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I have trouble solving the following problem, I have to find the Thevenin and Norton equivalent at the head of the R5 resistor:

schematic

simulate this circuit – Schematic created using CircuitLab

I have calculated successfully the equivalent resistance, I am having trouble finding Vth, I replaced R5 with an open circuit like this:

schematic

simulate this circuit

But I don't get the expected result, I realize that this problem isn't necessarily difficult, but I find that there is something I am missing while trying to solve it and I would appreciate some help, thank you !

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  • \$\begingroup\$ What is the result you get, and how did you get it? Here we won't do your homework for you, but if you show your thought process we can give hints and nudge you to come to the solution on your own. \$\endgroup\$ – Chris M. Jun 5 '17 at 19:22
  • \$\begingroup\$ I found the currents traveling along the first and second mesh, ( i1 the current that travels the first mesh is 4/3A and i2 is 5/3A) then I added togheter V1, the tension at R3 and R2, because I thought that whatever tension was on the branch that has R4 was Vth. It gave me 29V, the correct result was 27.6V. \$\endgroup\$ – Matthew Jun 5 '17 at 19:43
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This is easy to solve:

You keep combining sources until you are left with a single source. In this case, working with Thevenin sources (as apposed to Norton) seems easier.

V1,R1 is a Thevenin source you can see by inspection. So is V2,R2. Combine them.

Now add R3 to the resistance of that source. That is the Thevenin equivalent looking left into the right side of R3.

Add R4 to this and force I4 thru the result.

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  • \$\begingroup\$ Thank you, in the first edit you said I should convert the Norton source to Thevenin and combine them, but isn't the Norton source supposed to have a resistor in parallel ? \$\endgroup\$ – Matthew Jun 5 '17 at 20:16
  • \$\begingroup\$ @Mat: I realized I typed something wrong, then fixed it. \$\endgroup\$ – Olin Lathrop Jun 6 '17 at 10:51

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