The negative sense lead is connected to the North end of the sensing resistor. That is the only way the current flow in the dependent source would be the proper polarity.

 

JoeJester,

The negative sense lead is connected to the North end of the sensing resistor.

Ah, finally an answer. But is that what the problem says, or is that something you are definining out of thin air?

That is the only way the current (no, charge) flow in the dependent source would be the proper polarity.

Nope, the dependent current source can be defined to be in any direction you want it to be. Its definition does not depend on the orientation of the sensing leads, but its value does. Now, did the problem itself give a sensing direction for the voltage across the resistor VX? If not, then it is incomplete. Ratch

 

The problem is incomeplete. I used electron flow so that's how the +/- connections were determined.

I think the professor didn't mention the polarities because of the profileration of circuit simulators and this allows the student to wire them either direction and produce two different answers.

Either way, the OP hasn't returned to clarify.

 

Ratch,

Stop calling for unecessary information, I really can't see your difficulty.

Either you can help the original poster by providing at least a start with the Thevenin analysis as he requested and I suggested, or let someone else do it.

All the information required to solve the circuit is available. The polarities of all sources are clearly indicated by their respective arrows. The polarities are, of course, positive at the broad end of the arrow.

All other polarities are uniquely determined by these sources and the circuit configuration.

In the loop analysis I prepared I assumed current directions, but at least one will be wrong because I have all the currents entering node F and none leaving.
This does not matter because one will simple solve as a negative value. Reversal of that arrow will give me the correct direction.

The current in the 2 ohm resistor CF is no different in this respect from any other in this or other linear circuits. Once I know the sign of my I4 I will know the polarity of the resistor.

 

studiot,

Stop calling for unecessary information, I really can't see your difficulty.

Yes, that is a problem.

Either you can help the original poster by providing at least a start with the Thevenin analysis as he requested and I suggested, or let someone else do it.

Thevenin or not, we still need the polarity of the sense leads. I welcome someone else also do this problem as a cross check. Joe said that he is trying to get the information I asked about. In the mean time, I will compute two solutions with the sense leads one way and reversed if I get some time this evening.

All the information required to solve the circuit is available. The polarities of all sources are clearly indicated by their respective arrows. The polarities are, of course, positive at the broad end of the arrow.

Only for the independent current sources. The dependent current source depends on the polarity of the sense leads.

All other polarities are uniquely determined by these sources and the circuit configuration.

Not the polarity of the dependent current source. We need to know the polarity of its sense leads.

In the loop analysis I prepared I assumed current directions, but at least one will be wrong because I have all the currents entering node F and none leaving.
This does not matter because one will simple solve as a negative value. Reversal of that arrow will give me the correct direction.

You will get another different value if you reverse the sense leads.

The current in the 2 ohm resistor CF is no different in this respect from any other in this or other linear circuits. Once I know the sign of my I4 I will know the polarity of the resistor.

The current is the resistor is not at issue. We have to know how the dependent current source is sensing the voltage across that resistor. Ratch

 

JoeJester,

The problem is incomeplete. I used electron flow so that's how the +/- connections were determined.

I would advise you not to use the direction of the charge carriers as a direction. Use the mathematical convention as outlined in this thread http://forum.allaboutcircuits.com/showthread.php?t=11410&highlight=conventional&page=4 .

I will try to solve your problem both ways if I get some time this evening. Might be late before I post. Ratch

 

JoeJester,

OK, let's try to solve your problem. First define the loop currents and voltages across the dependent and independent current sources.

Northwest loop = I1
North loop = I2
Southwest loop = I3
South loop = I4
East loop = I5
Dependent current source voltage defined plus at arrow = x
Independent 2A current source voltage defined plus at arrow = w
Independent 8A current source voltage defined plus at arrow = y
Voltage across Rx assumed plus at north end = v
Negative sense lead for dependent source connected to south end of Rx, plus lead to north end

Nine variables need nine equations to solve.

-4*I1 + w -2(I1-I3) = 0 ; loop1
-2*I2 - w - v - 2*(I2-I4) = 0 ; loop2
x-2*(I3-I1) - 4(I3-I4) = 0 ; loop3
-4(I4-I3) - 2(I4-I2) - y = 0 ; loop4
y + v -6*I5 = 0 ; loop5
I1 - I2 = 2 ; loop6
I4 - I5 = -8 ; loop7
I3 = 2*v ; loop8
v = 2(I2-I5) ; loop9

Now we set up the equations in a matrix

I1   I2   I3   I4   I5    v    w    x    y  
-6    0    2    0    0    0    1    0    0   =  0
 0   -4    0    2    0   -1   -1    0    0   =  0
 2    0   -6    4    0    0    0    1    0   =  0
 0    2    4   -6    0    0    0    0   -1   =  0
 0    0    0    0   -6    1    0    0    1   =  0
 1   -1    0    0    0    0    0    0    0   =  2
 0    0    0    1   -1    0    0    0    0   = -8
 0    0    1    0    0   -2    0    0    0   =  0
 0    2    0    0   -2   -1    0    0    0   =  0

Using my hand calculator to solve, I get:

I1 = -3.16
I2 = -5.16
I3 = -13.28
I4 = -9.84
I5 = -1.84
v = -6.64
w = 7.60
x = -34
y = -4.40

So the voltage across Vo is -1.84*2 = -3.68 volts.

Now find the Thevenin impedance. First find the open circuit voltage which we can simulate by assigning a very high resistance in the I5 loop. Then the equations are

I1   I2   I3   I4   I5    v    w    x    y  
-6    0    2    0    0    0    1    0    0   =  0
 0   -4    0    2    0   -1   -1    0    0   =  0
 2    0   -6    4    0    0    0    1    0   =  0
 0    2    4   -6    0    0    0    0   -1   =  0
 0    0    0    0  1E8    1    0    0    1   =  0
 1   -1    0    0    0    0    0    0    0   =  2
 0    0    0    1   -1    0    0    0    0   = -8
 0    0    1    0    0   -2    0    0    0   =  0
 0    2    0    0   -2   -1    0    0    0   =  0

The solution is v = -14, y= -78 for a Thevenin voltage of -92

Then the short circuit current.

I1   I2   I3   I4   I5    v    w    x    y  
-6    0    2    0    0    0    1    0    0   =  0
 0   -4    0    2    0   -1   -1    0    0   =  0
 2    0   -6    4    0    0    0    1    0   =  0
 0    2    4   -6    0    0    0    0   -1   =  0
 0    0    0    0    0    1    0    0    1   =  0
 1   -1    0    0    0    0    0    0    0   =  2
 0    0    0    1   -1    0    0    0    0   = -8
 0    0    1    0    0   -2    0    0    0   =  0
 0    2    0    0   -2   -1    0    0    0   =  0

Which solves to I5 = -2.09 . Therefore the Thevenin impedance is -92/-2.09, which is a resistance of 44.02 ohms. Setting up the voltage divider, we get -92(2/(44.02+6) = -3.68, which agrees with the direct calculation.

studiot,

Now lets see what we get when we reverse the sense leads.

I1   I2   I3   I4   I5    v    w    x    y  
-6    0    2    0    0    0    1    0    0   =  0
 0   -4    0    2    0   -1   -1    0    0   =  0
 2    0   -6    4    0    0    0    1    0   =  0
 0    2    4   -6    0    0    0    0   -1   =  0
 0    0    0    0   -6    1    0    0    1   =  0
 1   -1    0    0    0    0    0    0    0   =  2
 0    0    0    1   -1    0    0    0    0   = -8
 0    0    1    0    0    2    0    0    0   =  0
 0    2    0    0   -2   -1    0    0    0   =  0

Notice I changed the sign of v in the eighth equation. Solving we get:

I1 = 8.08
I2 = 6.08
I3 = 25.54
I4 = 4.46
I5 = 12.46
v = -12.77
w = -2.62
x = 119.23
y = 87.54

Quite a difference. Anyone who ignores the sense leads does so at their own peril. Ratch

NOTICE: This morning I found a mistake in my equations. I have since corrected this mistake and what you see is my latest revision. Ratch

 

Thanks guys. You really did a wonderfull job at least to get me started. I am happy to associate with you all. (The bad news is that I do not have 24hrs access to internet).

Unfortunately, my prof is not helping matters. He has insisted that all information required to tackle the exercise were provided. Besides, the polarities of all sources (including the dependent current source pointing northwards) are clearly indicated by their respective arrows.

The fact is, I sincerely rely and depend on your expert advise and solution to this problem. Kindly bail me out.

Favour

 

Hello favour, glad to see you are back with us.

There is another thread with your problem - a fellow student?

http://forum.allaboutcircuits.com/showthread.php?t=13101

Anyway your lecturer is correct, Ratch is mistaken. You have all the information necessary, there are no 'sensing terminals'.

I am of course presuming that you have copied the question down correctly - Please check?

What you have posted is capable of solution and I have attached one to this post, based upon my earlier notation. Please note I have made a minor amendment to doc3, which is the version you should be following.
This solution has been tested for self-consistency.

In terms of the question of polarity of the driving voltage for the dependent current source.

By Definition The direction of the current (my I4) must be such that an increase in the voltage dropped across resistor CF causes an increase in the dependent sourced current, in the direction of the arrow in the diamond. Since we know the direction of I4 we can assign positive and negative ends to this resistor and calculate the voltage drop across it.

As I predicted, my direction for I4 was incorrect. However this has been 'taken care of' in the solution and the correct direction is evident.

 

Just looking at your numbers, Ratch

I1 = 2*v ; loop8

Shouldn't this read

I3=2*v ; loop 8

I think you have it correct in the matrix itself.

I note that the numbers and signs are both consistent for your first solution, but only the numbers match on the second, the signs being opposite.

 

Studiot,
Manny thanks for the solution.

Just for confirmation,the question as stated on the original attached is correct. This I have repeated as follows:

((a)Determine the voltage developed accross the resistor labeled Ro in the circuit of figure Q1 through an application of Thevenin's theorem. (b) Verify your solution by applying the loop analysis method).

Favour

 

It's the circuit, not the question I was querying.

Your original Word document misssed one of the resistor values (my DE). This was added in the pdf.

I did wonder what the little boxes by the diamond were for? And if you (or your lecturer) had that diamond current arrow the correct way round?

A further thought. The method I have shown is what I refer to as 'loop analysis'. That is based on Kirchoffs Voltage Law around a closed loop, using actual currents directly.

I do not know what stage your studies have reached, but the method chosen by Ratch is a more advanced method called mesh analysis. This will not make much sense if you have not yet covered this topic. Essentially you assign a ficticious loop current around each entire loop in the mesh and work on these. The actual currents are made up of contributions from each relevant loop.

 

Studiot, I think the method you have used for your analysis is called Branch analysis; see http://www.allaboutcircuits.com/vol_1/chpt_10/index.html.

Loop analysis and mesh analysis are the same thing, which is the method Ratch has used.

Branch analysis usually has to use a couple of loop equations to get the required number of independent equations, but it's still Branch analysis.

The original poster's problem requirements are: "((a)Determine the voltage developed accross the resistor labeled Ro in the circuit of figure Q1 through an application of Thevenin's theorem. (b) Verify your solution by applying the loop analysis method)."

You haven't applied Thevenin's theorem, have you? Maybe I missed it.

And, you used the Branch method instead of the loop method as the problem required, so even though you solved the problem, you haven't done it the way the problem specified.

 

Favour,

Unfortunately, my prof is not helping matters. He has insisted that all information required to tackle the exercise were provided. Besides, the polarities of all sources (including the dependent current source pointing northwards) are clearly indicated by their respective arrows.

Your prof is just flat out wrong. If the dependent current source were controlled by an independent current, then he would be correct. But it is a dependent current source controlled by a voltage, and the sense of the voltage must be specified, or different value will be obtained in the solution.

studiot,

Anyway your lecturer is correct, Ratch is mistaken. You have all the information necessary, there are no 'sensing terminals'.

Of course there is. How could there not be? The dependent current source is measuring the voltage across Rx, whether mathematically or physically. You have to know how to hook up or interpret the sense leads. Did you disregard the link I sent showing how a voltage controlled current is supposed to be defined? Your denial does not make "sense".

By Definition The direction of the current (my I4) must be such that an increase in the voltage dropped across resistor CF causes an increase in the dependent sourced current, in the direction of the arrow in the diamond. Since we know the direction of I4 we can assign positive and negative ends to this resistor and calculate the voltage drop across it

You are arbitrarily defining the polarity of the sense voltage. The problem could just as well have defined them in the opposite direction. In your case , you hooked the negative lead of sense lead to the north end of the resistor. That is OK, because there was no definition in the problem, but is a arbitary choice which was opposite to the choice I made. Neither of us is right or wrong.

Shouldn't this read

I3=2*v ; loop 8

I think you have it correct in the matrix itself.

I note that the numbers and signs are both consistent for your first solution, but only the numbers match on the second, the signs being opposite.

Yes, I corrected the matrix, but not the equations. Sorry for the confusion. Notice the last calculation was done assuming the negative sense lead was on the north side of the resistor.

The Electrician,

((a)Determine the voltage developed accross the resistor labeled Ro in the circuit of figure Q1 through an application of Thevenin's theorem. (b) Verify your solution by applying the loop analysis method).

And that is what I did to solve and verify the problem, right?

studiot,

I do not know what stage your studies have reached, but the method chosen by Ratch is a more advanced method called mesh analysis.

As far as I know, loop and mesh analysis are one and the same. Show me the difference.

The Electrician

Studiot, I think the method you have used for your analysis is called Branch analysis

I have always known it as node analysis.

You haven't applied Thevenin's theorem, have you? Maybe I missed it.

No, he did not. He was not trying to do so. But I did.

studiot,

Your node analysis is faulty in that it does not take into consideration of the components between the nodes. For instance, there is a 4 ohm resistor between node B and D. What if it were 8,10, or 16 ohms? Where is that accounted for the the solution matrix? How about the other components that are missing from the matrix. Anyway I did a node analysis on the circuit both ways with respect to the sense leads and it agrees with my loop analysis results.

For the negative sense lead on the north side of the resistor"

Vb = 86.92
Vc = 74.77
Vd = 119.23
Ve = 84.31
Vf = 87.54
Iv = 25.54

For the negative sense lead on the south side of the resistor:

Vb = -21.36
Vc = -11.04
Vd = -34.0
Ve = -13.96
Vf = -4.40
Iv = -13.28

If you are interested in the solution matrix, I can post it. Ratch

 

I will return to the question of terminology in another post as I do not wish the thread to degenerate into a squabble over definitions.

For the negative sense lead on the north side of the resistor"

Vb = 86.92
Vc = 74.77
Vd = 119.23
Ve = 84.31
Vf = 87.54
Iv = -13.28

For the negative sense lead on the south side of the resistor:

Vb = -21.36
Vc = -11.04
Vd = -34.0
Ve = -13.96
Vf = -4.40
Iv = -13.28

The second 'solution' is clearly impossible as all the currents at node D are inward because
Vb>Vd and Vb>Vd in the sense of more positive than.

Your node analysis is faulty in that it does not take into consideration of the components between the nodes. For instance, there is a 4 ohm resistor between node B and D.

If you look carefully at my first loop equation DBCHGED it starts with the term 4I1. This take into account the 4 ohm resistor and would be a different value for a different resistor.

Let's stop squabbling on this and co-operate.

I am still not convinced that the circuit has been copied correctly.

 

studiot,

The second 'solution' is clearly impossible as all the currents at node D are inward because
Vb>Vd and Vb>Vd in the sense of more positive than.

Would you like to read revise "Vb>Vd and Vb>Vd? Anyway, I show the following for the currents values and directions at node D:

For north negative sense lead:
I1 = 8.08 outward
I5 = 17.32 outward
Ix = 25.4 inward

For south negative sense lead:
I1 = 3.16 inward
I5 = 10.12 inward
Ix = 13.28 outward

So the node, loop, and Thevinen calculations agree with each other.

If you look carefully at my first loop equation DBCHGED it starts with the term 4I1. This take into account the 4 ohm resistor and would be a different value for a different resistor.

You are right, I did not notice that you were doing a hybrid analysis. However, your calculations don't agree with mine. Check Loop EFCHGE 2I7 -2I4 + I3 + 2I3 -4I6 = 0 . Shouldn't the I3 term have a coefficient of 6?

I am still not convinced that the circuit has been copied correctly.

Don't know about that, but I get numbers that match up. The circuit is definitely incomplete with regard to the sensing resistor, however. Ratch

 

Vb>Vd twice - yes a silly slip - one should have been Ve>Vd.

This then agrees with your current directions.

I take it that Ix is the current in the dependent current source.

If so do you think a solution which reverses the direction shown in the question is acceptable?

If you look at my matrix in solution.doc I have indeed got the coefficient of I3 as 6. I thought that I had managed to amend favourhomework.doc to reflect this. Obviously I was mistaken. I am still struggling with the mechanics of the forum. I asked to have the erroneous doc removed and was told that I could amend it myself, which I did try.

Incidentally if you look carefully at my matrix I too have reversed the direction of the dependent source, with 2VX= +4I4, instead of -4I4. If you put a coefficient of -4 instead you get your other solution.

This is the reason I am querying the original copying.

 

Here is the solution with the dependent current the other way round.

Note I3 is now -1.8 which make the voltage (= 2I3) across Ro 3.68 as you have.

 

I have always known it as node analysis.

On this page:
http://www.allaboutcircuits.com/vol_1/chpt_10/index.html

Compare the details of the Branch method with the Node method, and you will see that Studiot is using the Branch method.

 

studiot,

Note I3 is now -1.8 which make the voltage (= 2I3) across Ro 3.68 as you have.

Yes, things are starting to come together now. Ratch