Why does the mesh current equal for Thevenin's Equivalent

Question

The question wants us to find the thevenin's equivalent of this circuit. Vth is across R. The 5ohms and 6A was transformed into a series voltage and resistance.

The answer shows that the first loop is : -20 + 3i + Vth + 5i -30 = 0

The Above Loop is 2i - 10 - Vth = 0

My question is why does both mesh share the same current i, shouldn't they have their own respective currents i1 and i2 . Eg, -20 + 3i1 + Vth + 5i1 -30 = 0 and 2i2 - 10 - Vth = 0

Answer 1

In short:

To find \$V_{TH}\$ voltage you need to disconnect the "load" resistance.

^{simulate this circuit – Schematic created using CircuitLab}

And fins the \$V_{TH}\$ voltage.

But to be able to find \$R_{TH}\$. The Thevenin's equivalent resistance we need to short's all the voltage sources and open all the circuit current sources. So, for your circuit it will look like this:

^{simulate this circuit}