This is a follow up question of this question. Summarized: The question is solved there good, however I wish to solve the problem the using Kirchhoff laws.
Kirchhoff's voltage/mesh law is nice as it gives a good method to solve passive networks. It just boils down to defining a (linearly independent) meshes, and solving the equations that come from these meshes; probably using some easy matrix algebra.
However I wonder how to handle this in the following schematic
There can be three meshes easily "identified" (of course there are many more just take the obvious ones), and for the calculations let's name the loop currents \$I_a, I_b, I_c\$ from left to right, and let's assume all loop currents are counter clockwise. (Can't seem to draw those in the schematic..).
Now the to calculate the loops starting with the 8V voltage source: $$ 8 - I_aR_1 - 3 + (I_a- I_b)\cdot X = 0 $$ $$ (I_a- I_b)\cdot X - (I_b - I_c) \cdot R_2 = 0$$ $$ -(I_c - I_b) \cdot R_2 - I_c \cdot Y = 0$$
However as you can see I putted "X" and "Y" for the current sources - as I can't really see what a current source does with the voltage in a mesh. At first I would simply remove them from the equations (Ideal current source doesn't provide / remove voltage right?). And then add some extra equations: $$ I_a -I_b = 3 A$$ $$I_c = 1.25A$$
But can anyone do this? How to add current sources in the equations? Similar question: how would one add other passive components like inductors/capacitors (diodes?) to kirchhoff's laws? Or can I no longer use this law then?