Find Thevenin equivalent, but off by a small amount

[Color_of_Cyan]

Homework Statement

Apply source transformation to find the Thevenin equivalent circuit

Homework Equations

voltage division, source transformation

The Attempt at a Solution

Simplify the far left current source to get

Then Vth of this = (50V + 20V)*(100/200)

= 35V

and Rth of it = 100||100 = 50Ω

so now:Then Vth of this = (35V - 10V)*(100/150)

= 16.6667V

Rth of it = 50||100 = 33.333Ωso finally this:

And then the final Vth I got

= 16.6667*(100/133.3333)

V_Th = 12.5V

but from the given solution from my professor the answer is really 10V for the Thevenin voltage and she did the whole problem differently by instead making everything Norton equivalent instead (changing voltage sources to current sources and moving the resistors in front to parallel them instead). Shouldn't the answer still be the same though or did I just make a mistake somewhere?

I got the same RTh though which is

33.333||100 = 25Ω

R_Th = 25Ω

 

[szynkasz]

Then Vth of this = (50V + 20V)*(100/200)= 35V
and Rth of it = 100||100 = 50Ω

This is not correct: [itex]V_{th}=15V[/itex]

 

[gneill]

Did you consider the option of first converting all the voltage sources to current sources? Can you see any advantage that might bring?

 

[Color_of_Cyan]

This is not correct: [itex]V_{th}=15V[/itex]

So you can't add the voltage sources with each other in same direction here?

Here it's unique because even with that though I guess you would add 10V to the next simplification, thereby still getting 25V for the next simplification.

Did you consider the option of first converting all the voltage sources to current sources? Can you see any advantage that might bring?

No, but I see it; you do that and then just use KCL and still get Thevenin resistance / voltage. I just can't seem to get the same answer doing it this way though so I think I'm missing something.

 

[gneill]

No, but I see it; you do that and then just use KCL and still get Thevenin resistance / voltage. I just can't seem to get the same answer doing it this way though so I think I'm missing something.

Everything ends up in parallel. Just gather all the currents into one, gather all the resistors into one. You end up with a simple Norton model which can be converted to its Thevenin equivalent rather trivially.

 

[szynkasz]

So you can't add the voltage sources with each other in same direction here?

What you calculated is a voltage on resistor. To get [itex]V_{th}[/itex] you have to subtract the source voltage :[itex]35V-20V=15V[/itex]. The next simplification gives:

[itex]V_{th}=(15-10)\cdot\frac{100}{150}+10=\frac{40}{3}V[/itex]

 

[NascentOxygen]

So you can't add the voltage sources with each other in same direction here?

If in doubt, work it out a couple of ways. If the answers don't match up, determine where your thinking is wrong.

Output voltage = 50V - the drop across the 100 Ω
The loop current = 70V / 200Ω, so output voltage = ...

Good luck with your circuit analysis course!