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I've a hard time understanding Thevenin's theorem.

I know that it is just a theorem to simplify a complex circuit, and to easily analyse that. My hard understanding is in the mathematics of it.

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Given this circuit, we'd take out the R2, because it is the load resistance, right?(Can you explain me why we need to take it out?)

Also, a load resistence just a load, which is just the main thing of the circuit, that consumes the power, right? Is it the core subject, that the circuit is built for to power?

Now, to get the Vthev, we need to calculate the Rtotal, and here we'd calculate it if it were parallel, but here it is clearly in series, because we took out R2, and and there is no current in that branch anymore, so why we need to calculate it as parallel?

Also, when we need to get the Rthev, why we need to short the voltage sources, and why we don't need the R2?

If I imagine a black box: enter image description here Does the Rthev basically just the overall resistance of the blackbox between terminal A and B? Therefore this is why we doesn't include the Rload when calculating the Vthev and Rthev?

I don't know if my questions are good, because I'm really confused about it, but can you explain me in simple steps?


EDIT

enter image description here

In this example the solutions are: VTHEV = 3 V, RTHEV = 60 Ω, INORT = 0.05 A

Rthev = 1/(1/100+1/150): Why we ignore the parallel 200Ohms?

Vthev = (5/250)*150: Again why we ignore the 200Ohms, and here why we calculate the resistance as series and(250Ohms) and not as parallel, like in the case of Rthev(60Ohms)

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  • \$\begingroup\$ You can see that if you turn the 5-V source off and replace it by a short circuit in the schematic diagram, the 200-\$\Omega\$ naturally disappears. \$\endgroup\$ – Verbal Kint Jun 16 at 17:17
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Thévenin applies to linear circuits and with passive elements as shown in your example, you can obtain the output resistance by turning off all sources while "looking" through the connecting terminals across which you want \$R_{th}\$:

enter image description here

A 0-V voltage source is replaced by a short circuit while a 0-A current source is open-circuited. By inspecting the right-side picture, the output resistance is immediate:

enter image description here

For the voltage \$V_{th}\$, keep all the network as it is and apply superposition: alternately turn a voltage source off and calculate the intermediate voltage across \$R_2\$ then sum both results to obtain the Thévenin voltage:

enter image description here

A quick SPICE simulation tells us if this is ok: a dc operating point calculation followed by a .TF directive confirm we are good to go.

enter image description here

Now, regarding the question about the load: keep it or don't keep it? Thévenin offers a means to model a given linear circuit by a black box characterized by a voltage source \$V_{th}\$ affected by an output resistance \$R_{th}\$. In your case, if you keep \$R_2\$ in place, then you have the below box:

enter image description here

But you could also decide to consider \$R_2\$ as a load and disconnect it for your analysis. In this case, the equivalent box would be:

enter image description here

It provides the insight that \$R_2\$ is driven with a source affected by a 0-8\$\Omega\$ resistance.

What practical usage can you make out this mumbo jumbo? Well, look at the below circuit and try to determine the transfer function:

enter image description here

You apply Thévenin at resistance \$R_2\$ and reduce the circuit to a simple more familiar \$RC\$ circuit whose capacitor is driven by \$R_{th}\$. The transfer function is immediate:

enter image description here

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  • \$\begingroup\$ Oh, my god! Thanks for the clear explanation, now I understand it. :D I still have a hard time to think through it, but atleast now I understand how it works. In real life, I'd need to calculate these stuff, or is there any simulation program, which do it automatically? \$\endgroup\$ – kisroby Jun 16 at 12:25
  • \$\begingroup\$ It depends on what you look for. If you want a symbolic expression to see how a component tolerance affects the result, then you'll have to derive the expression. SPICE and Mathcad (or any other solver) are useful to confirm your results. \$\endgroup\$ – Verbal Kint Jun 16 at 17:19
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Thevenin's theory states that for a circuit consisting purely of fixed sources we can replace part of the circuit with a voltage source \$ (V_{th}) \$ and a series impedance \$ (R_{th}) \$ to have the same result.

To do this we first remove our load and calculate \$ (V_{th}) \$ as the open circuit voltage you would see across the load. Next we replace any sources with their output impedance (short circuit for voltage source, open circuit for current source) and calculate the impedance you would measure.

Taking your circuit as an example

$$V_{th} = 7 + (28-7) \cdot \dfrac{1}{4+1} = 11.2 \text{ volt }$$

$$ R_{th} = 4 || 1 = \dfrac{4 \times 1}{4+1} = 0.8 \text{ ohm}$$

We can now put our load back into the simplified circuit and from the loads point of view the circuit is equivalent.

The best way to verify this for yourself is to assume you have an output voltage between 0V and 11.2V (depending on load). Calculate the current in the load resistor for each voltage and the resistor value. Do this for both the original circuit and the Thevenin equivalent. Both sets of answers should be identical.


For a voltage source you replace it with a short because that is its output impedance. You can draw as much current as you want and the voltage does not change.

For a current source you replace it with an open circuit because that is its output impedance. Current does not change irrespective of voltage.

If you don't remove the load then you may not have simplified the circuit depending on how you create \$ V_{th} \$.

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  • \$\begingroup\$ But why we need to short the sources and take out the load? I edited my post, can u explain my new questions, please? Thanks for spending time for this! \$\endgroup\$ – kisroby Jun 16 at 11:52
  • \$\begingroup\$ @kisroby extended my answer. Hope this helps. \$\endgroup\$ – Warren Hill Jun 16 at 12:14
  • \$\begingroup\$ Thank you, @Warren Hill! I understand it now. \$\endgroup\$ – kisroby Jun 16 at 12:37
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The theorem uses the concept behind the LOAD LINE.

Plot a current_voltage diagram; horizontal is voltage axis with 0/5/10/15/20/25/30 volt markings; vertical is current axis with 0/1/2/3 .. 9/10 ampere markings.

  1. mark the voltage with 28 volts

  2. mark the current at 7 amps

  3. plot a 4 ohm resistor load line --- a straight line between 28 volts and 7 amps

  4. the Thevenin model is the 28 volt point

  5. the Norton model is the 7 amp point

  6. you may use either model in your circuit

  7. by picking the (7 amp || 4 ohms) point, you now have a parallel equivalent you can insert into your original circuit.

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You should verify the open-circuit Thevenin model is equivalent to the short-circuit Norton model, for ALL loads possible.

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I've a hard time understanding Thevenin's theorem.

enter image description here

And, basically what underpins Thevenin is the conversion of voltage sources to current sources like this: -

enter image description here

Can you see that R1 and R3 are in parallel yielding a single resistor of 0.8 ohms and that the two current sources are in parallel yielding a single current source of 14 amps: -

enter image description here

If you can then, you can back-convert that new current source (14 amps) in parallel with 0.8 ohms to the equivalent voltage source and Thevenin resistance i.e. 11.2 volts in series with 0.8 ohms.

Given this circuit, we'd take out the R2, because it is the load resistance, right?(Can you explain me why we need to take it out?)

I can't justify why it needs to be taken out - I've not taken it out in my solution so maybe you can justify why you think it should be taken out?

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  • \$\begingroup\$ Thank you for your answer! Many examples I've seen show that you need to take out the Rload. Thus, open circuit it. I don't know why. Also, in real life, can you tell me how much would I use Thevenin's and Norton's theories? Is it fundamental to know the maths, or is it enough to know the theories? \$\endgroup\$ – kisroby Jun 16 at 11:00
  • \$\begingroup\$ It crops up now and then and I just go back to basics - the rule is simplify then simplify some more and use the conversions I did above. That's all I ever use. Just get used to converting from voltage source to current sources (and vice versa) then paralleling the elements (simplification). \$\endgroup\$ – Andy aka Jun 16 at 11:04
  • \$\begingroup\$ @kisroby Andy's solution does not require you to remove R2 because he shorts it out to find the Norton equivalent. There is no current in R2 so it makes no difference if its in or out. He then goes back to a Thevenin equivalent. My solution doesn't short out the load so does require you to remove the load. Both solutions are equally valid. You should verify Andy's intermediate Norton equivalent, My Thevenin equivalent and the original circuit all give the same output voltage and load current for a range of loads. \$\endgroup\$ – Warren Hill Jun 16 at 11:31
  • \$\begingroup\$ @WarrenHill I didn't short R2 out - I did a direct conversion from a voltage source in series with a resistor to a current source in parallel with that same resistance. No shorts were harmed (or created) during the process. If you wished to take baby steps through that process then shorting out R2 is a possible baby-step but nobody who uses this technique more than just seldomly, realistically needs to take that particular baby step. \$\endgroup\$ – Andy aka Jun 16 at 11:40
  • \$\begingroup\$ You did in effect short out the load to find the currents \$ \dfrac{7 \text{ V} }{1 \text{ ohm}} = 7 \text{ amp} \$, etc. It is a baby step but one you need to take when first learning. \$\endgroup\$ – Warren Hill Jun 16 at 12:05

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