# Help with thevenin problem

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#### babyboang boang

Joined Sep 12, 2016
18
can anyone help me or explain me how to get the voltage in my circuits ,, i get the RL =8 ohms and 3.24324V ,, and my RTH is =60/11 ... now i put the voltage on the circuits and i dont know to get all the voltage .>> help me guys thanks

#### DGElder

Joined Apr 3, 2016
351

1 State and show the problem as presented to you. Uniquely identify each element in the circuit.

2. Then show, step by step, how you arrived at a solution - as far as you got anyway.

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• babyboang boang

#### babyboang boang

Joined Sep 12, 2016
18
at that point am using superposition to get the VTH or ETH , , And my RL=8 ohms and , RTH= 60/11 .. and i get the voltage at RL and i dont know how to get all the voltage in every single resistor in the circuits the VTH =4v +1.45454 = 5.45454 ,,,RL voltage is 3.24324V

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#### babyboang boang

Joined Sep 12, 2016
18

1 State and show the problem as presented to you. Uniquely identify each element in the circuit.

2. Then show, step by step, how you arrived at a solution - as far as you got anyway.
helppp

#### The Electrician

Joined Oct 9, 2007
2,848
I'm puzzled by the statement of the problem. Is it your understanding that you are to find Vth and Rth, and then use that result to find the voltages across all the other resistors in the circuit? Are they asking for those other voltages when RL is in place, or not?

• babyboang boang

#### babyboang boang

Joined Sep 12, 2016
18
I'm puzzled by the statement of the problem. Is it your understanding that you are to find Vth and Rth, and then use that result to find the voltages across all the other resistors in the circuit? Are they asking for those other voltages when RL is in place, or not?
yes .. all across voltage . they are not asking to replace the RL but the voltage across all resistor ... i dont know how to find those voltages when i get the RL voltage help

#### MrAl

Joined Jun 17, 2014
8,475
Hi,

Do you know Nodal Analysis or Loop Analysis or anything else like that?
You could use Nodal to get all the voltages for example.
You write one equation for each node, then solve them simultaneously.

Also, you need to label each part and it helps to label each node also as in the diagram attached.

#### Attachments

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• babyboang boang

#### babyboang boang

Joined Sep 12, 2016
18
Hi,

Do you know Nodal Analysis or Loop Analysis or anything else like that?
You could use Nodal to get all the voltages for example.
You write one equation for each node, then solve them simultaneously.
thank you ,,, but i dont understand why is that i will get RL voltage if i can use loop or nodal analysis. if for example i replace RL with another resistor value ? and get the voltage of it. after that i will use nodal and lopp analysis to get all the voltages??? thanks help

#### MrAl

Joined Jun 17, 2014
8,475
thank you ,,, but i dont understand why is that i will get RL voltage if i can use loop or nodal analysis. if for example i replace RL with another resistor value ? and get the voltage of it. after that i will use nodal and lopp analysis to get all the voltages??? thanks help

Hi,

You can start with Nodal, you dont have to do that later.
But you need to label everything as in the attachment shown below.
See if that helps.

When you solve, you'll get all the voltages shown, then you can subtract to get the voltages across each resistor.

For example, the equation at v2 could be:
(v1-v2)/R1+(v4-v2)/R3=(v2-v3)/R2

and that is just the sum of currents entering the node at v2 equated to the sum of currents leaving the node at v2.
You can also do it as the sum of all currents entering the node.

Also note that voltages v1, v3, and v5 are already known because of the placement of the ground (GND) which is taken to be at zero volts and all the sources are connected in series. This means only three voltages are unknown so you only need three equations, for v2, v4, and v6.

#### Attachments

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• babyboang boang

#### babyboang boang

Joined Sep 12, 2016
18
Hi,

You can start with Nodal, you dont have to do that later.
But you need to label everything as in the attachment shown below.
See if that helps.

When you solve, you'll get all the voltages shown, then you can subtract to get the voltages across each resistor.

For example, the equation at v2 could be:
(v1-v2)/R1+(v4-v2)/R3=(v2-v3)/R2

and that is just the sum of currents entering the node at v2 equated to the sum of currents leaving the node at v2.
You can also do it as the sum of all currents entering the node.

Also note that voltages v1, v3, and v5 are already known because of the placement of the ground (GND) which is taken to be at zero volts and all the sources are connected in series. This means only three voltages are unknown so you only need three equations, for v2, v4, and v6.
thank you so much ..

#### babyboang boang

Joined Sep 12, 2016
18
Hi,

You can start with Nodal, you dont have to do that later.
But you need to label everything as in the attachment shown below.
See if that helps.

When you solve, you'll get all the voltages shown, then you can subtract to get the voltages across each resistor.

For example, the equation at v2 could be:
(v1-v2)/R1+(v4-v2)/R3=(v2-v3)/R2

and that is just the sum of currents entering the node at v2 equated to the sum of currents leaving the node at v2.
You can also do it as the sum of all currents entering the node.

Also note that voltages v1, v3, and v5 are already known because of the placement of the ground (GND) which is taken to be at zero volts and all the sources are connected in series. This means only three voltages are unknown so you only need three equations, for v2, v4, and v6.
Can it only solve by nodal ? or any theorem can solve it? am not familiar in nodal theorem . help thanks very musch

#### MrAl

Joined Jun 17, 2014
8,475
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.

• babyboang boang

#### WBahn

Joined Mar 31, 2012
26,398
can anyone help me or explain me how to get the voltage in my circuits ,, i get the RL =8 ohms and 3.24324V ,, and my RTH is =60/11 ... now i put the voltage on the circuits and i dont know to get all the voltage .>> help me guys thanks
You say that your RTH is 60/11. I have to assume that that is a resistance and therefore 60/11 Ω. In your image, however, you are adding that 60/11 to a voltage. So is it a voltage or is it a resistance? If it's a resistance, then how can you add it to a voltage. TRACK YOUR UNITS!!!

What is it you are SUPPOSED to do with this problem? Is the point to use Thevenin's Theorem? Or are you just choosing to do that?

You might start by showing the work you did to get your value of RTH, because it is wrong. But unless you show your work, we can't point out what you did wrong.

#### DGElder

Joined Apr 3, 2016
351
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.

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.

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• babyboang boang

#### The Electrician

Joined Oct 9, 2007
2,848
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.
Using the TS's image, here's how I get Rth: 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.

• babyboang boang

#### DGElder

Joined Apr 3, 2016
351
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 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.

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• babyboang boang

#### MrAl

Joined Jun 17, 2014
8,475
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.

• babyboang boang

#### DGElder

Joined Apr 3, 2016
351
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.
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.

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• babyboang boang

#### The Electrician

Joined Oct 9, 2007
2,848
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.

His text almost seems like it's asking to find the voltages across all the resistors using Thevenin's theorem. I don't see how one would do that. Would it be done by finding Thevenin equivalents for various subsections of the network to calculate each resistor's voltage drop? Wow! That would be a lot of work, but it could be done.

What would be more sensible would be to find the Thevenin equivalent at the A/B terminals, which he did, and then find the voltages across all the resistors by any method.

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.

• babyboang boang

#### JoeJester

Joined Apr 26, 2005
4,390
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.

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• babyboang boang
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