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Mesh Analysis

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

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  • \$\begingroup\$ 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/… \$\endgroup\$
    – G36
    Aug 24, 2020 at 12:49
  • \$\begingroup\$ I did that already. Just wondering why the two mesh have the same current \$\endgroup\$
    – Dugong98
    Aug 24, 2020 at 13:07
  • \$\begingroup\$ Because now you have only one path for a current to flow (series circuit). \$-20V+I_13Ω+I_12Ω-10V+I_16Ω-30V = 0\$ \$\endgroup\$
    – G36
    Aug 24, 2020 at 13:20
  • \$\begingroup\$ I've updated the diagram. But the solution shows that the 2ohms parallel to (3+5) ohms. shouldn't they have different currents? \$\endgroup\$
    – Dugong98
    Aug 24, 2020 at 14:08
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    \$\begingroup\$ 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) \$\endgroup\$
    – user215805
    Aug 24, 2020 at 14:49

1 Answer 1

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In short:

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

schematic

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:

schematic

simulate this circuit

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  • \$\begingroup\$ thank you so much sir \$\endgroup\$
    – Dugong98
    Aug 24, 2020 at 15:05

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