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How can you set up a system of equations for branch current analysis if there is a current source in a branch, such as the current source with the source current i02? The values of the components​​can be seen in the picture. The task is to find all branch currents using Kirchhoff's branch current analysis.

My question is how can one apply the branch current analysis in this case if there is a current source in the lower right mesh.

I have already set up the system of equations. I just need the voltage at the power source and then it would work. But how can I use the method without the voltage at the power source?

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I would be grateful for further tips.

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  • \$\begingroup\$ What's stopping you from doing this? \$\endgroup\$
    – Andy aka
    Commented Apr 6 at 10:44
  • \$\begingroup\$ The voltage at the current source is missing for the system of equations. I have already set up all the equations and I just don't know exactly how to do the branch current analysis when a current source is present in a mesh (like the one below right). \$\endgroup\$
    – Omar
    Commented Apr 6 at 12:03

3 Answers 3

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Since \$I_2\$ is known, you don't need to include it in the variable vector. Instead, you can add \$U_{02}\$ to the variable list. You have 6 equations and 6 unknowns \$I_1, I_3\$ to \$I_6\$ and \$ U_{02} \$ so it's solvable.
I assume your system of equations is correct, so you can express them as follows: \begin{array}{r@{}r@{}r@{}r@{}r@{}r@{}r@{}r@{}r@{}r@{}r@{}r@{}r@{}} 1 * I1 & {}-{} & 1 * I3 & {}+{} & 0 * I4 & {}+{} & 0 * I5 & {}-{} & 1 * I6 & {}+{} & 0 * U_{02} & {}={} & 0 \\ 0 * I1 & {}+{} & 1 * I3 & {}-{} & 1 * I4 & {}+{} & 1 * I5 & {}+{} & 0 * I5 & {}+{} & 0 * U_{02} & {}={} & 0 \\ 0 * I1 & {}+{} & 0 * I3 & {}+{} & 0 * I4 & {}-{} & 1 * I5 & {}+{} & 1 * I6 & {}+{} & 0 * U_{02} & {}={} & -1 * I2 \\ R1 * I1 & {}+{} &R3 * I3 & {}+{} & R4 * I4 & {}+{} & 0 * I5 & {}+{} & 0 * I6 & {}+{} & 0 * U_{02} & {}={} & U01 \\ 0 * I1 & {}+{} & 0 * I3 & {}-{} & R4 * I4 & {}-{} & R5 * I5 & {}+{} & 0 * I6 & {}+{} & 1 * U_{02} & {}={} & R2 * I2 \\ 0 * I1 & {}-{} & R3 * I3 & {}+{} & 0 * I4 & {}+{} & R5 * I5 & {}+{} & R6 * I6 & {}+{} & 0 * U_{02} & {}={} & 0 \\ \end{array} Or in matrix form: \$ \begin{bmatrix} 1 & -1 & 0 & 0 & -1 & 0 \\ 0 & 1 & -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 1 & 0 \\ R_{1} & R_{3} & R_{4} & 0 & 0 & 0 \\ 0 & 0 & -R_{4} & -R_{5} & 0 & 1 \\ 0 & -R_{3} & 0 & R_{5} & R_{6} & 0 \\ \end{bmatrix} \begin{bmatrix} I_{1} \\ I_{3} \\ I_{4} \\ I_{5} \\ I_{6} \\ U_{02} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ -I_{2} \\ U_{01} \\ R_{2} \cdot I_{2} \\ 0 \\ \end{bmatrix} \$

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  • \$\begingroup\$ Thank you for your answer. Shouldn't there be a voltage in the vector on the right-hand side in the third line or can there also be a current there? \$\endgroup\$
    – Omar
    Commented Apr 6 at 17:00
  • \$\begingroup\$ @Omar no, your third equation is \$I_2 + I_6 = I_5\$, so I rearranged it as \$-1 * I_5 + 1 * I_6 = -I_2\$. You can verify this in the equation or matrix form above. \$\endgroup\$
    – internet
    Commented Apr 6 at 17:10
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initial thoughts from reading your question

It's a phrase that I don't commonly see used here -- branch current analysis. But if I remember correctly (and if a quick google search as interpreted by me to enhance my memory is worth much), then branch current analysis is the application of directed graph theory and Schur's complement. It uses KVL/Ohm's Law and KCL, so this is definitely a graph theoretic approach. (See also here and elsewhere.)

Directed graph theory captures all of this as KCL is found in the rowspace and right nullspace and Ohm's Law and KVL is found in the columspace and left nullspace. (And here is a description of the four fundamental subspaces, for a short summary in review.)

That said, I also see you using mesh -- or, at least, setting up loop currents. And I see a failure to get up a ground node. So this looks more to me as though it is less branch current analysis than it is mesh (KVL) analysis. Besides, KCL would dictate setting up a ground node and you didn't do that. (It's necessary to choose a ground node -- or set some one node to some known voltage -- for graph-theoretic reasons.)

You could approach this with extremely powerful and broadly applicable tools -- directed graph theory -- which would include all of KCL and KVL and provide you with all the key mesh current loops as a simple and automatic precision output. Or you could just take this as one of two simplified slices through the more general graph theory by setting up either nodal (KCL) or mesh (KVL) and eventually solve all the branch currents, either way.

In the end, I'm unclear about precisely what you need.

current source

For mesh analysis, if the current source is isolated in just one loop then that current source simply sets that loop current for you. You don't need to solve for that loop current, as you know what it is given the specified current source.

For nodal analysis, just split the current source into two parts (coming from nowhere and entering or leaving the appropriate nodes), and apply that to your KCL equations.

Either way, this is trivial.

So what's the problem? Just set up your mesh (or nodal) equations and solve.

summary

I had to write this as an answer as there was too much to do in comments. If you clarify your problem/difficulty then I will try and address it. I could just solve the problem for you using any of three or more techniques. But I may wind up completely missing providing the answer you are struggling at. So I'll hold short until you write more.

final note

I've never been able to figure out some of the European conventions with respect to your voltage sources. I've seen both of the following:

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So you will also need to clarify the meaning of that notation, as well.

Sorry. But I've frankly just totally given up trying to decide whether the question comes from Italy, Spain, or somewhere else. Too many conventions, too little time to parse them.

So you need to clarify this for me, as well, if I am to proceed.

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  • \$\begingroup\$ Thank you for your answer. The circuit comes from Germany. There is a current source in the bottom right of the mesh. I actually have to analyze the circuit using branch current analysis and have already set up a system of equations, but I'm still missing the voltage at the current source or maybe there is another possible solution for the branch current analysis if there is a current source in the mesh. \$\endgroup\$
    – Omar
    Commented Apr 6 at 12:09
  • \$\begingroup\$ @Omar Can you add to your question (provide an additional edit at the end) to help me see what you are struggling over? That would help me a lot to see. \$\endgroup\$ Commented Apr 6 at 12:36
  • \$\begingroup\$ I have added additional information and a picture of the system of equations. I hope it helps you. \$\endgroup\$
    – Omar
    Commented Apr 6 at 13:54
  • \$\begingroup\$ @Omar Looks like I'm not needed. Best wishes! \$\endgroup\$ Commented Apr 6 at 19:28
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Analysis of the network based on pairs of independent nodes and solution with the TABLEAU matrix.

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