- #26

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It's an old thread which appears to have spanned several years.Now that this problem has been substantially worked on, here's another thread for the same circuit: https://www.physicsforums.com/threads/thevenins-theorem.775385/

- Thread starter Jason-Li
- Start date

- #26

- 1,790

- 748

It's an old thread which appears to have spanned several years.Now that this problem has been substantially worked on, here's another thread for the same circuit: https://www.physicsforums.com/threads/thevenins-theorem.775385/

- #27

The Electrician

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Where did you get that value (39.9 + 40.7 i) in the denominator of the middle term of your equation? The correct value is (35 + 35.70714 i) as given in post #1.All, here is another way of going about this. The Nodal method basically says to model the currents leaving or entering a particular node, in this case the node on top of the load, which I call here as V3. If a current is entering a node, it is positive. If leaving, it is negative. Summing all the currents should equal zero. It is fair to assume for the moment that V1 provides a current into the node, V2 provides a current into the node, and then both of these current sum and leave the node, going to the load. If you convert the currents in its equivalent Ohm's Law of voltage over impedance, you will be left with an equation that has only one variable to solve. It's a lot of algebra to isolate it, but you could use Symbolab.com to crunch complex numbers and just let Symbolab solve for V3 or x in this equation solver. Of course, you won't likely be able to use Symbolab during a test.

View attachment 237649

x is found as 167.9+236.9i.

View attachment 237650

If you take this voltage and divide it by the impedance of the load, you get a load current of 5.0 + .8i

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