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Calculating Thevenin Equivalent Doesn't Seem Practical

  1. Nov 13, 2015 #1
    Ok, so I know the point of the Thevenin is to make a simpler circuit so that that it results in the same current being sent to the other connected circuit. But why do we go through all the hassle of computing the Thevenin Voltage when we could choose any arbitary Voltage and a Resistor that would result in the same current?

    For example: http://imgur.com/YOHCMr5

    On the left circuit, the circuit is open on one end, so only the current goes through the R1 and R2 resistors. If you calculate the current, it is 2.67mA. The voltage drop across the R2 resistor is then 2.67V. Ok, so we use 2.67V as the Thevenin Voltage.

    Now, we connect the open part of the circuit (still left hand figure), and calculate the current through R3 using KVL. The current through R3 is 1.6mA. If you recalculate it now, the current through R2 has changed: it is now 1.6mA, and the voltage drop across R2 is now 1.6V. But when the circuit was OPEN, the voltage drop across R2 was 2.67V <-- this was the Thevenin Voltage

    So, why do we calculate Thevenin Voltage when the circuit is open (according to left diagram)? Why don't we use the voltage across R2 (1.6V) as Thevenin when everything is connected? You can still get the same current of 1.6mA by calculating the right resistor value.

    If we use 2.67V as Thevenin Voltage, then to get 1.6mA current, Thevenin Resistance is 1.67kΩ.
    If we use 1.6V as "Thevenin Voltage", then to get 1.6mA current, "Thevenin Resistance" is about 1kΩ.

    As you can see, if the goal of Thevenin Equivalent circuit is to get the right resulting current of 1.6mA, why can't we just use ANY arbitrary voltage, and calculate the right resistance and call those Thevenin Voltage/Resistance?
  2. jcsd
  3. Nov 13, 2015 #2


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    The goal of the thevenin equiv. is to duplicate the behavior over all loads from short to open.
  4. Nov 13, 2015 #3
    Doesn't that just mean ensuring the right current is being passed through? You can do this with an infinite combination of Voltage and Resistance values.
    What is it exactly about calculating the Thevenin Equivalent Voltage when the circuit is open that is so special?
  5. Nov 13, 2015 #4


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    read this
    https://en.wikipedia.org/wiki/Thévenin's_theorem[/PLAIN] [Broken]

    You have to design a 2 component circuit that duplicates the behavior of the original circuit AT ALL LOADS without using different components at different loads. SO it must have the same open circuit voltage and the same short circuit current as the original circuit.

    If you sealed the original circuit in one black box and the thevenin in another, you could not detect any difference at the 2 terminals FOR ANY ARBITRARY LOAD. They will track exactly.
    Last edited by a moderator: May 7, 2017
  6. Nov 13, 2015 #5

    Yes, but let's assume Thevenin Resistance is R and the current we want is I. We will not use different components at different loads, meaning R is always the same.
    So to satisfy this situation, our voltage must be V=IR. Even if R is constant, for any current I we want, we can calculate the required load V and modify the power supply accordingly. No components have been changed.

    Still, it doesn't explain why Thevenin Voltage is the way to do things. Why can't we can't use any arbitary Voltage and Resistance to obtain the right current.
    Last edited by a moderator: May 7, 2017
  7. Nov 13, 2015 #6


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    Why do you think you can compute a different voltage for every load? Those are not thevenin equiv's that you are computing. They are just some arbitrary circuit that matches at that single load. It is meaningless.

    You are totally missing the whole point.

    The thevenin equiv. circuit gives the correct current and voltage for ALL LOADS, all the time, without changing the two components (V and R).
    Please read that last sentence over and over until it changes your mind.

    Then go back and read post 2.
  8. Nov 13, 2015 #7
    thanks, i understand what you mean now. i was able to show myself that the thevenin voltage, which is calculated when the circuit is open, is just for convenience because it allows Thevenin Resistance to not be dependent on the test load.
  9. Nov 13, 2015 #8


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    You still don't quite get it.
    Some arbitrary resistor that you compute to match currents from an arbitrary voltage has nothing to do with Thevenin's Theorem and cannot be called a "Thevenin resistor".

    If the goal is to create a 2 port network that indistinguishable from the original network from the viewpoint of the network terminals (called a thevenin equivalent network) , it is not "just for convienience" to use the open circuit voltage. It is by definition.

    There is a thevenin-equivalent resistor, and a thevenin-equivalent voltage within the network. BOTH are required for THE thevenin-equivalent network. If you want to isolate or separate or consider them independently for some reason, feel free to so so. You can even call them "thevenin resistor" and "thevenin voltage" for short and people will understand that you mean the Req and Veq from the thevenin equivalent network. There is only 1 possible thevenin-equivalent resistor and only 1 possible thevenin-equivalent voltage.
  10. Nov 13, 2015 #9


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    Yeah. What he said !
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