# Thevenin Equivalent

Discussion in 'Homework Help' started by RoKr93, Jul 31, 2013.

1. ### RoKr93 Thread Starter New Member

Jun 17, 2013
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I'm studying for my final exam and going over old problems. Something doesn't quite sit well with me about this one; did I redraw it wrong?

2. ### joeyd999 AAC Fanatic!

Jun 6, 2011
3,111
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The redrawn circuit is not equivalent. Further, there is no need to "reduce" the circuit in the manner you have shown.

Remember that the Thevenin equivalent voltage is the open circuit voltage (from nodes A to B), and the resistance is the open circuit voltage divided by the short circuit current.

3. ### wayneh Expert

Sep 9, 2010
13,625
4,416
Look at your first drawing. It's not equivalent.

4. ### WBahn Moderator

Mar 31, 2012
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Always ask if the answer makes sense, which you are admittedly trying to do even to be saying that something isn't sitting well with you. The next step is to try to identify what doesn't sit well with you, so that you can focus on either finding out why what you have done is wrong or why you thought it didn't sit well was wrong.

In this case, consider the two 20Ω resistors. In the original circuit the are connected together at node A, but the other ends are NOT connected together and one of them is connected to the negative side of the source and the other is connected to the positive side. But in your first redraw circuit, the are connected together at both ends and only connect to the negative side of the battery. Why did you choose the negative side? Is someone had redrawn the circuit exactly as you did exempt had the source turned around, what would you have said to them to convince them that yours what right and theies' was wrong?

One way to determine if two circuits are equivalent is to ask what that various paths are to get from one node to another node. Another way is to put one finger on a node in one circuit and another finger on the corresponding node in the other circuit and see if that node connects to exactly the same components in both circuits, making allowance for substitutions such as replacing two series components with an equivalent component. In this case, you can draw a box around the components that are being combined and think of it as one component.

Any time you redraw or simplify a circuit, it is good to get in the habit of doing these checks because this is a common place for mistakes to creep in and a hard place to track them down to.

In your circuit redrawn circuit, consider the paths from A to B. The paths branch at

5. ### RoKr93 Thread Starter New Member

Jun 17, 2013
24
0
I should have been more specific, WBahn, but that was precisely what didn't sit well with me- after looking at it for a bit I realized I'd assigned the ends of the voltage source pretty much arbitrarily.

I took another look at this problem today and I think it clicked for me- I ended up using mesh analysis around the three loops for I_sc=0/V_ab=V_oc and I_a=I_sc, V_ab = 0 and found V_oc = 3 V, I_sc = 0.15 A, and R_th = 20 ohms.

Seems correct to me.

6. ### LDC3 Active Member

Apr 27, 2013
920
161
I calculated that VAB is -3V. Is that correct?

Edit: Sorry, I had the resistors reversed in my mind.

7. ### WBahn Moderator

Mar 31, 2012
20,237
5,758
Your answer is correct. You went about it a long way, but any way that works is fine.

Below is a redrawn circuit. I've defined the negative terminal of the source to be ground, just so that it is easier to talk about Va and Vb separately.

From this view, is it obvious by inspection that Va=9V and Vb=6V, so Vab=3V.

By inspection, Rth=(20Ω||20Ω)+(30Ω||15Ω)

The first of these is obviously 10Ω. The second is (30Ω)(15Ω)/(45Ω)=10Ω.

So Rth=20Ω

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8. ### chri\$topher New Member

Aug 3, 2013
1
0
Hi,

Regarding the voltage 18V. that's E or V ? Because if I make this operation: E = V+RI, I will got Vth. Because E is the value of the Thevenin's voltage source. Am I right ?
(I try to do this exercise).

9. ### WBahn Moderator

Mar 31, 2012
20,237
5,758
Where does E=V+RI come from?

The Thevenin voltage is simply the open-circuit voltage between whatever two nodes consitute your terminals.

10. ### screen1988 Member

Mar 7, 2013
310
3
There are two ways to calculate Rth.
1. Thevenin resistance R is the resistance seen at AB with all voltage sources replaced by short circuits and all current sources replaced by open circuits.
2. Rth = Uoc/Isc
Where: Uoc : open circuit voltage
Isc : short circuit current

Are there ways to prove these statements and prove that two ways are equivalent?

11. ### WBahn Moderator

Mar 31, 2012
20,237
5,758
Sure. Show that both approaches yield the same result for an arbitrary linear circuit.

12. ### screen1988 Member

Mar 7, 2013
310
3
I have tried with many specific circuits and they give the same result. However, is there a general way to prove it?

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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One needs to be careful when applying method 1 if there are dependent sources in the mix.

14. ### WBahn Moderator

Mar 31, 2012
20,237
5,758
Yeah, I was sidestepping that issue, at least for now.

The equivalent resistance is the resistance seen at the terminals of interest when all independent sources are turned to zero (voltage sources replaced with shorts to enforce zero voltage and current sources replaced by opens to enforce zero current). But dependent sources remain, which makes finding the resistance a bit tricky. The normal approach here is to apply a test voltage (or current) to the terminals of interest and calculate the resulting current (or voltage) and then take the ratio of voltage to current to compute the effective resistance (or impedance, more generally).

15. ### WBahn Moderator

Mar 31, 2012
20,237
5,758
Normally you do it in a few steps. The first step is to show that, for any linear circuit, that the voltage and current at any particular point is a linear superposition of the voltages and currents due to each of the independent supplies taken separately.

This means that if you pick two nodes as your "output" terminals and you place an independent source (let's just use a voltage source) across it and treat it as just one more independent source in the circuit that the voltage across it is just whatever voltage it is set to and the current through it is a linear combination of the voltage across it and the linear combination of the contributions of all of the other independent sources. Since all of the other independent sources are static (let's keep it simple and assume DC sources, at least for now), the current at the output will be of the form:

Iout = Isc - Gout * Vout

Where Isc is the contribution due to all of the other independent sources which will be the total output current when the source across the output, Vout, is set to zero and so we call is the short-circuit output current. Gout is simply the coefficient that comes out of the analysis of the circuit and it has units of conductance.

By setting Iout equal to zero, we discover that the voltage that the supply at the output needs to be set to in order to produce zero output current (i.e., effectively create an open circuit at the output) is

Vout = Voc = Isc/Gout

We then note that these equations are identical to the equations we would have if our entire circuit, except the test source we added at the output terminals, were replaced with either a single independent voltage source in series with a resistance equal to 1/Gout or a single independent current source in parallel with a resistance equal to 1/Gout.

This pretty well covers your Method 2. For Method 1, you need to look at the form of the expression for Gout and how it relates to a circuit in which all of the independent sources have been set to zero.