I am wondering if this is possible.
The Thevenin voltage is the open circuit voltage across the terminals.
How is VS measured?
VS is 40, given, i just need to know if its possible or not
You did not answer Simon's question. He asked the question for a reason...
And is your source impedance reactive or resistive...? Is your load impedance reactive or resistive?
So VS is just some "given value" for some arbitrary letters?
Note VS cannot be "40" ... it has to have some units :)
I know that sounds nit-picky but I think it is related to your understanding of voltage.
Even a "given voltage" had to have been measured at some place some time. Maybe it is measured across the terminals of the PSU, or maybe it is measured between the VS and GND rails on the circuit board? What it means tells you it's relationship to the Thevenin voltage and that will tell you the answer to your question.
What are the circumstances in which VS may be given?
and thanks berkeman - sometimes there is a misunderstanding of what we do here. The idea is that the person posing the question figures out the answer with our guidance. This approach maximizes learning for the questioner and for the person answering at the same time. Otherwise why should we bother? It's not like we are getting paid for this.
40 angle 0 volts,
what do you mean how is VS measured?
it is an ac circuit, so I am guessing its impedance,
I didn't answer your question because I don't know how to answer them, I am just a beginner, vs is provided through a question, and my answer for calculating VTH is higher than the source, I think you are assuming I'm doing something else, its just a general question..i don't understand your question lol
Hi beginner. Attach a pic of your circuit. Or if it is very simple, describe in words.
Well, if you don't know what VS is then that will be a problem.
If you are only guessing if the source is impedence reactive or resistive, that is a problem too.
In your question you must have a circuit diagram with an arrow on it labelled "VS".
That is the place where VS is measured in your case.
In general, VS can be anything - perhaps the open-circuit voltage across a battery or the supply rail voltage to an Op-Amp... anything.
When you understand te role that VS plays in the circuit and how that can turn into a V-Thevenin, you will be able to answer your own question.
Since it is a general question, you should be able to answer generally - in a circuit diagram, where can VS be? What sort of thing has a "VS" value? What role does that play in the circuit? How does the VS usually factor in to an arbitrary Thevenin equivalent circuit?
I have a suspicion that you don't understand what the Thevenin equivalent actually is, or what the various voltages labelled in your circuits are for... instead you are applying formulas and equations and hoping you've done it right. That's what the question is for: so you see it is general, relevant, and important.
Probably the best way forward is to do an example - your example is handy :)
Alright, let's take this slowly, I will try to respond to every sentence you wrote. I will tell you what I know which may or may not be wrong.
VS is the voltage source, which is where power comes from.
I don't know how to tell if the source is impedance reactive or resistive. Can you explain to me in simple words? I am very slow, so it takes me a while to actually understand it intuitively.
So you tell me VS is measured through the circuit diagram, which is what you wanted to know, I still don't understand how this relates to "Can the voltage thevenin be higher than VS?"
Yes, VS can be anything, since its a source. I don't understand how VS plays in the circuit other than it is the source, so I don't think it is possible for VTH to be higher(my initial thought, but I ask the question hoping to get some insight on this).
Where can VS be?
Anywhere in the circuit? lol
What sort of thing has a "VS" value?
Batteries, PSU, etc.
What role does that play in the circuit?
It supplies voltages to the circuit
How does the VS usually factor in to an arbitrary Thevenin equivalent circuit?
I don't know, which is why I ask my original question to see if I can figure out some of the relationships between the two. All I know is that Thevenin is a simplified circuit, and when you want to add loads, Thevenin makes it easier to calculate.
I actually don't understand what Thevenin equivalent actually is, again, which is why I ask my original question hoping to gain some intuitive insight on this.
The various voltages labelled in the circuit(I am assuming you mean on the different elements), are voltage drops, which means when voltage passes through an element, the passing voltage is divided into the elements, which then the sum is equal to the voltage source.
And no, I don't "see", because you don't explain anything,all you did was just ask me questions to things I don't understand, I'm not sure if you are assuming I know these things or what but, to me, it just feels like your asking questions that I am trying to figure out as well. So you see, I am very slow, berkeman should know this since he answer one of my other questions lol.
Yup, still clueless.
EDIT: NASCENTOXYGEN, I don't have the circuit anymore, it was just a random circuit I was practicing on. But I just remember getting VTH higher than VS. It was an AC circuit.
There could be more than one voltage source in the circuit though. There could also be current sources. There could be op-amps ... all kinds of things.
Note: not all voltage sources are ideal - the battery, for eg, would be modelled as an ideal voltage source in series with a resistance. More complicated non-ideal voltage sources may have a small reactance as well as a resistance there. Thus: resistive or reactive source.
But if you google "reactive source", you'll see other definitions.
Anyway - I think you are starting to get the idea about the role played by the voltage source.
Not all physics questions are hard, and, sometimes, the extremely obvious can yield insights. Don't neglect to see what is in front of you ;)
In a linear network with a single ideal voltage source - which I suspect is what you had - can you set one up so the open-circuit voltage between the terminals is higher than the only supply voltage? Wouldn't it mean that a charge carrier could arrive at terminal A with an energy higher that what could be supplied by the actual power supply?
The only thing I got out of that is voltage sources may not be ideal, nvm, I give up, what ever comes out of you does not help me at all.
I still know the same amount of things from the beginning of my post, so no, I did not "start to get", I said that in my previous post. I'm starting to wonder if we are even on the same wave here.
Apparently simple words is to much to ask for.
Guess this forum isn't meant for beginners, might as well talk to a bowl of sliced potatoes.
Thanks for trying though, thumbs up for your effort.
You were starting to say that the role of the voltage source was to supply energy.
Of course you still know as much as you started out with: that is sufficient. As you explore, you also come to understand what you know.
t is this understanding that is holding you back.
The questions I have been asking you are about understanding the work you are trying to do ... namely: figure out Thevenin equivalent circuits. At that level, a simple yes or no will not serve you even though you are sure it is what you want. Apart from that - the words have been quite simple and you'll find all of them in your course notes and in whatever other references you are using so it is fair to expect you to be familiar with them.
One of the beautiful things about providing help for free is that you get to give people the help they need even when it is not what they ask for. To get the help you ask for instead of the help you need - you usually have to pay for that ... one way or another.
Sorry to see you are giving up - maybe another forum will be more suited to you.
You are getting responses and patient help from some of the best folks on the Internet forums here, so please don't get too frustrated, and know that you are getting very good quality help.
The short answer to your question is no. If the source impedance and the load impedance are resistive and not reactive (no inductors or capacitors), then there is no way to get a load AC voltage higher than the source AC voltage. If there are reactive components in the circuit (like an LC resonant "tank" circuit), then the voltage source can help build up a larger resonant voltage on the reactive load circuit.
We understand that you are a beginner and trying to learn. We are just trying to ask for more information to try to give you an accurate answer. If you can post the simulation that you did (or the hand calculations that you did), it would help us to help your understanding.
Separate names with a comma.