So above given is the circuit I am supposed to find the Thevenin Equivalent of with respect to the 1/5Ω resistor load.

I've sort of managed to put something together, however, it seems totally wrong...
Setting all other non-included sources to 0, I get:

Is this on the right track at all or have I completely lost it?

Thanks!

 

Why does it seem totally wrong?

If you take the time to express why you think something seems wrong, you will often walk right into either the realization that it is not wrong or you will see exactly why it is wrong and how to proceed to fix it.

Barring that, show how you arrived at your left hand circuit so that we can see what your reasoning was.

In particular, consider what the Req should be in terms of looking at the resistance seen between the terminals with all of the sources turned off.

You also need to be explicit about the polarity of your Thevenin equivalent terminals and how they map to the terminals in the real circuit. It doesn't appear you are being consistent.

Work on that a bit and revise your work.

Once we get through it, I'll show you how you can get the answer to this particular problem by inspection. As a teaser, the Vth is between -9V and -10V and the Req is between 1/8 ohm and 1/4 ohm.

 

As a teaser, the Vth is between -9V and -10V and the Req is between 1/8 ohm and 1/4 ohm.

I get Vth = -28/3 = -9.33... and Req = 1/11. Not sure how can you estimate these numbers. To me it is more difficult than directly solve for their exact numbers.

 

I get Vth = -28/3 = -9.33... and Req = 1/11. Not sure how can you estimate these numbers. To me it is more difficult than directly solve for their exact numbers.

Req is the value without the 1/5 ohm resistor.

 

Thanks a lot for the help guys, unfortunately, browing through my notes and scouring through a few videos didn't quite get me to understand on how to obtain the Vthevenin.

However, if I'm not mistaken:

Because no current flows through the 1/3 resistor it becomes obsolete, thus Rthevenin = 1/6

Looking through the rest of these posts, I'll try find out how anhnha arrived at -9.33V

EDIT:
Using Microcap, using Dynamic DC analysis + an animated meter, I managed to 'somewhat' simplify the circuit a little bit and got the same voltage...
Since I assume that the 2A current source and the 1/3 resistor are in series, using Ohm's Law, it becomes +0.667V thus sum of voltages being -9.33V.

Or is this just coincidence?

 

To find Thevenin voltage you need to find the voltage across two terminals of 1/5 ohm when the resistor is removed.

Since I assume that the 2A current source and the 1/3 resistor are in series, using Ohm's Law, it becomes +0.667V thus sum of voltages being -9.33V.

No, they are not in series. To find the current through 1/3 ohm resistor, you can use KCL at the node between two current sources.
What is the current through 1/6 ohm as 1/5 ohm resistor is removed?
Then what is the Thevenin voltage?

 

Thanks a lot for the help guys, unfortunately, browing through my notes and scouring through a few videos didn't quite get me to understand on how to obtain the Vthevenin.

However, if I'm not mistaken:

Because no current flows through the 1/3 resistor it becomes obsolete, thus Rthevenin = 1/6

Correct.

See how redrawing the circuit really helps? After a bit of experience you'll be able to do much of that redrawing mentally. But even after over three decades I still physically redraw circuits as soon as they start getting a bit complex or unusual.

Looking through the rest of these posts, I'll try find out how anhnha arrived at -9.33V

EDIT:
Using Microcap, using Dynamic DC analysis + an animated meter, I managed to 'somewhat' simplify the circuit a little bit and got the same voltage...
Since I assume that the 2A current source and the 1/3 resistor are in series, using Ohm's Law, it becomes +0.667V thus sum of voltages being -9.33V.

Or is this just coincidence?

It's just coincidence. I could change the 2A resistor to 20A and the 1/3 ohm resistor to 1000Ω and it would not change the Thevenin equivalent circuit at all.

Remember, the only thing that matters is what voltage is produced across the terminals where the load sits.

 

Alright, well, I think I have made a little bit of progress but I'm still not certain.

So the 2A + 4A + 1/3 Resistor makes 2V
The 1/6 Resistor + 4A makes 0.667V
Then you have the 10V Voltage Source

Are these the right figures to be working with? If so, I don't know how the 2V affects it. But I sort of understand how -9.33V comes from the other values.

 

Alright, well, I think I have made a little bit of progress but I'm still not certain.

So the 2A + 4A + 1/3 Resistor makes 2V
The 1/6 Resistor + 4A makes 0.667V
Then you have the 10V Voltage Source

Are these the right figures to be working with? If so, I don't know how the 2V affects it. But I sort of understand how -9.33V comes from the other values.

Yes, you are making progress, but you are still floundering and flaying a lot.

Let's see if this helps.

Consider the whole circuit with the 1/5 Ω resistor removed (i.e., all the supplied in place and turned on).

Q1) What does Io have to be?

Q2) What does the voltage across the 1/6 Ω resistor have to be?

Q3) If the voltage at the bottom of the circuit is 0V (as defined by the ground symbol), what is the voltage at the top of the 1/6 Ω resistor have to be?

Q4) If you know the voltage at the top of the 1/6 Ω resistor, what does the voltage at the left side of the 10V supply have to be?

Q5) If you know the voltage at the left of the 10V supply, what is the voltage at the right side of the supply?

Q6) If you know the voltage at the right side of the supply, what is the voltage at the top of the load terminals?

Q7) What is the voltage at the bottom of the load terminals?

Q8) What is the open-circuit voltage of the circuit?

 

Ahh, this is most useful.

1. 4A, current takes the 'shortest' path.
2. 4A * 1/6 = 0.667V
3&4. This question, I honestly don't quite understand how it's 0.667V as well... does it have something to do with the fact that the two nodes surrounding the voltage source are in a Supernode?
5. -10 + 0.667 = -9.33V
6. -9.33V
7. 0V
8. -9.33V - 0V = -9.33V

 

Ahh, this is most useful.

1. 4A, current takes the 'shortest' path.

Has nothing to do with "shortest' path. It is the ONLY path that the 4A can take.

2. 4A * 1/6 = 0.667V

Correct, except you are dropping units.

4A * 1/6 Ω = 0.667V

3&4. This question, I honestly don't quite understand how it's 0.667V as well... does it have something to do with the fact that the two nodes surrounding the voltage source are in a Supernode?

Has nothing to do with a supernode.

Q3 has to do with the fact that if you have a resistor between nodes A and B that the voltage across the resistor is the voltage at Node A minus the voltage at Node B. In other words, Vab = Va - Vb. Ohm's law says that the voltage across a resistor is equal to the current in the resistor (flowing in the direction of the voltage difference) times the resistance. So Vab = Iab*R. Hence

Vab = Va - Vb = Iab*R

Va = Vb + Iab*R

In other words, if you know the voltage on one side of a component and you know the voltage across that component, then you know the voltage on the other side of the component.

Q4 merely reflects the fact that (ideal) wires are always at the same voltage anywhere along the wire.

5. -10 + 0.667 = -9.33V

Correct, except dropping units again.

-10V + 0.667V = -9.33V

The better answer to the question that was asked is that if we know the voltage on the left side of the 10V supply, that the voltage on the right side is whatever that voltage is minus 10V. But your answer captured that fact.

6. -9.33V
7. 0V
8. -9.33V - 0V = -9.33V

Correct.