thevenin equivalent example problem

[PG1995]

Hi

In the following linked pictures I have marked the sections which I have found more troublesome with numbers (1, 2, 3, 4, 5). The aim is to find Thevenin equivalent. Please help me. It would be really kind and nice of you. Thank you

1: https://img843.imageshack.us/img843/4292/norton1.jpg
2: https://img847.imageshack.us/img847/2450/norton2h.jpg

At "1" in link #1, it says (Vo - 0)/2. It is for the branch in which Ix is flowing. Ix is flowing upward - from the ground toward Vo which simply means that Vo is at lower potential than the ground. Therefore, it should be (in my view): (0 - Vo)/2. Do you see the problem? I understand that if we do that the nodal equation will also change.

At "2" it says Vo = -4 V. What does this mean? In what situation does the current flows from ground to some other point (at -ve potential) in real world?

At "3" having Rth = -4 ohm is completely incomprehensible for me. Please help me with it. I know the author gives some explanation. But one ohm resistance is equivalent one volt used per coulomb. So having a -ve sign with resistance doesn't make sense because it's directionless quantity.

At "4", can we take the direction of current CCW instead of CW? (To see what is author doing here in this circuit please have a look on link #2).

At "5", why has the author used -4i when he has used +ve sign for +9i?

 

[KeepItSimpleStupid]

1. I'm skipping

2. This basically means V is opposite the way it was assumed.

3. -4 Ohms. P = V*I, but P being positive means dissapated power and P being negative means generated power. The problem is if we had a 10 mW power plant and 10 million customers were drawing a megawatt each, mathematically it doesn't make sense, unless 10 MW = k * 1 MW where k is the number of customers.
Re-arranging 0= -10 MW = k * 1 MW. You don't normally think of power plants creating -10 MW of power, nor a battery supplying -100 W, but the fact is you have to reverese the sign to analyize the circuit.

4. Yes you can use CCW or CW. When doing these equations be VERY careful keeping track of the signs.

5. I'm skipping

 

[MrAl]

Hi

In the following linked pictures I have marked the sections which I have found more troublesome with numbers (1, 2, 3, 4, 5). The aim is to find Thevenin equivalent. Please help me. It would be really kind and nice of you. Thank you

1: https://img843.imageshack.us/img843/4292/norton1.jpg
2: https://img847.imageshack.us/img847/2450/norton2h.jpg

At "1" in link #1, it says (Vo - 0)/2. It is for the branch in which Ix is flowing. Ix is flowing upward - from the ground toward Vo which simply means that Vo is at lower potential than the ground. Therefore, it should be (in my view): (0 - Vo)/2. Do you see the problem? I understand that if we do that the nodal equation will also change.

At "2" it says Vo = -4 V. What does this mean? In what situation does the current flows from ground to some other point (at -ve potential) in real world?

At "3" having Rth = -4 ohm is completely incomprehensible for me. Please help me with it. I know the author gives some explanation. But one ohm resistance is equivalent one volt used per coulomb. So having a -ve sign with resistance doesn't make sense because it's directionless quantity.

At "4", can we take the direction of current CCW instead of CW? (To see what is author doing here in this circuit please have a look on link #2).

At "5", why has the author used -4i when he has used +ve sign for +9i?

Hi again,

It's nice to see you trying to learn all this stuff, and i believe you are making good progress.

If you look at a resistors current vs voltage curve you'll see that as the current rises the voltage also rises, so the slope is positive.
If you look at a negative resistance current vs voltage curve you'll see that as the current rises the voltage decreases, so the slope is negative. It's a different kind of thing than an actual resistance like a 10 ohm resistor, and so has different characteristics. The units might be the same but since the sign is negative it causes different effects than a regular resistor would.

Also, the task in that exercise is basically asking for an input impedance. Dont be surprised to find unusual impedance values for circuits that contain active elements, as they might really be surprising sometimes For example, we can make any resistance you choose look like an infinite resistance simply by modulating both terminals in a particular way. If a circuit had this resistance in series with its input, it might look like the input impedance was equal to that value or somewhere in that neighborhood yet we could design a circuit so that the input impedance is infinite.

In your #5, they are simply following the rule of voltage drops around a close path where the current has an assumed particular direction of flow. There's nothing unusual here. If you follow the path around the loop and do your algebra (multiply the current times each resistance) you'll come out with that very same equation. It's that simple. If you can do this with 'regular' resistors than you can do it with negative resistances too, but the sign will be different that's all.

You also have to realize that sometimes we do things in circuits on paper that we dont do in real life. The most recent theory on this kind of thing is called the "Too much information theory". It basically states that in this universe we have more information than we have reality. A 'consequence' of this is that we can create things on paper that could never possibly exist in real life. Might as well start getting used to this happening quite a bit

 

[PG1995]

Hi

In the following linked pictures I have marked the sections which I have found more troublesome with numbers (1, 2, 3, 4, 5). The aim is to find Thevenin equivalent. Please help me. It would be really kind and nice of you. Thank you

1: https://img843.imageshack.us/img843/4292/norton1.jpg
2: https://img847.imageshack.us/img847/2450/norton2h.jpg

At "1" in link #1, it says (Vo - 0)/2. It is for the branch in which Ix is flowing. Ix is flowing upward - from the ground toward Vo which simply means that Vo is at lower potential than the ground. Therefore, it should be (in my view): (0 - Vo)/2. Do you see the problem? I understand that if we do that the nodal equation will also change.

At "2" it says Vo = -4 V. What does this mean? In what situation does the current flows from ground to some other point (at -ve potential) in real world?

At "3" having Rth = -4 ohm is completely incomprehensible for me. Please help me with it. I know the author gives some explanation. But one ohm resistance is equivalent one volt used per coulomb. So having a -ve sign with resistance doesn't make sense because it's directionless quantity.

Thanks a lot, everyone.

Let me rephrase some of my misunderstandings which are hampering my understanding...

Please have a look on the linked diagrams:
https://img810.imageshack.us/img810/1671/imgef.jpg

1:
The nodal equation used by the author in link #1 in my first post is the one for the Diagram 1. Right?

But in the circuit diagram Ix is flowing upward as shown in Diagram 2. So, in my humble opinion (which I'm almost sure is wrong ) the equation I have written below the Diagram 2 should be applicable because both Io and Ix are flowing toward the same junction.

Do you see the problem I'm having? How can we assume some other direction for the current when we have explicitly been told the direction of the current (which is "Ix" in this case)? Yes, in the case of the branch contain 4 ohm the current will flow downward toward the ground.

2:
The Vo has been found to be -4V. Vo is equivalent to Vab ( where, Vab = Va - Vb ). So Vo = -4V simply tells us that Va is at lower potential for the circuit shown.

3:
Okay, the Rth = -4 ohm.

As the author says the -ve sign tells us that the circuit is supplying power.

(1) P = VI, (2) P=I^2.R, (3) P=V^2/R

So, for the power to be -ve the R should take -ve values because otherwise for (2) and (3) power can never be -ve because I^2 and V^2 can never be negative. Perhaps, when the current I flows into the resistor R from lower potential point and comes out of the higher potential point the resistor is said to be -ve (and perhaps this is only done for the same of mathematical computations!). Generally, the current I enters the resistors from the point of higher potential.

Thank you very much for all your help and time.

 

[MrAl]

Hi,

If i understand your first problem correctly you are asking a question more or less about the polarity of Ix.
You have to realize that although Ix has a polarity of its own, it is not the direction of current flow through that resistor, it's the 'take off' polarity, which is really not the current itself but the direction we want to assume Ix is flowing in for the sake only of the dependent current source to the left. In other words, it is the 'input' polarity to the dependent current source, not any true current flow direction. In other words, Ix is a 'measurement', not a current itself.
Perhaps you could look at Ix as a current sensor that is sensing the true current through that 2 ohm resistor. The current sensor is sort of connected backwards so we have to reverse the polarity for the input to the dependent source. If you look at it that way, if we have say 3 amps flowing through the 2 ohm resistor then Ix has to be made equal to -3 amps. It's just a measurement, and if we turned the sensor (or current meter) around we would get a reading with a different sign.

This is the way dependent sources work. They have the generator part and the measuring part, usually two leads for each part so that's a total of 4 leads. Two of the leads generate a current, but the two sense leads simply measure another current and dont generate anything. So we have one current we are generating with the source, and another current that we are measuring with the input to the source. Something else usually generates the current that we are measuring, and in the case of that circuit there is also an independent source that starts the current flowing. Note the generator part may in fact change the measured current too, but the input side is still just a measurement not a generation of anything.

Sometimes a dependent source that measures current and generates either current or voltage is drawn with four leads, or just three leads where two are the generator part and one sense lead, and sometimes with just two leads as your circuit does and then the measurement is indicated by a variable such as Ix, and that is considered the input sense lead(s).

Here are some examples of dependent sources (see Figs 8e and 8f):
https://www.electro-tech-online.com/attachments/ee3-0008-gif.51874/

 

[PG1995]

MrAl, I offer my belated thanks for all your help.

This is confusing me since I started this thread. I would try to ask it now and try my best to convey my confusion. Please have a look on the following link: https://img708.imageshack.us/img708/3581/img0002ri.jpg

The scan also has my comments there which are an attempt to explain what is troubling me.

I have taken the Circuit #2 from the my post #1; specifically the Circuit #2 could be found here: https://img843.imageshack.us/img843/4292/norton1.jpg (Original link)

As you see the author in the "Original link" says that the circuit on left of the terminals a-b is supplying power but I don't get it. From the viewpoint of current source Io the terminal "a" should be +ve because current source is pointing toward it; that means its +ve terminal is connected with "a". But from the viewpoint of the circuit on the left of terminal a-b the terminal "a" is at -4V potential which I don't get. Do you see my confusion?

Thank you very much for all the help.

 

[Jony130]

I think that there in no confusion here. You simply must remember that you deal with ideal constant current source.
And Ideal constant current source "produces" the current in the direction show by the arrow.Whatever load or network of elements is connected to source, the current pumped by the source into the load always remains same.
And this current is independent of the voltage across the current source.
So even when you short the current source ( no voltage drop) the current still by flow through the current source.

 

[PG1995]

how to nortonize this?

Thank you, Jony.

Please have a see on the following link: https://img339.imageshack.us/img339/3438/imgmw.jpg

You also see my question there. There would be potential drop for Vo. I have also read that when the loop is traversed in the direction of loop current, then the IR term is subtracted; if the loop is traversed in the direction opposite to the direction of current, then IR term is added. Please help me. Thanks.

 

[KeepItSimpleStupid]

Your ENTERING V on the positive side <assuming V flows pos to neg), so the term is subtracted.

So it's really simple. I*R is a voltage. It typically has loop currents in various directions). When you collied with the tip of a curent, that loop current is negative.

If you encounter a voltage source, insert an imaginary arrow from + to -. The rule above applies. Enter on + side, added. Enter on - side, subtracted.

If there is a defined current that you were supposed to find and it's in the opposite direction of your assumed direction, then....?