Equivalent resistance and Dependent Sources

m237b

Joined Apr 8, 2013
18
Sorry for confusion, I thought it is a common thing...I mean I learned from some on-line course that when doing nodal analysis with KCL you start by default, defining all currents as coming out of the node in question. Now, if some of the currents do have direction already defined then you reflect it accordingly but those that don't you define as the coming of the node current. I understand that there are obviously more ways of doing it. I have to see how to inject the diagram images in here so I could make it more visual, apologize again for confusing the matters here.
 

WBahn

Joined Mar 31, 2012
29,932
Sorry for confusion, I thought it is a common thing...I mean I learned from some on-line course that when doing nodal analysis with KCL you start by default, defining all currents as coming out of the node in question. Now, if some of the currents do have direction already defined then you reflect it accordingly but those that don't you define as the coming of the node current. I understand that there are obviously more ways of doing it. I have to see how to inject the diagram images in here so I could make it more visual, apologize again for confusing the matters here.
The problem with just letting the equation define it is what happens when you get to the node on the other side of that component and now write I10 as the current leaving that node? You now have I10 defined as going in both directions. It HAS to go into one node and out of the other, so it needs to be indicated which it is.

You can either annotate the diagram or just write out, like I did, how you are defining the direction of things. But let's get out of this rabbit hole. Your initial equation is correct and we are agreed on what the definited direction of I10 is.

Do you understand the problem with your next step involving the sign error associated with Ix and Vab/30Ω?
 

m237b

Joined Apr 8, 2013
18
I think I'm getting it for Ix....so if Ix is negative (-Vab/R1) it means in reality it is actually leaving the node, then Iab is getting into the node ? Now the third one to understand is the one generated by the Isource...which is on the diagram going into the node but still will have to be positive
 

m237b

Joined Apr 8, 2013
18
Still confusing.....just looked again....since the direction for Ix is defined in assignment it still has to be into the node but with negative value. Then Isource defined as well into the node...but with positive value (since it is a source?) and Iab then has to have positive value and out of the node. All that is when we have Vab connected to A & B terminals with + to A and - to B ....is it better picture now ?
 

screen1988

Joined Mar 7, 2013
310
If you apply a 30V voltage across A and B with the positive pole connected to B and negative connected to A.
ix= 30V/30Ω = 1A
The current through current source will be 2ix = 2A
The current through 10Ω resistor will be 30V/10Ω = 3A
Apply KCL at B node:
The current from B go into the circuit will be sum of three currents above and have value 6A
=> ZAB = VAB/ I = 30V/6A = 5Ω
 

m237b

Joined Apr 8, 2013
18
@screen1998....the thing is that I don't get 6A as you say I would. That is because if I apply Vab=30V (+B/-A) the currents are as follows Ix=1A, 2Ix=-2A, I10=-1A (Ix+2Ix=-1)
 

WBahn

Joined Mar 31, 2012
29,932
Still confusing.....just looked again....since the direction for Ix is defined in assignment it still has to be into the node but with negative value. Then Isource defined as well into the node...but with positive value (since it is a source?) and Iab then has to have positive value and out of the node. All that is when we have Vab connected to A & B terminals with + to A and - to B ....is it better picture now ?
Let's take it step by step.

Defining I10 to be the current going downward in the 10Ω resistor and applying KCL to node A, we have

- Ix - 2Ix + I10 = 0

Which is what you started with. Life is good, so far.

However, as soon as we connect an external source, Vab, we have added another path that has to be taken into account. If we apply source Vab (i.e., Va - Vb) then the current flowing out of it will be Iab. Note that, by the passive sign convention, Iab flows OUT of the source at Node A and back INTO it at Node B. Hence your node equation becomes

- Ix - 2Ix + I10 - Iab = 0 <== THIS IS THE FIRST MISTAKE YOU MADE

It's critical to note that the equation you came up with above applies to the circuit when there is NO test source attached to it. As soon as you apply the test source you've changed the circuit and you need to analyze the circuit you have, not the one you had.

If we apply Vab (i.e., Va - Vb) then the current in the 30Ω resistor will be Vab/30Ω flowing RIGHT to LEFT because Va is the voltage on the RIGHT side of the resistor and Vb is the voltage on the LEFT side of the resistor. But Ix is defined as the current flowing LEFT to RIGHT in the resistor, hence

Ix = -Vab/30Ω <== THIS IS THE SECOND MISTAKE YOU MADE

Vab is also the voltage applied across the 10Ω resistor and Vab/10Ω will be the current flowing DOWN in the resistor because Va is the voltage at the TOP of the resistor and Vb is the voltage at the BOTTOM of the resistor. Since this is how I10 is defined

I10 = Vab/10Ω

Plugging these into the original expression, we have

-[Ix] - 2[Ix] + [I10] - Iab = 0
-[-Vab/30Ω] - 2[-Vab/30Ω] + [Vab/10Ω] - Iab = 0
Vab/30Ω + 2Vab/30Ω + Vab/10Ω - Iab = 0

Vab(1/30Ω + 2/30Ω + 1/10Ω) = Iab
Vab(1 + 2 + 3) = 30Ω*Iab
Vab(6) = 30Ω*Iab

Rab = Vab/Iab = (30Ω/6) = 5Ω

Be sure you understand each step in the above solution.
 

WBahn

Joined Mar 31, 2012
29,932
@screen1998....the thing is that I don't get 6A as you say I would. That is because if I apply Vab=30V (+B/-A) the currents are as follows Ix=1A, 2Ix=-2A, I10=-1A (Ix+2Ix=-1)
If Vab=30V, then the A node is 30V greater than the B node. The very notation Vab means "the voltage at Node A minus the voltage at node B" (assuming we are case-insensitive).

If Ix=1A, then how is it remotely possible that 2Ix=-2A?

If there is 30V across a 10Ω resistor, then how can the current be 1A, regardless of sign?
 

m237b

Joined Apr 8, 2013
18
Wow....trivial looking circuit took so long to understand. Thanks so much. I can see my mistakes now. One would be treating two DIFFERENT circuits (second one when we attach Vab) as one ...since I operateded the same quotation. Second grave mistake was with signs when -[Ix] doesn't just mean -Vab/30 but actually -(-Vab/30) ...important difference...so looks like first "-" would be applied for direction of a current and a second "-" for a value based on the Voltage signs applied ? not sure how to define it properly or formulate a rule out of it. Thanks a lot for your help !!
 

WBahn

Joined Mar 31, 2012
29,932
Is the stuff in the right hand generated by your simulator or is it stuff you typed in?

Clearly Ix and Isource have to have the same sign (if k is positive). If that pane is generatd by the simulator, then I'm suspicious of whether they got their dependent current source modelled correctly.
 

m237b

Joined Apr 8, 2013
18
It is generated by the simulator. And K is defined as positive value. As well this dependent source has a value it is controlled by and as you can see it is Ix.
 

WBahn

Joined Mar 31, 2012
29,932
Im not sure if I follow your meaning completely, but I think you are at least in the ballpark.

The minus sign in

-Ix

in the KCL equation mearly means that Ix is defined as a current flowing into the node we are summing and we want to sum up the currents flowing out of the node, so we need that minus sign.

Next, I added the square brackets

-[Ix]

just to clearly separate the minus sign from the variable Ix.

Then, we had

Ix = -Vab/30Ω

because Vab will produce a current in the opposite direction of the direction that Ix is defined in. When you then substitute that into the previous expression you get

-[-Va/30Ω]

which reduces to

Va/30Ω

But it is important to realize and understand that each minus sign means something different and is there for a different reason.

If the arrow on the diagram for Ix been turned the other way around, then neither minus sign would be there. Of course, if the arrow for Ix was going the other direction, then the coefficient for the current source would have needed to be -2 or, alternatively, the direction of the current source flipped.

If the arrow was the way it is but we had decided to sum the currents into the node instead of the currents out of the node (KCL doesn't care which) then the first minus sign would not have been there but the second would have.
 
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