imaginary roots. Thus in normalized S^2 +b*S + 1, 2*zeta must equal b, or 2 and obviously zeta = 1 for all real roots and critical damping !!Also, should have thought a little before posting, my badBest

author: SteveThackery
author_rank: Super Contributor
timestamp: April 27, 2025, 08:26:35 pm
content: @MrAI: thank you for showing the step-by-step method.I'm embarrassed and apologetic for not getting it yet. Steps B to G seem to result in evaluation of the initial conditions in terms of the voltages. How do we take that forward to give the answer at any timet? It appears that is the term at the top of your diagram?

author: IanB
author_rank: Super Contributor
timestamp: April 27, 2025, 10:34:38 pm
content: Quote from: SteveThackery on April 27, 2025, 08:26:35 pmIt seems like we find the time domain solution for steps D and E separately, and then add them together. But I'm not sure how finding those separate solutions is easier than solving the original problem? Is there a trick to it?

author: MrAl
author_rank: Super Contributor
timestamp: April 28, 2025, 12:59:56 pm
content: Quote from: IanB on April 27, 2025, 10:34:38 pmHi,Well it's simpler if a person does not know how to solve circuits with two sources yet. They can solve the circuit twice, once for each source, then simply add the results.As a counter example, if they already know something like Nodal analysis then they don't really have to use superposition. It's just there for anyone who needs itI use it sometimes anyway because it makes it easier to do some circuits in my head, and also get some extra intuition about how the circuit works.On the somewhat comical side: "The circuit analysis method is in the brain of the beholder" (ha ha).

author: MrAl
author_rank: Super Contributor
timestamp: April 28, 2025, 01:31:35 pm
content: Quote from: SteveThackery on April 27, 2025, 08:26:35 pmHi,No need to feel embarrassed it's just the process of learning, possibly a new method? I am not sure what methods you have used in the past but I assumed you can analyze a circuit with just one source in it. If not we can certainly go over that until you understand it as much as possible.If you refer to circuit diagram "C" you will see the circuit has been transformed into a circuit with two regular constant sources. This is useful for solving this in a bunch of different ways and one of them includes using superposition which we can start to do next.Looking at diagram "D", the current source has been disconnected so that we can start to analyze with just the capacitor voltage source. I will use Laplace Transform techniques because that's the easiest. We can use regular differential equations also if you would like to look at that kind of solution.Ok, so starting with diagram "D" we first transform the components into their respective impedances:zR=RzL=s*LzC=1/(s*C)and the voltage source v0 becomes a step voltage source:V0=v0/sNext, we can use the voltage divider formula where we want the voltage across R. That simply means adding up all the impedances for the denominator, and then placing R in the numerator:D=zR+zL+zCN=Rand forming the expression:N/D=R/(zR+zL+zC)and then multiplying by V0:vRas=V0*N/D=V0*R/(zR+zL+zC)which using the original impedance and source transformations gives us:vRas=(v0/s)*R/(R+s*L+1/(s*C))and simplifying that we get:vRas=(v0*C*R)/(s*C*R+s^2*C*L+1)Now at this point you have the choice of transforming that into the time domain first, or wait until later until we have the solution for vRbs also.To transform that into the time domain right away, we would either use a table of Laplace Transforms or use a math software package like Maxima which is free. Math software is easier to use of course. Just before we do that however, if we are not looking for a very general solution we first replace the variables with their actual values:vRas=(30*(1/4)*2)/(s*(1/4)*2+s^2*(1/4)*(1/2)+1)and after simplifying:vRas=120/(s^2+4*s+8)Now all we do is look that up in a table or use math software to transform into the time domain. The Maxima function for that is "ilt(vRas,s,t)" and we get:vRat=60*e^(-2*t)*sin(2*t)That is the time domain response to the initial capacitor voltage.Note that if there was no initial inductor current with this problem that would be the final solution, but because we have that inductor current as well we have to solve that and then later add that to vRat. That would mean we would have solver for vRbt also, and then the total solution would be vRat+vRbt. That way may be a little simpler than adding the two frequency domain solutions and then transforming back into the time domain, but you can try both methods.If that does not make sense yet we can look at this in more detail. The other method is to use differential equations which we often refer to as Ordinary Differential Equations (ODE's) and you can look into that also. The way I like to do it is more like in the form of a State Vector Differential Equation which without using a matrix simplifies the notation quite a bit. We can look into that also, but first see if the above makes sense to you so far.

author: MrAl
author_rank: Super Contributor
timestamp: April 28, 2025, 01:40:45 pm
content: Quote from: mawyatt on April 27, 2025, 02:55:14 pmHi again,Yes, the discriminate makes the difference between the three possible solutions.There's no problem here though as we are all on the same page nowYou may also find it interesting if you never looked at this before, but if you form the symbolic solution for the overdamped case (which would imply certain set values for R, L and C) and then replace those variables with values that end up in the underdamped case, then simplify, you would get the solution for the underdamped case. The sinh() and cosh() forms would change into sin() and cos() forms. It's pretty cool how that happens I think.It works the other way too. If we solve for the solution with forms sin() and cos() and then use values for R, L, and C that really give us the overdamped case, we end up with sinh() and cosh() in the solution.If we use values for R, L, and C that give us the critically damped case, we end up with that even though we started with a symbolic solution for another case.This means we can use either solution as long as we keep it symbolic, and that means we can have one general solution as long as we keep it symbolic. We can then later insert the right values for R, L, and C.The main idea is that then we don't have to keep track of three separate solutions for the three different cases.This would work with any circuit because it's the underlying math at work not the circuit itself.

author: mawyatt
author_rank: Super Contributor
timestamp: April 28, 2025, 02:55:05 pm
content: Quote from: MrAl on April 28, 2025, 01:40:45 pmRelating the mathematical and circuit behavior seems to be a lost art today, most just dive into a simulator for the result rather than thinking about what's going on. Contrary to the usual thinking where an "Engineering Solution" doesn't care about the path to the answer, only the correct answer matters, which leads to a limited understanding of what's behind the answer.We were just somewhat guilty of this by throwing out the answer (no Google or Sim) to this post before thinking about it, relying on memory (which is fading now). Also after your post realized why we threw out the incorrect answer for zeta as critical damping, a little more "thinking" on our part would have shown this is the zeta for a classic Butterworth 2nd Order Filter which is not critically damped but slightly underdamped. Since we deal with filters often, and especially Butterworths (s^2+0.707s+1 for 2nd order and s^3+2s^2+2s+1 for 3rd order) this is what initially came to mind as critically damping!!It's interesting that the Butterworth (zeta of 0.707) is flat (no peaking) in Frequency Domain, but not in Time Domain (has step overshoot), as this is critical damping (zeta = 1).Best

author: SteveThackery
author_rank: Super Contributor
timestamp: April 28, 2025, 11:24:39 pm
content: Once again, thank you, @MrAI.When it comes to electronics I am a hobbyist, so have missed out on a lot of the underlying theory. Nevertheless I'm going to work through your latest post and see how far I get. It is good of you to give so much time.

author: PGPG
author_rank: Frequent Contributor
timestamp: April 28, 2025, 11:59:06 pm
content: I didn't tried to check any analysis done in this thread (I'd like, but now totally have no time to spend it remembering the details) but two of your sentences:Quote from: MrAl on April 28, 2025, 12:59:56 pmQuote from: MrAl on April 28, 2025, 01:40:45 pmmake me want to add the caveat that they are only true for linear circuits.Quote from: MrAl on April 28, 2025, 12:59:56 pmI prefer to say that "The best circuit analysis tool you have between your ears."

author: mawyatt
author_rank: Super Contributor
timestamp: April 29, 2025, 12:11:49 am
content: Quote from: PGPG on April 28, 2025, 11:59:06 pmExactlyBest

author: ledtester
author_rank: Super Contributor
timestamp: April 29, 2025, 11:04:35 pm
content: From the attached figure you can elect to represent an initial inductor and capacitor condition as a series element as well as a parallel element. Therefore it is possible to represent the circuit with both initial conditions as one series circuit requiring you to only solve it once.Figure taken from:https://jazapka.people.ysu.edu/ECEN%202633%20Chapter%2013.pdf

author: MrAl
author_rank: Super Contributor
timestamp: April 30, 2025, 09:55:15 am
content: Quote from: SteveThackery on April 28, 2025, 11:24:39 pmHi again,Oh you're welcome I hope it helps you understand this stuff better.If you are so new to this then maybe you could start off with circuits that only have one resistor and either one inductor or one capacitor. That way you can get used to how to analyze circuits with inductors and capacitors and resistors, which is in a lot of circuits.For example, a resistor in series with an inductor, and a different circuit with a resistor in series with a capacitor (each circuit also with a constant voltage source). That would be just three circuit components per circuit, which would get you started. Say the voltage is 10 volts, then what does the time domain expression look like. Then you can add components like another resistor to each circuit and analyze, then compare and see how things changed. This starts to give you a lot of insight into these circuits.

author: MrAl
author_rank: Super Contributor
timestamp: April 30, 2025, 10:06:28 am
content: Quote from: mawyatt on April 28, 2025, 02:55:05 pmHi,I am not sure how lost it really is though because circuit simulators and not yet advanced enough to derive the total behavior of a circuit they just provide numerical results. A numerical result is different than a symbolic result with a symbolic result being closer to pure theory. Once we have a symbolic result we can examine it for all kinds of behaviors. We could not do that with a numerical result simply because all of the necessary information is not yet present in that. We would have to do many cases and compare the numerical results of every case in order to develop a theory for some things. That's not impossible I guess, but it's not yet automatic, at least not any simulators that I know of.Some time ago this prompted me to create my own 'spice' simulator, but it's not a numerical simulator it is a symbolic simulator. It will provide a result like:V(s)=R/(s^2+A*s+B)or other expressions like that. This allows a different kind of analysis altogether that can reveal important concepts about circuits like RLC circuits and much more complicated ones too. It's great for study. Instead of entering R1=10, L1=100uH, C1=1.2uf, you just enter R1, L1, and C1, and let the program do the rest. You can always replace with numerical values later. Some circuits get really complicated where we might have several resistors and other elements. In the end we get more complicated expressions, but because it's symbolic sometimes it simplifies into much simpler expressions with gives us insight into those circuits. There are cases where the results can run a couple pages though, but then it's still useful because it can be used in a program for a microcontroller that has to calculate a lot of different expressions in real time.In short, it ends up very interesting and useful.

author: MrAl
author_rank: Super Contributor
timestamp: April 30, 2025, 10:07:45 am
content: Quote from: PGPG on April 28, 2025, 11:59:06 pmHi,Didn't I mention the idea about linearity before this? If not, thanks for clarifying.Once a person studies superposition, it necessarily must include the idea of linearity.

author: MrAl
author_rank: Super Contributor
timestamp: April 30, 2025, 10:10:30 am
content: Quote from: ledtester on April 29, 2025, 11:04:35 pmHi there,That looks like a good summary of how to use initial condition generators and also shows the two main types: the impulse type and the stepped type.Nice of you to post that for everyone to check out.