[SOLVED] Thevenin's law - tricky example.

[piotrekn95]

Hello everyone,

I've been trying to solve this circuit. I have to use Thevenin's method to find the I5 current - so I tried to calculate the Thevenin's voltage and Thevenin resistance.

I tried to change the triangle settings of resistors to star, but I don't know if it will help. Do you have any pieces of advice on how to start solving it?

 

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Why is there a current source shown but a voltage is given as if it was supposed to be a voltage source?

 

[piotrekn95]

It's just the symbol which are used at my university to draw a "voltage source".

 

[ads-ee]

It's just the symbol which are used at my university to draw a "voltage source".

Well it's the wrong symbol...says a lot about your university.

 

[CataM]

1) Change R1,R2,R3 to star. Leave the other triangle as is because you will lose track of I5 otherwise.
2) Then, you will have another triangle formed by R7, R6 and 2 other resistors that come from the star you just formed. Change again to star.
3) You are left with simple parallel/series connections. You are done.

 

[andre_luis]

It's just the symbol which are used at my university to draw a "voltage source".

It is indeed a current source, having an unknown current (to be calculated), but with a given voltage across its terminals for the above equivalent thevenin load.

 

[piotrekn95]

What do you think about that? Could it be correct? Of course Ra, Rb, Rc will be calculated from triangle/star formulas.

 

[CataM]

Yes, looks good.

 

[andre_luis]

Each one has their own standpoints and all of them would give the same result; I would particularly prefer to perform the Δ-Y conversion in both sets R1/R2/R3 and with R4/R5/R6. At the end you will have two arms of series resistors in parallel, with two others in series, namelly:

[B]R[SUB]EQ[/SUB] = { R[SUB]2Y[/SUB] + [ ( R[SUB]3Y[/SUB] + R[SUB]7[/SUB] + R[SUB]4Y[/SUB] ) || ( R[SUB]1Y[/SUB] + R[SUB]5Y[/SUB] ) ] + R[SUB]6Y[/SUB] }[/B]

 

[piotrekn95]

Ok, I know what i should do with resistors. But now, for example I want to calculate the Thevenin's voltage on A B terminals. Is it possible to use the node potential method ?
It will be a problem that I have lonely voltage source in the branch ?

Regards

 

[CataM]

No, just calculate the voltage between A-B with whatever method you want.

You can see that you have another triangle there. It might be easier to convert it to star.

 

[Ratch]

Hello everyone,

I've been trying to solve this circuit. I have to use Thevenin's method to find the I5 current - so I tried to calculate the Thevenin's voltage and Thevenin resistance.

I tried to change the triangle settings of resistors to star, but I don't know if it will help. Do you have any pieces of advice on how to start solving it?

View attachment 147660

Nothing tricky about this problem. Using Thevenin is the hard way of doing it. Mesh analysis easily givers the current of 42/5 amps. If you insist of using the Thevenin method, the open circuit voltage with R5 removed is 84 volts. The shorted current where R5 was connected is 14 amps. Therefore the Thevenin resistance is 6 ohms. The current with R5 connected is 84/(6+R5) = 42/5. If you need help calculating the open voltage and shorted current, let me know.

Ratch