Why does current source define the current to all loop?

Question

Lets say I have this circuit below and I want to solve it using the Mesh Method.

For every essential loop I define the current loops \$i_a\$ (left), \$i_b\$ (right), and \$i_c\$ (below) clockwise. The mesh equations are:

$$-V_2 + R_2i_a + R_1(i_a-i_b) + R_5(i_a-i_c) = 0$$ $$V_3 + R_4(i_b-i_c) + R_1(i_b-i_a) + R_3i_b = 0$$ $$i_c = -I_1 = -1$$

When it comes to the \$i_c\$ loop I know that \$I = -1\$, but how is it true that the whole \$i_c = -1\$? Isn't the current splitting at node 2?

Why is it true that the current passing through the wires conflicting with the opposing effect of loops \$i_a\$ and \$i_b\$ are exactly -1 A?

It is not intuitive to me why it is the case and/or if it is always the case. It is really bugging me.

Answer 1

It has to do with superposition, which is the mathematical "engine" under the hood of loop analysis. The resistor doesn't care where the each individual electron moving through it comes from, it will develop the same voltage at its terminals in accordance with Ohm's Law. The contributions also stack, so the effect of one current loop in isolation can be added to the effect of another current loop that happens to pass through the same device to come to the correct answer.

Answer 2

Think about how that would show up in the equations. The current through R4 (right-to-left) isn't just \$-i_c\$, it's \$i_b - i_c\$. And the current through V3 (top-to-bottom) is \$i_b\$. So the equations do allow for the current to split or merge at node 2, according to the value of \$i_b\$.

Or in other words, \$-i_b\$ is the amount of current that doesn't go through R4, and goes the long-way around the right-hand loop instead. (Then some of that current may be diverted away from R1 by \$-i_a\$)