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?
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.
Best Answer
The currents split up as shown in the following 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.