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Superposition Theorem (basic query)

  1. Jul 25, 2016 #1
    This relates to a homework question which I have spent considerable time on and although I understand the concepts, the process of getting to the answer is difficult because of several different 'versions' of the right answer I see.

    The relevant threads are https://www.physicsforums.com/threads/thevenins-theorem.775385/ and https://www.physicsforums.com/threads/superposition-theorem.789030/#post-4955422

    My textbooks (for Thevenin) demonstrate that we would find the open terminal voltage (where the load connects), and then calculate the current through the load. That's clear.

    In the second part of the question, we are asked to use superposition to prove that the algebraic sum of the current through the load is equal to that supplied by the individual voltage sources.

    Looking at the diagram in the above thread, when we remove either V1 or V2 then we create a parallel combination with the load and one of the inductors. However, I don't believe that we should be calculating the Thevenin voltage with the load attached.

    So where we might use the voltage divider formula to calculate the voltage at the junction of the two inductors, the value of the load has also been included in the calculations that I have seen.

    Can someone help me out please?
  2. jcsd
  3. Jul 25, 2016 #2
    It has just occurred to me that perhaps I should not be thinking in terms of Thevenin (and in terms of the unloaded terminals), and just see it for what it is in relation tot he question. i.e. what is the current through the load with V1, and what is it with just V2?

    Couldn't see how to edit my previous post.
  4. Jul 25, 2016 #3


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    Right. You can use Thevenin's theorem for that (if asked to) but generally, it can simply be done with KCL, KVL and current/voltage division.
  5. Jul 25, 2016 #4
    I still can't quite figure out the answer. Do we calculate the Thevenin equivalent impedance (series resistor constant voltage source) as a way to model the circuit only?

    The definitive question I have is once we have the Thevenin equivalent voltage at the load terminals, it is simply I=V / Z to calculate the load current? I'm trying to figure out if the Thevenin equivalent impedance plays any part in the load current.

    In the circuit which I linked to above, I'm not sure if the Thevenin (j2.4 Ohms) equivalent series impedance should be added in series with the load resistor. I have a feeling not because there is not a obvious (theoretical way) to do this when comparing for load currents when using the superposition theorem.

    Hopefully I've explained the question clearly!
  6. Jul 25, 2016 #5


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    You can solve the above problem using both Thevenin and Superposition. If you want to use Thevenin's theorem, remove the load and calculate Vth using KVL. Now, short the voltage sources (and open the current sources, if any) and get the Thevenin impedance Zth between the load terminals. Once you have Vth and Zth, draw a circuit with Zth in series with the load, connected across a source Vth.
    Load current will be Vth/(Zth+Zload).
  7. Jul 25, 2016 #6


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    Zth is added to Zload for calculating the load current. In Thevenin's theorem, you compute Vth means you compute the voltage across the load terminals due the rest of the network (sources and impedances). Zth is the impedance of the circuit viewed from the load terminals. So a voltage source Vth in series with impedance Zth is your Thevenin equivalent circuit viewed from the load terminals.
    download (1).png
    To get the load current, put the load back between the load terminals (A and B in the above diagram) so that Zth and ZL are in series and I=Vth/(Zth+ZL).
  8. Jul 25, 2016 #7
    Thank you for the response.

    If I understand correctly then, ZTh has been calculated and incorporated into VTh ; my VTh agrees with that of other people. However, if I then add ZTh in series when using Superposition, I get a subtly different answer. If I remove ZTh from in series with RL, I get the correct answer.

    Do you think I should make a separate post in the homework help forum?
    Last edited: Jul 25, 2016
  9. Jul 25, 2016 #8


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    You have to use the original voltage sources for superposition theorem, not the Thevenin voltage. Thevenin voltage considers the effect of both the sources at the same time. In superposition method, you consider one source at a time and then add the responses of the individual sources.
    That would be better. Use the HW template and show your attempt in detail. Or you can request a mentor to move this thread into the HW forum.
  10. Jul 25, 2016 #9
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