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?
When at the node you writeAt "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.
Dependant sources are active circuits.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?
In a negative resistance, applying voltage the current flows in the opposite direction.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.
Why not? He is free to choose their preferred direction while follow the sign conventions.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).
Is negative but not by the current, R is negative.At "5", why has the author used -4i when he has used +ve sign for +9i?
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.
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.
Generally, the current I enters the resistors from the point of higher potential.
At diagram 1, you wrote: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.
When you assume a direction, it's only for the proper equation writing.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.
Wait, wait...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.
Just out of curiosity I was trying to solve the circuit the opposite way. The answer was the same. Given below is the link to the scan of my attempt to solve the circuit the other way around and of my failed attempt at using mesh analysis. Can't I use mesh analysis there? It seems so.
Link:
https://img853.imageshack.us/img853/5296/imgil.jpg
In your mesh analysis, you have 4 unknowns, I1, I2, I3 and Ix. The equations of the second and the third mesh should not include Ix. Then you add the fourth equation:
Ix = I3 - I2. Try to remember that Ix which is the 'resultant' current flowing in the 2 Ohm branch and its direction as indicated is an 'unknown' term, much like I1, I2, I3 and Vo. I think this point is at the center of most of your confusion.
So try to re-write the 3 equations without without using Ix beside of the current source 2Ix (which is used in the first equation). Then add the fourth equation:
Ix= I3 - I2
And let us see what you will get. I am lazy to do it... If the result will not be as expected then... I will join your club
Note: First line above is edited
I think I should write all equations by myself, one by one then write them to you.... because... I am lost
Added:
The thing I discovered why I am lost is that we don't have 3 meshes! Do we?
There is only one NODE (other than the ground) in the circuit.
I think I need to update my glasses soon :???:
So we can apply here only the nodal equation. OK?
Ok... let us take the inverse case... suppose we have a series of voltage sources and resistors forming ONE mesh (loop). Can we write a nodal equation?!
We have here only parallel branches between a and b... so we need at least another node as c to have a loop.
---------- Post added at 21:29 ---------- Previous post was at 21:24 ----------
I should see this since long ... but you know... you are not the only one to miss things
Of course, you can go back to previous problems and check... the meshes and the nodes.
Not at all.1: To apply nodal analysis we must have at least two nodes. Am I correct?
It applies to a single loop --> Only 1 unknown.2: To apply mesh analysis we must have at least two meshes.
1: To apply nodal analysis we must have at least two nodes. Am I correct?
2: To apply mesh analysis we must have at least two meshes.
I think you are correct; though it took me some time to realize that!
Here:
https://img853.imageshack.us/img853/5296/imgil.jpg
We have 4Ω and 2Ω resistors in parallel configuration; though we cannot proceed to find the equivalent resistance because of Ix flowing through 2Ω resistor but it IS still a single loop circuit. Correct?
If I'm correct, then I think it is not possible to apply mesh analysis on a single loop.
So, what do I learn?
1: To apply nodal analysis we must have at least two nodes. Am I correct?
2: To apply mesh analysis we must have at least two meshes.
Not at all.
If you have a single node, there is nothing to solve, all are null.
You need at least two nodes to have something to solve.
It applies to a single loop --> Only 1 unknown.
You are correct, but between these two nodes there must be more than 1 branch, otherwise we don't have a circuit to analyse with one branch ;-)
And even if there are two branches between the two nodes we don't really need to talk about nodal or mesh analysis. There will be just 1 loop current and one branch voltage and we learnt how to solve this situation before we heard of these two analysis.But obviously we can still apply:
I_branch1 + I_branch2 = 0
Vab + Vba = 0
This depends on how we see our analysis. Even if there is one mesh, we will likely need to write the equation of its loop if it has more than 2 nodes. Again and in case the circuit is formed by a single loop that has two nodes only, solving it becomes rather simple.
EDIT: In this post Kerim said: Let us assume A is its positive terminal and B its negative one. If one of these two terminals is connected to the ground of a circuit (its reference node), Va=+6V tells us that the terminal B is the one which is connected to ground. And Va=-6V tells us it is A instead.
I think it should be " Vb=-6V tells us it is A instead which is connected to the ground". Is my thinking correct? I think it's a typo.
3: It applies to a single loop --> Only 1 unknown - In mesh analysis we need to subtract the current of loop from the current of the other. So, if we have a single loop, then we have only one current. What am I missing here?
4: between these two nodes there must be more than 1 branch - What does "between" mean in this context?
2: To apply nodal analysis we must have at least two nodes - In nodal analysis we also need one extra node which could be used as a reference node.
1:Is it a single loop circuit? Please let me know.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?