# Why does the mesh current equal for Thevenin's Equivalent  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

• To be able to find the Vth voltage you need to disconnect the R resistor from the circuit and then you can find Vth in a circuit without R resistor. electronics.stackexchange.com/questions/345594/…
– G36
Aug 24, 2020 at 12:49
• I did that already. Just wondering why the two mesh have the same current Aug 24, 2020 at 13:07
• Because now you have only one path for a current to flow (series circuit). $-20V+I_13Ω+I_12Ω-10V+I_16Ω-30V = 0$
– G36
Aug 24, 2020 at 13:20
• I've updated the diagram. But the solution shows that the 2ohms parallel to (3+5) ohms. shouldn't they have different currents? Aug 24, 2020 at 14:08
• 2ohm parllel to (3+5) if there is an actual source instead of just a notation (vth) and if you are confused by equation written on book , just add both equation and vth term gets cancelled and you got equation of just 1loop , while your equation work when you calculate thevenin resistance because now you have to replace vth by a source voltage and then 2ohm parllel to (3+8) Aug 24, 2020 at 14:49

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

• thank you so much sir Aug 24, 2020 at 15:05