# thevenin equivalent example problem

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#### PG1995

##### Full Member level 5 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: http://img843.imageshack.us/img843/4292/norton1.jpg
2: http://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?

#### Miguel Gaspar

##### Advanced Member level 1 Please don't confuse about the setting of the equation:
i the three first branches the consideration is te current flowa from a to b: so Is + (Vo - 0)/4 + (Vo - 0)/2
The last one flows from b to a it is opposites a to b so it is negative.

That is OK
The beauty of Kirchoff LAW is that you don't need to think laike you does.
Only follow the procedure.
In 2 the result says to you that Va = Vo is opposite than you believe and current realy flows form b to a.
In 3 is a mathematical result it is OK and for your information ther are negative resistances. They exist.
Whe the result gives negative current it means that the current is in the opposite way than you stablish in your calculus.

5) he uses the circuit 4.35 d to write the equation

• milan.sorsa and PG1995

### PG1995

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### milan.sorsa

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#### KerimF

##### Advanced Member level 4 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.
Let us remember what a nodal equation is. It says that the sum of all currents of its branches is equal to zero. But for this to be true we have to decide first on the positive direction of these currents; inward or outward. I noticed that in equation 4.10.1, the first term shows the current 2x is added as it is (hence positive) and its direction is outward (that is going out from the node a ). So in the branch of 2 Ω, the positive current (in the equation) should be from the node a to ground (in this equation). This means (Vo-0)/2Ω. This has nothing to do with the direction of Ix.

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?
The sign itself (alone) doesn't give useful information. For example one can say that the voltage of a battery is +6V or -6V. The sign will tell us something new only if we know also how the battery is connected to a circuit. 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.

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.
Me too, the first time I heard of a negative resistance I thought it denotes an unreal resistance (Like an unreal person, Santa Claus ) . I had to wait many years before understanding how a negative resistance could exist. So I will try my best to explain it to you now ;-) First, let us see how a positive resistance acts. If the voltage increases on a (positive) resistor, its current increases as well. And the inverse is true, if the resistor current increases, its positive voltage increases too. But you will learn later that a device may start to act as a positive resistance (though not necessarily constant) till some voltage. Then, its current increases (by itself) while its voltage decreases!!! And when its current reaches a certain limit, the device continues acting as a positive resistance again; that is both its voltage and current continue to increase together as expected normally. This could be seen as a sort of transient breakdown in which the device acts as a negative resistance; so the slope of its V/I curve becomes negative in this region. An example of such a device (having a V-I region with a negative slope) is the DIAC and the unijunction transistor (UJT). If you are curious, you can see on figure 6, the negative resistance region on the curve V versus I of the later at:
Unijunction Transistors

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).
Of course we can. And the equation would be (if CCW):
-4i + 9i -10 = 0
But in this case i=+2A because the direction of the 'real' current is also CCW and that is why we got i=-2A when we assume i be CW.

To write the equation, i of the loop is assumed positive for the chosen direction (though its end value could be positive or negative according to its real direction). When it (the current i) enters a resistor, a term iR is added. If it enters a voltage source from its positive terminal then +V is added otherwise it will be –V.

At "5", why has the author used -4i when he has used +ve sign for +9i?
As it is pointed out earlier -4i is in fact an iR term. But the value of R in this case happens to be -4 (not positive as in the usual case)

Kerim

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• PG1995

### PG1995

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#### _Eduardo_

##### Full Member level 5 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.
When at the node you write
Σ Ik = 0​
all the currents must have the same convention, all coming into the node or all leaving the node. Solving the system, the current sign will tell you what was the positive direction

Then, that equation can be written as:
2 Ix + (vo-0)/4 + (vo-0)/2 + (-1) = 0​
or
- 2 Ix + (0-vo)/4 + (0-vo)/2 + 1 = 0​
They are the same

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?
Dependant sources are active circuits.
Negative impedance converter - Wikipedia, the free encyclopedia

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.
In a negative resistance, applying voltage the current flows in the opposite direction.
Impossible with passive components ... But these are actives.

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).
Why not? He is free to choose their preferred direction while follow the sign conventions.

At "5", why has the author used -4i when he has used +ve sign for +9i?
Is negative but not by the current, R is negative.

• PG1995

### PG1995

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#### PG1995

##### Full Member level 5 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: http://img843.imageshack.us/img843/4292/norton1.jpg
2: http://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:
http://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.

#### KerimF

##### Advanced Member level 4 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.
So the question now is to find out the difference between the term (Vo-0)/2Ω and -Ix (the minus sign is added since it is defined as inward).
I am not sure why my brain refuses to find the direct and clear answer that says the latter is not allowed to be a term in the equation. I hope you will find it for me after you read my answer.
So here is the indirect explanation:
When we write Ix in the equation, we mean that it is a current source. In other words, we say that Ix is independent of the resistor 2Ω. On the other hand, we are sure that Ix in the 2Ω branch is not meant to be constant (like the dependent current source 2Ix) but actually depends on both Vo and the branch resistance. So naturally, we have to choose (Vo-0)/2Ω and not -Ix to confirm that Ix in this branch is indeed not a current source (like the current source 2Ix).

Generally, the current I enters the resistors from the point of higher potential.
As you said P=VI. So P is positive when V and I have the same polarity (same sign) and in this case it is said that the power is consumed. Otherwise, P is negative hence generated.
In case of a real resistor, V and I have always the same sign that is a positive current produces a positive voltage on its two terminals and a negative current produces a negative voltage as well. So a real resistor (modelled as a positive resistance) always consumes power. As you known the inverse applies to DC voltage and current sources (that are power suppliers/generators) on which the polarities of their voltage and current are opposite. Sorry for repeating all this, since I believe that it is already clear to you so far... My idea is to tell you to remember this (and others) while your study in electricity and later in electronics deals with new circuits and devices. Now you feel somehow confused because your study is oriented to be more mathematical/abstract (like when we had to learn addition, substraction, multiplication...etc, without knowing why). So the real fun starts when your abstract skills will reach someday a level that helps you model real circuits (you will design) and alter any part on them to get what you expect from them. Meanwhile I am afraid that you need to be patient and keep asking questions while learning (even to yourself if no one hears you) because this is the sign that you understand what is presented to you (for instance, who look to just memorize what they learn don't need to ask any question... any data recorder, small or huge, never asks ).
I am sure that someday you will be able to clarify to others how a negative resistor could be understood in a way better than mine (because I didn't have the chance to ask this question in the first place, then later it became one of the many notions I use in my equations or a circuit simulator takes care of it ;-) ).

For instance, do you know when you will study how to analyze AC circuits? Now your circuits are DC and all their values are also not function of time. I think this is the reason why it is hard now (actually it is impossible, so I hope one can correct me if I am wrong) to give a real example of a 'passive' device having a negative resistance when all voltages and currents are constant with time. So you have a very good reason to ask on how R=-4Ω could be real.

Kerim

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• PG1995

### PG1995

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#### _Eduardo_

##### Full Member level 5 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.
At diagram 1, you wrote:
Io = (Vo-0)/2 + (Vo-0)/4 + 2Ix ; Good!​

At diagram 2, you wrote:
Io + Ix = (Vo-0)/4 + 2Ix ; Good2!​

But you know: Ix = (0-Vo)/2 ==> Both eq1 and eq2 are the same, just replace (Vo-0)/2 by -Ix

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.
When you assume a direction, it's only for the proper equation writing.
After solving the system, if the current value is positive means that the real direction is which had been assigned, and if negative the opposite.
Both sollutions are valids.

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.

OK

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.
Wait, wait...
P = VI is not an unsigned multiplication If the current and voltage are positives according the image, P will be positive and the is absorbing power.
Following this convention, in your circuit the current is positive and the voltage is negative ==> P < 0 ==> the block is supplying power.

• PG1995

### PG1995

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#### PG1995

##### Full Member level 5 Thank you, Kerim, Eduardo.

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:
http://img853.imageshack.us/img853/5296/imgil.jpg

It also prompts me ask a new question from practical point of view. Suppose I have the circuit "a" of Figure 4.35. I connect the +ve terminal of the current source to "a" terminal of the circuit and -ve to the "b" terminal. The current source is set at 10A. I understand that the circuit is designed to deliver the power. But won't the current source push the 'current' through "a" terminal and the pushed 'current' will get out of "b" terminal of the circuit. Hence, "a" terminal would be at higher potential. I know I'm not making much sense. Perhaps, you would understand what I'm trying to say. If one pump (let's say circuit "a") is pushing water toward the right of the screen through a pipe, which means that the pump is delivering power. If I attach another pump can push the water toward the left of the screen more powerfully, then I think overall flow would be toward the left.

#### KerimF

##### Advanced Member level 4 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

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#### PG1995

##### Full Member level 5 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:
http://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
Thank you. I understand I have been told not to use Ix but let's play with it for some time.

By the way, I had the wrong equations for loop 2 and loop 3. For loop 2 it should be:
4(I2 - (-2Ix)) + 2(I2 - I3) = 0

For loop 3 it should be:
2(I3 - I2) = 0
I3 - I2 = 0 [We know the value of I3 because it's equal to Io which is 1A in this case]
Using I3 = Io = 1A
1 - I2 = 0
ERROR...

We have these equations now:
For loop 1: 4(-2Ix - I2) = 0

For loop 2: 4(I2 - (-2Ix)) + 2(I2 - I3) = 0 [But we have been given the value of current flowing through 2 ohm resistor which is Ix which means that (I2 - I3) = Ix ERROR, it would mean that (I2 - 1 = Ix) ]

Do you also find those "Errors" above to correct?

#### KerimF

##### Advanced Member level 4 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?

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#### PG1995

##### Full Member level 5 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?
Thanks a lot, Kerim.

To me, it seems we do have three meshes. We have three loops and those loops do not contain any other loop(s) within them, so they are meshes. What am I missing now?:roll: Help!!!

#### KerimF

##### Advanced Member level 4 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.

#### PG1995

##### Full Member level 5 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.
I think you are correct; though it took me some time to realize that! Here:
http://img853.imageshack.us/img853/5296/imgil.jpg

We have 4Ω and 2Ω resistors in parallel configuration; though we cannot find 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.

#### KerimF

##### Advanced Member level 4 As far as I remember the rule in mesh analysis, it is:
The sum of all voltages in a loop equals zero. (I think it is better written on your book)
This implies that a mesh should have 3 nodes in the least.

For nodal analysis:
The sum of all current at a node equals zero.
This implies that a node should have 3 branches in the least.

So if we have in a circuit 2 nodes and 2 branches only, there would be no need to apply either analysis Last edited:

#### _Eduardo_

##### Full Member level 5 1: To apply nodal analysis we must have at least two nodes. Am I correct?
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.

2: To apply mesh analysis we must have at least two meshes.
It applies to a single loop --> Only 1 unknown.

#### KerimF

##### Advanced Member level 4 1: To apply nodal analysis we must have at least two nodes. Am I correct?
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

2: To apply mesh analysis we must have at least two meshes.
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.

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#### potdarsagar

##### Newbie level 6 i may help u.listen first assume direction of currents.then refer arrow direction in ur example.if its in ur same direction its positive , if its opposite its negative.if u r getting opposite to it...then could be misprint in book.

#### PG1995

##### Full Member level 5 I think you are correct; though it took me some time to realize that! Here:
http://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.
Hi again, Kerim, Eduardo,

Actually I have been quite busy for the last some days, therefore didn't get time to finish this stuff. I would request you both to help me understand this completely. Actually I'm just curious to know when and where we can apply mesh and nodal analyses. Perhaps, you can help me to understand this with some simple circuit diagrams. Thanks.

I have highlighted the parts which I found more troublesome after reviewing the material.

1: I don't think I'm correct where I say this: but it IS still a single loop circuit

It is not a single loop circuit even if we were able to find the equivalent resistance of 4Ω and 2Ω. The circuit would look like this:
http://img231.imageshack.us/img231/6141/imgayr.jpg
Is it a single loop circuit? Please let me know.

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.

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?

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.

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#### KerimF

##### Advanced Member level 4 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.
You are right... it's a typo and this means that you have a great talent to analyse... circuits and texts as well.
Now I will re-read the posts with the hope I can get exactly what you are looking for as answers 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?
This means that the currents of the outside meshes have a zero value. This is how I see it :wink:

4: between these two nodes there must be more than 1 branch - What does "between" mean in this context?
Two nodes means there is a potential difference. A branch between the two nodes may be seen as a current path. But for 'DC circuits', if there is one path only its current has to be zero. Therefore, to have a current there must be one more path (branch) in the least between the two nodes.

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.
Yes two nodes at least, and one of them could be a reference node. The circuit of your question (1 is a good example of this.

1:Is it a single loop circuit? Please let me know.
You are right... as usual :wink: And it will be up to you which equations you like to work with. And you will get always the same result. The difference is that the road you choose will be longer or shorter than the other one(s). For instance, now you are learning many ways to solve even the same circuit, later no one will stop you in using any of them... hmmmm... or none of them as well... by using a simulator :grin:

Kerim

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