Help with thevenin problem

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WBahn

Joined Mar 31, 2012
29,976
Using the TS's image, here's how I get Rth:

View attachment 111864

Removing Rl, replacing the voltage sources with shorts and starting from the left: I have 2Ω in parallel with 4Ω; then 4Ω in series; then that's in parallel with 2Ω and finally that's in series with 4Ω. It can be shown as (2Ω||4Ω+4Ω)||2Ω+4Ω; I get 60/11 Ω for this.
I got the same expression but I got a value of 4-2/3 Ω. But I did it in my head and now realize that I made a mental math error (namely thinking that 8/6 is 3/2 and not 4/3). So I agree that it is 60/11 Ω.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

When i read the problem in 'red' in my post #9 i figured that he only had to use Thevenin/Norton for the final source and resistance, and nodal seemed like a good choice for finding the other voltages because it did not seem likely that he woudl have to use T/N for everything. It could be so however, so we just have to WTHB (wait to hear back) from the OP.

BTW, i got a different result for Rth then the rest here, i got 600/110.
 

Thread Starter

babyboang boang

Joined Sep 12, 2016
18
Hi,

No there are more methods, nodal is one of them.

The example i gave you for the node at v2 was:
(v1-v2)/R1+(v4-v2)/R3=(v2-v3)/R2

and the equation for the node at v4 would be:
(v2-v4)/R3+(v6-v4)/R5=(v4-v5)/R4

and if you already knew what voltage v6 was, then you just solve those two equations simultaneously for voltages v2 and v4. There are various methods for doing that too.
If you dont know what v6 is yet, then just write one more equation for that node and then solve the three equations simultaneously for v2, v4, and v6.

If this is very unfamiliar to you then perhaps you should learn nodal first. You'll probably have to look that up.
thank you need to learn nodal .
 

Thread Starter

babyboang boang

Joined Sep 12, 2016
18
From what you have said, your class has not studied nodal analysis yet, but you have studied Thevenin equivalents and superposition. Then you should find all the voltages by using superposition - that is by adding the 3 voltages you get across each resistor due to each of the 3 voltage sources in succession. Finding the Thevinin equivalent circuit is a separate problem.

I did not get the same Thevinin values that you got so I think you did it wrong and your notation is impossible to follow so I don't have any idea what you did wrong. Edit: correction, see my post below

To get a Thevenin equivalent circuit you can simplify the circuit in about 8 steps, starting from the left side and moving to the right. You should redraw the circuit after each simplification, labeling the resistance and voltages as you go.
i dont know if that possible, ... i think your using norton theorem? am using thevenin theorem to get the voltage RL and its easy to get all voltage if i will replace RL and GET again all voltage across the circuits > > and i dont know to get all voltage after i get the voltage at RL
After your first step the 20V source and 2 ohm resistor in series will be replaced by a 10A current source and 2 ohm resistor in parallel. After your second step (combining the 2 ohm resistor with the 4 ohm in parallel) you will have a 10A current source in parallel with an 8/6 ohm resistor. Next find the equivalent voltage source and thevinin series resistance....... continue creating Thevenin equivalents and combining resistances and voltages where possible as you work, step by step, toward the right hand side. Show your work.
 

Thread Starter

babyboang boang

Joined Sep 12, 2016
18
I got the same Rth.

Actually, I now see that babyboang mentions the correct (my values at least) Rth and Vth in post #3, but in the confusing attachment is where I saw that Vth (-0.364 V) was wrong. So I guess the OP did get the correct values afterall.

So, babyboang, now find all the voltages using superposition.
i stated that the voltage of ,RL voltage is 3.24324V .. on the chat not on the attachment . help
 

Thread Starter

babyboang boang

Joined Sep 12, 2016
18
Hi,

Poor Ed gets overlooked again. Sorry Mr. Norton, but we honor your memory here now :)

To be perfectly succinct, this circuit can not be solved in that manner using only Thevenin. In fact, doing it from left to right means the very first step uses Norton's theory not Thevenin. Too bad these two often get lumped into Thevenin so he gets all the credit and Norton almost none unless someone happens to mention him.

To be clear, Norton basically says that a resistance in series with a voltage source is equivalent to a current source in parallel with a resistance.
Thevenin goes the other way. A current source in parallel with a resistance is equivalent to a voltage source in series with a resistance.

Although they seem the same, they are not exactly the same they are called "duals", and circuits in general may have entire duals that replace current sources with voltage sources and vice versa, and resistances replaced with conductances and vice versa.

The above is not the original wording however, but it boils down to that in it's simplicity. There could be more than one resistor involved.
i solve the RL voltage ,RL voltage is 3.24324V and i dont know where to start after i got this ?
 

Thread Starter

babyboang boang

Joined Sep 12, 2016
18
One tends to remember principles and not the names of those principles over the years. You are correct of course. I assumed that both had been taught together. But if not and/or the instructor is expecting the student to only use superposition and Thevenin equivalents, then it would be best to start with the Rth per The Electricians method and use superposition to get Vth. In any case the OP has already solved that portion of the problem.
what should i do after i get the voltage ,RL voltage is 3.24324V.. how to start
 

Thread Starter

babyboang boang

Joined Sep 12, 2016
18
Here we have a situation that often occurs on the forum. The TS did not take a picture of the problem in the text (or whatever) and post it. He redrew the problem by hand and posted that, so his text may not be exactly what the problem says.

He found the voltages across all but one of the resistors using superposition with RL missing. It wouldn't be much harder to do it again with RL in place as DGElder suggested.

So, babyboang boang, if you can, please take a picture of the problem in your book and post that.
My instructor said use thevenin theorem to get the RL voltage , and he said it would be simplier to get the other voltage after you get the voltage at RL ,,, and i dont know what method i use after i get the ,RL voltage is 3.24324V .. RTH=60/11 ,,, ETH=5.454545 V
i need to findd the resistor voltage across the circuits help
 

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Thread Starter

babyboang boang

Joined Sep 12, 2016
18
I see it as an application of Thevinin's Theorem. The tasks are:

1. Find Vth
2. Find Rth
3. Calculate the Voltage drop across each resistor using Thevinin Theorum. The each resistor in this case is the Rth and the RLoad.

I agree with the others. The TS should have provided an accurate problem statement i.e. a picture of the assignment.
help get all resistor voltage after i get the ,RL voltage is 3.24324V RTH 60/11 ETH 5.454545V
 

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My instructor said use thevenin theorem to get the RL voltage , and he said it would be simplier to get the other voltage after you get the voltage at RL ,,, and i dont know what method i use after i get the ,RL voltage is 3.24324V .. RTH=60/11 ,,, ETH=5.454545 V
i need to findd the resistor voltage across the circuits help
Is your instructor commanding you to use Thevenin's theorem, or do you have a choice to use another method?
 

JoeJester

Joined Apr 26, 2005
4,390
If you are reduce the circuit to it's Thevenin equivalent, all you have is Vth, Rth, and Load.

If you must find all the voltage drops of all the resistors, you could use nodal analysis or superposition. The whole point of the Thevenin equivalent is to replace that whole circuit to it's two port equivalent.

Now, your professor might want you to do the whole circuit for their convenience.

As far as your answer, Rth is 5.45 Ohms. Vth is 5.45 Volts and the VRL is 3.24 Volts; rounded to two decimal places.
 
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MrAl has annotated your circuit, so I reproduce that image here for reference:

Cirx.jpg

You must repeat the process you used to find the voltage across RL.

Consider the resistor labeled R1. You must remove R1 and in its place there will be a pair of terminals labeled A and B. To find Rth at that pair of terminals, replace the voltage sources with short circuits and calculate the equivalent resistance seen at those terminals by reducing the other resistances in the circuit by means of parallel and series transformations.

Then to find Vth at the pair of terminals where you removed R1, restore the voltage sources and solve for the current in R2 using superposition. Once you have the current in R2 you will also have the voltage across R2. Then Vth for this case will be the series combination of the voltage across R2 and the 20 volt E1 source.

Now that you have Rth and Vth at these terminals, you can replace R1 and calculate the voltage across R1 when it is driven by Vth in series with Rth. In other words, Rth and R1 form a voltage divider across Vth.

This is the same thing you did to find the voltage across Rth.

Now you must repeat this process for each resistor in the circuit, except for Rl which was the first thing you did.
 

DGElder

Joined Apr 3, 2016
351
Once you have solved for the voltage on RL you can find all the voltages on each node, except v2, by inspection. Then you use superposition to find the current through R1 to get the voltage across R1 and the voltage at v2. You then have all the node voltages and can use them to get all the resistor voltages.

You can see that....
v6 = VRL
v4 = VRL + 0.5 VRL
v5 = 4V
v3 = -4V
v1 = 16V

VR5 = 1/2 VRL
VR4 = V4 - V5
etc.

To assign voltages to the resistors you need to assign a polarity to each resistor consistent with your node voltage differences.




An easier way to find v2 is to realize that all the current that goes into node 4 must come out of node 4. Since you can easily calculate the current through 2 of the 3 branches you can find the current in the third branch, R3. Then you know the voltage across R3 and can get v2.
 
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WBahn

Joined Mar 31, 2012
29,976
My instructor said use thevenin theorem to get the RL voltage , and he said it would be simplier to get the other voltage after you get the voltage at RL ,,, and i dont know what method i use after i get the ,RL voltage is 3.24324V .. RTH=60/11 ,,, ETH=5.454545 V
i need to findd the resistor voltage across the circuits help
What have YOU done to even attempt to do this? All I'm seeing is you keep asking someone to do it for you over and over and over.

If you know the voltage across a known load resistance, can you not find the current in that resistance?

If you know the current in the load resistance, do you not know the current in the 4 Ω resistor that is in series with it?

If you know the current in that 4 Ω resistor, do you not know the voltage across it.

Can you not determine the voltage across that right-most vertical 2 Ω resistor and, hence, the current through that.

Keep working your way back one component/node at a time.
 
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