I am asking for help in solving the following tasks. In task 1 I counted the currents flow (probably unnecessarily). I don't know what source power is. Is it about the current in the branch with the voltage source? If so, then I1 = 15 A. I did not continue but decided to do the task again....

15A is a bad answer. Thevenin theory is not needed for task 1. 10A is the correct answer and the solution is correct. And the source current is the source current. Do you have any questions about task 1? If not, this 2nd task is waiting

No and no. So far, we have not yet come to the use of Thevenin's Theorem to the end. Now count the I2 current in your configuration. Multiply it by the resistance R4 and you have a voltage drop on the element that interests us (i.e. you are able to calculate the Uab voltage). Having the Uab voltage, having a substitute resistance, you have to make a substitute circuit to which you attach the previously cut branch, here: This shows that it's bad You have marked current I3 at the moment.

Unfortunately .... it's not good. The equivalent resistance seen from the viewpoint of terminals A and B is 6/5 ohms. As you stated in post # 5. In post # 9 is good. Because in post 9 you count the second system, with the 6V source cut out. You have calculated a replacement circuit with a 10V source (where the 3A source current comes out). You just need to develop it to know the voltage drop at the extreme right resistance: If you know this voltage drop, this voltage drop is your substitute voltage in the Thevenin system. Only after that you create another layout to count, the one from post number # 10. PS In the circuit from post # 9 you have the error that you marked the current I3 as a current in the branch with a 10V source. Only that in the task, the current I3 (the correct one to count) is the current in the cut branch with a 6V source. From here you can get mixed up.

Regarding task 1 - the final solution will look like this, right? If so, how did I count the current flow (this first test in the previous post), then I1 = 15 A came out in the branch with the source. Is it because of a calculation error or was it simply not the point? The source current (source current) is not equal to the current in the branch with the source? Edit: And here is my attempt to solve task 2:

Thevenin's theorem is that you cut the branch in which you want to count something (in this case I3 current). From the point of view of this cut branch, you count the resistance for shorted voltage sources in the remaining circuit after cutting (in this case, one will remain ...). Then you calculate the voltage at the terminals from which you cut the branch (i.e. the voltage drop from the only source - on the resistance R4). So we roll the circuit and develop. Having the voltage on the terminals, the replacement resistance, you substitute it as a new system with series connected: voltage source and resistance that you have just calculated. You now add a cut branch to their terminals and you have a simple circuit with a source voltage difference and two resistances.

I can solve Zad2 from the first post so far I corrected Rz counting but I still don't know how to count Ez. I could count it by the eyelet method or Kirchoff equations for 3 points and I would calculate I3 but this is to be done from tw. Thevenin. Please help what's next.

You have calculated the replacement resistance, Now it's time to count the voltage in this system.

I mean, I don't know how to count this tension. Could you tell in detail how to count them or it's best to count and paste them yourself?

With all due respect, but you must be kidding. Should I write the exam for you? I do not understand what you have a problem with when counting currents and voltages from one source. Since you only have E2, you add the resistances R4 and R5, then connect them in parallel with R1, and then in series with R2. Then pure oma law, and again we develop the circuit to find currents I1 and I2 (although here instead of currents, you immediately have the voltage on the corresponding elements).

equivalent resistance seen from terminals A, C: $$R_{45} = 2 + 2 = 4 Ω$$ $$R_{145} = \frac{2 * 2}{2+4} = \frac{8}{6} Ω$$ $$R_{Z2} = R_{1245} = 2 + \frac{8}{6} = 3\frac{1}{3} Ω$$ from Ohm's law: $$ I = \frac{E_{2}}{R_{Z2}} = \frac{10V}{3.33Ω} = 3 A $$ And now what is this 3 A current? Is this the current I3 and the end of the task? Will it only be used to calculate the substitute voltage? (I remind you that in the command they only ask for the current value I3 (the command is in the first picture in the first post - Zad2))

First, I did it like this (Kirchhoff equations): But I found that something is wrong because from the above system I counted Uab and I could even do I3. And then I don't use Thevenian's theorem at all. So I did it again, analogously to the previous current (what came out 3A): Different results came out. Are any of these calculations good? Is there something else?

I really mixed up with this I3. I will exchange I3 (the one in the branch with E2) for I4. If I understand correctly that he had those 3 A numbers, then So Ez = Uab = 2 V, right? If so: Right?

Near! Just look at the direction of tensions ... they won't add up.

Circuit Theory: Clarifying Source Power & Calculating No-Load Voltage with Thevenin's Theorem

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How do I determine the source-branch current and use Thevenin’s theorem to find the no-load voltage and branch current in this circuit?

For task 1, the source current is just the current through the source branch, and the correct value is 10 A; Thevenin’s theorem is not needed there [#16898183] For task 2, cut the branch with the 6 V source, find the equivalent resistance seen from terminals A–B as 6/5 Ω, and then solve the reduced circuit with the remaining 10 V source [#16906525][#16901995] After that, calculate the open-circuit voltage at the cut terminals from the reduced circuit, taking care with the voltage polarity because the directions do not simply add [#16904836][#16908741][#16908856] With the correct Thevenin voltage, the final result is I3 = 1.25 A [#16908926]

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#1 16898128 17 Dec 2017 23:20

Anonymous Anonymous

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Post #1
16898128 17 Dec 2017 23:20

I am asking for help in solving the following tasks.

In task 1 I counted the currents flow (probably unnecessarily). I don't know what source power is. Is it about the current in the branch with the voltage source? If so, then I1 = 15 A.

I did not continue but decided to do the task again. This time I counted the circuit's equivalent resistance (seen from the nodes that are the ends of the branch with the R2 resistor). I don't know how to calculate the no-load voltage E0 (I also don't know what it is but I think it needs to be counted). If you take the initial voltage U = 30V, then from Ohm's law I = 10A
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#2 16898183 17 Dec 2017 23:44

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Post #2
16898183 17 Dec 2017 23:44

15A is a bad answer. Thevenin theory is not needed for task 1. 10A is the correct answer and the solution is correct.
And the source current is the source current.
Do you have any questions about task 1?
If not, this 2nd task is waiting
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#3 16898205 18 Dec 2017 00:06

Anonymous Anonymous

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Post #3
16898205 18 Dec 2017 00:06

Regarding task 1 - the final solution will look like this, right?

If so, how did I count the current flow (this first test in the previous post), then I1 = 15 A came out in the branch with the source. Is it because of a calculation error or was it simply not the point? The source current (source current) is not equal to the current in the branch with the source?

Edit: And here is my attempt to solve task 2:
#4 16901995 19 Dec 2017 21:48

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16901995 19 Dec 2017 21:48

Thevenin's theorem is that you "cut" the branch in which you want to count something (in this case I3 current).
From the point of view of this cut branch, you count the resistance for shorted voltage sources in the remaining circuit "after cutting" (in this case, one will remain ...).
Then you calculate the voltage at the terminals from which you cut the branch (i.e. the voltage drop from the only source - on the resistance R4). So "we roll the circuit and develop."
Having the voltage on the terminals, the replacement resistance, you substitute it as a new system with series connected: voltage source and resistance that you have just calculated.
You now "add" a cut branch to their terminals and you have a simple circuit with a source voltage difference and two resistances.
#5 16904424 21 Dec 2017 01:15

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Post #5
16904424 21 Dec 2017 01:15

I can solve Zad2 from the first post so far

I corrected Rz counting but I still don't know how to count Ez. I could count it by the eyelet method or Kirchoff equations for 3 points and I would calculate I3 but this is to be done from tw. Thevenin. Please help what's next.
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#6 16904434 21 Dec 2017 01:33

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16904434 21 Dec 2017 01:33

You have calculated the replacement resistance,
Now it's time to count the voltage in this system.
#7 16904471 21 Dec 2017 03:56

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16904471 21 Dec 2017 03:56

I mean, I don't know how to count this tension. Could you tell in detail how to count them or it's best to count and paste them yourself?
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#8 16904670 21 Dec 2017 09:51

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16904670 21 Dec 2017 09:51

Erwin14 wrote:
is it best to count and paste it here?

With all due respect, but you must be kidding. Should I write the exam for you?

I do not understand what you have a problem with when counting currents and voltages from one source.
Since you only have E2, you add the resistances R4 and R5, then connect them in parallel with R1, and then in series with R2. Then pure oma law, and again we develop the circuit to find currents I1 and I2 (although here instead of currents, you immediately have the voltage on the corresponding elements).
#9 16904771 21 Dec 2017 10:41

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16904771 21 Dec 2017 10:41

equivalent resistance seen from terminals A, C: $$R_{45} = 2 + 2 = 4 Ω$$ $$R_{145} = \frac{2 * 2}{2+4} = \frac{8}{6} Ω$$ $$R_{Z2} = R_{1245} = 2 + \frac{8}{6} = 3\frac{1}{3} Ω$$
from Ohm's law:
$$ I = \frac{E_{2}}{R_{Z2}} = \frac{10V}{3.33Ω} = 3 A $$
And now what is this 3 A current? Is this the current I3 and the end of the task? Will it only be used to calculate the substitute voltage?
(I remind you that in the command they only ask for the current value I3 (the command is in the first picture in the first post - Zad2))
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#10 16904836 21 Dec 2017 11:22

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16904836 21 Dec 2017 11:22

Erwin14 wrote:
Is this the current I3 and the end of the task?

No and no.
So far, we have not yet come to the use of Thevenin's Theorem to the end.
Now count the I2 current in your configuration. Multiply it by the resistance R4 and you have a voltage drop on the element that interests us (i.e. you are able to calculate the Uab voltage).
Having the Uab voltage, having a substitute resistance, you have to make a substitute circuit to which you attach the previously cut branch, here:

This shows that it's bad You have marked current I3 at the moment.
#11 16906351 22 Dec 2017 02:34

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16906351 22 Dec 2017 02:34

First, I did it like this (Kirchhoff equations):

But I found that something is wrong because from the above system I counted Uab and I could even do I3. And then I don't use Thevenian's theorem at all. So I did it again, analogously to the previous current (what came out 3A):

Different results came out. Are any of these calculations good? Is there something else?
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#12 16906525 22 Dec 2017 09:19

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16906525 22 Dec 2017 09:19

Unfortunately .... it's not good. The equivalent resistance seen from the viewpoint of terminals A and B is 6/5 ohms. As you stated in post # 5.
In post # 9 is good. Because in post 9 you count the "second" system, with the 6V source cut out.
You have calculated a replacement circuit with a 10V source (where the 3A source current comes out). You just need to develop it to know the voltage drop at the extreme right resistance:

If you know this voltage drop, this voltage drop is your substitute voltage in the Thevenin system.

Only after that you create another layout to count, the one from post number # 10.

PS
In the circuit from post # 9 you have the error that you marked the current I3 as a current in the branch with a 10V source. Only that in the task, the current I3 (the correct one to count) is the current in the cut branch with a 6V source. From here you can get "mixed up".
#13 16908741 23 Dec 2017 12:07

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Post #13
16908741 23 Dec 2017 12:07

I really mixed up with this I3. I will exchange I3 (the one in the branch with E2) for I4. If I understand correctly that he had those 3 A numbers, then

So Ez = Uab = 2 V, right? If so:

Right?
#14 16908856 23 Dec 2017 13:12

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Post #14
16908856 23 Dec 2017 13:12

Near!
Just look at the direction of tensions ... they won't add up.
#15 16908889 23 Dec 2017 13:28

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Post #15
16908889 23 Dec 2017 13:28

Right, I corrected

so the final solution is I3 = 1.25 A
#16 16908926 23 Dec 2017 13:48

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16908926 23 Dec 2017 13:48

That's right
#17 16909148 23 Dec 2017 15:31

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Post #17
16909148 23 Dec 2017 15:31

Thank you very much for help.
For future readers I have gathered posts into one complete solution:
REF1:

REF2:
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Topic summary

✨ The discussion revolves around solving circuit theory problems using Thevenin's Theorem, specifically focusing on calculating source power and no-load voltage. The user initially miscalculated the current in a branch with a voltage source, leading to confusion about the source current. Participants clarified that the source current is distinct from the current in the branch with the source. The conversation progressed to calculating equivalent resistance and no-load voltage, with various methods discussed, including Kirchhoff's equations. Ultimately, the correct current (I3) was determined to be 1.25 A, and the no-load voltage (Ez) was calculated as 2 V. The final solutions were compiled for future reference.
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FAQ

TL;DR: For this two-source exercise, the Thevenin resistance seen at A–B is 6/5 Ω; and “this voltage drop is your substitute voltage.” Apply Thevenin to get the target branch current. [Elektroda, kozi966, post #16906525]

Why it matters: This FAQ shows how to turn a messy mixed-source network into a one-source, two-resistor snap calculation—fast and exam-proof.

Quick-Facts

Task 1 answer: source current = 10 A; Thevenin not required. [Elektroda, Anonymous, post #16898183]
Thevenin R_th at A–B for Task 2: 6/5 Ω (≈1.2 Ω). [Elektroda, kozi966, post #16906525]
No‑load voltage Ez equals Uab with the branch cut (“substitute voltage”). [Elektroda, kozi966, post #16906525]
Final result confirmed: I3 = 1.25 A (correct orientation fixed). [Elektroda, kozi966, post #16908926]

Quick Facts

Task 1 answer: source current = 10 A; Thevenin not required. [Elektroda, Anonymous, post #16898183]
Thevenin R_th at A–B for Task 2: 6/5 Ω (≈1.2 Ω). [Elektroda, kozi966, post #16906525]
No‑load voltage Ez equals Uab with the branch cut (“substitute voltage”). [Elektroda, kozi966, post #16906525]
Final result confirmed: I3 = 1.25 A (correct orientation fixed). [Elektroda, kozi966, post #16908926]

What did “source power” mean in Task 1?

The forum clarified they wanted the source current, not power. Compute source power later as P = U × I if needed. In Task 1, the accepted result is 10 A and Thevenin is unnecessary. “And the source current is the source current.” [Elektroda, Anonymous, post #16898183]

How do I get the source current in Task 1 quickly?

Use Ohm’s law directly on the given source and its connected resistance. Thevenin reduction is not required here. The checked answer in the thread is 10 A for Task 1, matching a direct calculation approach. [Elektroda, Anonymous, post #16898183]

What is Thevenin’s theorem in plain words?

“You cut the branch” where you want the current, deactivate other sources appropriately, find the equivalent resistance and the open‑circuit voltage at the cut terminals, then reconnect the branch to this Thevenin pair to solve. [Elektroda, kozi966, post #16901995]

How do I compute the Thevenin resistance R_th at A–B here?

Cut the target branch, short the ideal voltage sources, and look into the network from A–B. For this circuit, R_th = 6/5 Ω. That numeric result was explicitly corrected and confirmed in the discussion. [Elektroda, kozi966, post #16906525]

What exactly is the no‑load (open‑circuit) voltage Ez/E0?

Ez is the terminal voltage Uab across A–B with the branch removed. In this problem, compute the drop at the rightmost resistor from the remaining active source; “this voltage drop is your substitute voltage.” [Elektroda, kozi966, post #16906525]

How do I calculate Uab (Ez) in this setup?

With the 6 V branch cut, keep the 10 V source active. Reduce the network, find current I2 through R4, then Ez = I2 × R4. The moderator’s guidance was to multiply I2 by R4 to obtain Uab. [Elektroda, kozi966, post #16904836]

Once I have Ez and R_th, how do I get I3?

Build the Thevenin equivalent (Ez in series with R_th), then reconnect the 6 V branch. Solve the resulting simple series network with correct polarity. The validated final result is I3 = 1.25 A. [Elektroda, kozi966, post #16908926]

Where did that 3 A come from in intermediate steps?

That 3 A is the source current when only the 10 V source drives the reduced network (R_Z2 ≈ 3.33 Ω). It’s an intermediate check, not I3. The helper notes “the 3A source current comes out” in that reduced case. [Elektroda, kozi966, post #16906525]

How do I avoid sign mistakes with voltages and currents?

Track source polarity and assumed current directions consistently. A reversed sign causes sums not to match. The helper flagged a misalignment with “look at the direction of tensions ... they won’t add up,” which required correction. [Elektroda, kozi966, post #16908856]

Can I solve Task 2 without Thevenin (e.g., with Kirchhoff)?

Yes. You can write node or loop equations for the full two‑source network and compute Uab and I3 directly. A participant did this, got results, then reworked to align with the requested Thevenin method. [Elektroda, Anonymous, post #16906351]

Give me a 3‑step How‑To for this exact circuit using Thevenin.

Cut the 6 V branch; short other ideal voltage sources; compute R_th at A–B.
With only the 10 V source active, find Uab = Ez at A–B.
Build V_th = Ez in series with R_th, reconnect the 6 V branch, and solve for I3. [Elektroda, kozi966, post #16901995]

What is the target branch and current orientation for I3?

I3 is the current in the previously cut 6 V source branch, not in the 10 V source branch. Early confusion came from labeling I3 in the wrong branch; later posts corrected the assignment before confirming I3 = 1.25 A. [Elektroda, kozi966, post #16906525]

What edge case can flip my final answer?

If you reverse a source polarity relative to Ez, the Thevenin source and the branch source oppose or aid differently. That flips the sign of the net driving voltage and reverses I3’s direction, which the thread highlighted when fixing tension directions. [Elektroda, kozi966, post #16908856]

How can I sanity‑check numbers fast during an exam?

Estimate: with R_th ≈ 1.2 Ω and a small Ez, expect amperes only if Ez notably exceeds the branch source. Here, consistent orientation yielded I3 = 1.25 A, matching the discussed correction and confirmation. [Elektroda, kozi966, post #16908926]