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Is it possible to use mesh analysis when there's a current source in one of the mesh? (i.e the current source is not between the two mesh to form a supermesh)

In the following circuit, I was able to get rid of the 8A and 2A sources by using source transformation to voltage sources. However, I am not sure what to do with the 4A current source in mesh 3?

enter image description here

What would be a good approach to this problem?

EDIT i am looking for branch currents i1,i2,i3,i4.

EDIT2 After source transformation, this is what i get but I am still not able to deal with 4A current source.

enter image description here

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  • \$\begingroup\$ Well I would start by reducing the 30 Ohm in parallel with 10 Ohm to a single resistor and then do the same with 40 and 30 Ohm resistors. \$\endgroup\$ – JIm Dearden Oct 21 '16 at 11:28
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    \$\begingroup\$ Well, then you have two voltage sources at either end, with two equivalent series resistors for them, with that current source in the middle: the whole thing making a single loop. Right? What is the current? It's entirely determined by the current source, now. So you can compute the drops across these equivalent resistors and work that out. Are you trying to figure out the two left and right node voltages (assuming the bottom is the reference node?) \$\endgroup\$ – jonk Oct 21 '16 at 11:30
  • \$\begingroup\$ @jonk please see the edit. \$\endgroup\$ – Chris Aung Oct 21 '16 at 11:43
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    \$\begingroup\$ By inspection and using your arrow directions to mean 'positive', \$i_1=3\:\textrm{A}\$, \$i_2=9\:\textrm{A}\$, \$i_4=-\tfrac{8}{7}\:\textrm{A}\$, and \$i_3=-\tfrac{6}{7}\:\textrm{A}\$. \$\endgroup\$ – jonk Oct 21 '16 at 11:43
  • \$\begingroup\$ Look, you have a total of 12 A entering the left, top node from two current sources. There's only two exits for that current. Clearly, the left resistor of the two is three times harder for current to move through, so if there is one part through it then there must be three more parts through the other. This means you divide the current into four parts (3 A per part) and that tells you right away what the currents through those two resistors must be. The same logic applies to the other node. But now you have 2 A in, and 4 A out, so -2 A, which must come from two resistors. Do you need mesh? \$\endgroup\$ – jonk Oct 21 '16 at 12:02
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The currents split up as shown in the following schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

You could easily combine \$R_1\$ and \$R_2\$ into a single effective value and work out the voltage at their shared top node... and from that work out their currents. But you can also just as well see that \$R_1\$ is 3 times less conductive than \$R_2\$, so this means that if one part of the current flows through \$R_1\$ then it must be that three more parts of the current must flow through \$R_2\$. From this argument, you should consider dividing the total current of 12 A into four parts of 3 A each. And thus, \$i_1=3\:\textrm{A}\$ and \$i_2=9\:\textrm{A}\$.

Similar logic then applies to the other two resistors. Here, it's not hard to imagine a total of 7 parts of the -2 A net total current in \$R_3\$ and \$R_4\$, where now three parts go through \$R_3\$ and four parts through \$R_4\$. Each part is \$-\tfrac{2}{7}\:\textrm{A}\$, so this means \$i_3=-\tfrac{6}{7}\:\textrm{A}\$ and \$i_4=-\tfrac{8}{7}\:\textrm{A}\$.

Looking at the bottom node, we know the following must be true:

$$ -8\:\textrm{A} - 2\:\textrm{A} + 3\:\textrm{A}+9\:\textrm{A}+-\tfrac{6}{7}\:\textrm{A}+-\tfrac{8}{7}\:\textrm{A}=0$$

And it is.

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  • \$\begingroup\$ This is basically telling the OP to solve the problem with nodal analysis, not explaining how to deal with the source in mesh analysis like s/he asked for. \$\endgroup\$ – The Photon Oct 21 '16 at 16:18
  • \$\begingroup\$ @ThePhoton Well, in my defense, when asked the OP replied that they just wanted the currents and then asked me to post the answer as given as a comment. I hadn't considered it an answer until both those elements arrived. Glad to see you answered. \$\endgroup\$ – jonk Oct 21 '16 at 16:25
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Other answers have explained alternate methods to solve the circuit, rather than answered what you asked for, which is how to deal with a current source in mesh analysis.

The answer is very simple. \$I_3\$ (the current in mesh 3) is -4 A.

Similarly, \$I_1\$ and \$I_5\$ can be trivially determined because they also contain current sources, not shared with any other meshes.

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