Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thevenin equivalent of a network

  1. Dec 7, 2004 #1

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sorry for such a basic question, but I'm having trouble grasping the idea of why the Thevenin equivalent of a network (as seen from two terminals) almost always has a Norton equivalent in the form in which it has been defined. More specifically, I've been trying to work out from staring at the diagram why an ideal voltage source in series with a resistor is equivalent to an ideal current source in parallel with a resistor. (why specifically in parallel in the the latter case and in series in the former?) Everytime I stare at it, I think I'm close to an answer, but I can't put it into words resoundingly...to make a statment like..."the current source has to be in parallel with the resistor, NOT in series or in any other configuration, in order for blah blah blah such and such to be the same as it is in the Thevenin version!"

    Any help would be appreciated.
     
  2. jcsd
  3. Dec 7, 2004 #2
    Look at the open circuit and short circuit parameters of both the thevenin and norton equivialents. Thevenin with voltage source V and Resistor R in series: It has an open circuit voltage of V and a short circuit current of V/R.

    Now look at Norton equivalent with current source V/R in parallel with Resistor R. The open circuit voltage is V and the short circuit current is V/R.

    The two would no be equivalent if open circuit voltage and short circuit current were not equivalent.
     
  4. May 17, 2005 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sorry to dredge up such an old thread, but there is something I still don't understand. I do understand that the open circuit voltage and closed circuit current are the same, so these "equivalents" represent networks that must be identical, from the point of view of whatever is being connected to the port. My first question, is why are those two things (short circuit current and open circuit voltage) the determining factors for "equivalency?" Also, another thing that confuses the heck out of me is that I get the same results for open circuit voltage and closed circuit current when I have a current source with current V/R connected in series with a resistor of resistance R. So why is that equivalent never mentioned? Or did I make a mistake and is that not equivalent to both the Thevenin and Norton versions?
     
  5. May 17, 2005 #4
    Because the short circuit current and open circuit voltage are the two extreams which you can subject the two ideal sources to ..
    as for your second question , are you saying that an ideal current source // R1+ R2 is equivalent to an ideal voltage in series with R1 + R2..?? they are.
    this is wrong .. because an ideal current source has a current of I , thats it..!!
     
  6. Jun 5, 2005 #5
    Hi cepheid,
    The Norton and Thevenin equivalent circuits are based on the property of linearity, and apply only to circuits that are linear in this sense. Without worrying too much about linearity, the result is that these models have an equivalent impedance (or resistance) which you will have noticed is the same for equivalent models. This is the "output impedance" of the circuit and is the basis of matching it to another circuit for maximum power transfer.

    If you measure the open circuit voltage of a one port circuit (without worrying about what is in it) and then measure the short circuit current (you don't really want to do this, but imagine you did), then you get two points on a V-I plot ... one on the vertical voltage axis and one on the horizontal current axis. The slope of the line that connects these two points is the effective resistance of the circuit (V/I).

    If two circuits are to be equivalent, they must share all three characteristics - open circuit voltage, short circuit current, and effective (or output) resistance. The thevenin model (voltage source in series with resistance) has all three - open circuit voltage, effective resistance and short circuit current (v/R). The Norton model (current source in parallel with resistance) also has all three.

    BUT, a voltage source in parallel with a resistance does NOT have a finite short circuit current. If you short the output terminals you get infinite current and thus zero resistance [tex]\Rightarrow R=\frac{V}{I}==\frac{V}{\infty}=0[/tex]. This is the VI plot for an ideal voltage source, not a real circuit.

    Similarly, a current source in series with a resistance must have an infinite open circuit voltage (imagine holding this circuit in your hand, and watching the continuous lightning from one terminal to the other as the ideal current source kept pumping out electrons). This gives a vertical line on the VI graph, with infinite resistance.

    Therefore, of the four possible combinations, only two represent real circuit models, and the other two are simply ideal sources with resistors that do not affect the operation of the source.

    Hope this helps.
     
  7. Aug 17, 2005 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thanks, that does help somewhat more. But I'm still not clear on a certain number of things (haha...2 months later. What can I say? I was enjoying my summer.) First of all, as far as linearity goes, all of the circuit elements we've looked at so far in class have a linear relationship between current and voltage either described by linear algebraic or linear differential equations. I'm assuming that is what you meant? I'm also assuming your claim is that the Thevenin theorem only applies to a network with such linear elements, for that is a prerequisite for the required simplification to a source in series with an impedance. I can buy that: carrying out the simplification of a resistor network sort of illustrates the point.

    About the output or source impedance: it is what determines how the voltage decreases as current increases (the current actually drawn depends on the load on the circuit, but that is a separate outside factor that determines I in the V-I plot.) So intermediate values on the V-I plot represent different loads?

    Yeah, I'm basically just looking to see if my understanding of what you've been saying is correct. Looking at a real world example helps. The source signal you connect to a voltage amplifier for example. It could be coming from all sorts of circuits. But I can sort of see how if all of them can be reduced to the same Thevenin equivalent, then the amplifier will respond in the same way to all of them, as long as those three parameters, open circuit voltage, short circuit current, and output impedance of the sources are equivalent. Is that the gist of it?
     
  8. Aug 17, 2005 #7
    That's the gist of it.
     
  9. Aug 19, 2005 #8
    Argh. I am trying to figure these two out right now. I have no idea what Im doing, my lecturer isnt the greatest. :(

    Anyone else have trouble learning these at the start? A few guys I know are also struggling with this and its only 4th week in.
     
  10. Aug 20, 2005 #9

    If you sit down and do some brief visualizations it will start to make sense.

    Imagine a current source which is shorted with a resistor R1. As you increase the resistance
    of the short, the voltage across the resistor increases. If you stop
    increasing the resistance, you will have stopped at some voltage [tex]v=ir_1[/tex]

    That V is the same V of the equivalent voltage generator. But this new
    equivalent circuit also has a series resitance included, R2. Why? Because if
    you short this new circuit without the series resistor, you will get infinite
    current. But you need to get a current flowing which is the same I
    as the value of the original current source. This new series resistor R2 does
    that current limiting for you.

    Now imagine I have both circuits on the bench. I bring wires out of each
    circuit. The voltages are the same for both, and there is no current flowing
    out of either one. Then I short the wires, first the current generator circuit
    followed by the voltage generator circuit. I measure the voltage and
    current I flowing through this short. The voltage is zero of course and
    the current is the same I for both circuits.

    It turns out that if I connect any values of resistance to either circuit,
    the voltages and currents supplied by the two circuits will be the same.

    This is what it means for them to be equivalent.
     
    Last edited: Aug 20, 2005
  11. Aug 20, 2005 #10
    Great post Antiphon, I understood a bit of that. :) Ill try reading it again later, maybe Ill absorb more.
     
  12. Aug 22, 2005 #11
    ohh..that's our topic for this week! I'll post again when I learn the basic concepts..
     
  13. Aug 23, 2005 #12
    Waht textbook do you use? Is the textbook poor? Usually, EE texts are surprisingly god at explaining these concepts.
     
  14. Aug 25, 2005 #13
    "EE texts are surprisingly god at explaining these concepts." LOL
     
  15. Aug 26, 2005 #14
    c'mon! its just a typo error...hehe.. we've dicussed the norton and thevenin last week..we've not yet discussed norton and thevenin with dependent sources..how do you solve that?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thevenin equivalent of a network
  1. Thevenin Equivallent (Replies: 7)

Loading...