Node-voltage Method, some misconception?

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The discussion centers on the Node-voltage Method and its misconceptions, particularly regarding the necessity of essential nodes. It emphasizes that any node can be chosen as a reference point (GND), even if it connects only two components, contradicting the belief that essential nodes must connect at least three devices. Participants clarify that the node-voltage method and nodal analysis are essentially the same, both relying on Kirchhoff's Current Law (KCL) to solve for unknown voltages. An example illustrates how to apply KCL to find voltages at different nodes, reinforcing the concept of measuring voltages relative to the chosen reference point. Overall, the conversation highlights the flexibility of the node-voltage method in circuit analysis.
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Node-voltage Method, some misconception??

Please check the attachment:
The question is written upon it

Sorry bad drawing, the circuit is closed from the left hand side but not from the right hand side.
 

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But I don't see any problem. We can choose any node we want as a reference point (GND).
 


But the essential node must connect at least 3 appliances! Here we have only two :/
 


But you can solve the circuit using nodal analysis without any "essential node".
And I still don't understand why you need "essential nodes"?
 


What's nodal analysis? How does it differ from node-voltage method. Elaborate please.
[I only learned node-voltage method :p]
If we are talking only node-voltage method, how in the world did the Doctor consider the node below a reference when it only connects two appliances[see figure].
That's my question.
 
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I just checked, and they mean the same. Yet again, this is against the rule.
 
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Nodal analysis is exactly the same think as node-voltage method.
And we can choose any node we want as a reference point (GND).

See the example

attachment.php?attachmentid=47008&stc=1&d=1336291275.png


We have a four nodes in our circuit. I pick as a reference point (GND) node 4.
So node 4 by de definition has a potential of 0 V.
So we left with three nodes. but we know that voltage at node 1 is equal 9V.
So we only has two unknown nodal voltages 2 and 3.
Now we apply KCL to the nodes where the unknown voltages appears.

For node 2 (I assume that all current flow out from the node)

\frac{V2}{R2} + \frac{V2 - V1}{R1} + \frac{V2 - V3}{R3} = 0

And we notice that V1 = E1 = 10V

Now we write KCL for node 3.

\frac{V3 - V2}{R3} + \frac{V3 - V1}{R4} = 0

And now all we have to do is to solve for V2 and V3

http://www.wolframalpha.com/input/?i=A/3+++(A+-+9)/9+++(A+-+B)/9+=0,+(B-A)/9+++(B+-+9)/9=0

V2 = 3V and V3 = Vth = 6V
 

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Lots of thanks Jhony, now I get it..
 


We always measure all the voltage respect to this common point (reference point), also known as a "ground" (GND). And we assume that GND have zero voltage.
Look ta this examples

attachment.php?attachmentid=47009&stc=1&d=1336293980.png


attachment.php?attachmentid=42873&d=1327068105.jpg
 

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Thanks, that's very kind of you.
You really were helpful
 
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