Ah, finally an answer. But is that what the problem says, or is that something you are definining out of thin air?The negative sense lead is connected to the North end of the sensing resistor.
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. RatchThat is the only way the current (no, charge) flow in the dependent source would be the proper polarity.
Yes, that is a problem.Stop calling for unecessary information, I really can't see your difficulty.
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.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.
Only for the independent current sources. The dependent current source depends on the polarity of the sense leads.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.
Not the polarity of the dependent current source. We need to know the polarity of its sense leads.All other polarities are uniquely determined by these sources and the circuit configuration.
You will get another different value if you reverse the 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.
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. RatchThe 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.
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 .The problem is incomeplete. I used electron flow so that's how the +/- connections were determined.
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
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
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
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
Shouldn't this readI1 = 2*v ; loop8
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.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.
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".Anyway your lecturer is correct, Ratch is mistaken. You have all the information necessary, there are no 'sensing terminals'.
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.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
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.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.
And that is what I did to solve and verify the problem, right?((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).
As far as I know, loop and mesh analysis are one and the same. Show me the difference.I do not know what stage your studies have reached, but the method chosen by Ratch is a more advanced method called mesh analysis.
I have always known it as node analysis.Studiot, I think the method you have used for your analysis is called Branch analysis
No, he did not. He was not trying to do so. But I did.You haven't applied Thevenin's theorem, have you? Maybe I missed it.
The second 'solution' is clearly impossible as all the currents at node D are inward becauseFor 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
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.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.
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: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.
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?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.
Don't know about that, but I get numbers that match up. The circuit is definitely incomplete with regard to the sensing resistor, however. RatchI am still not convinced that the circuit has been copied correctly.
On this page:I have always known it as node analysis.
Yes, things are starting to come together now. RatchNote I3 is now -1.8 which make the voltage (= 2I3) across Ro 3.68 as you have.
by Jake Hertz
by Aaron Carman
by Jake Hertz
by Jake Hertz