# Node-voltage Method, some misconception?

1. May 5, 2012

### M. next

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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2. May 5, 2012

### Jony130

Re: Node-voltage Method, some misconception??

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

3. May 5, 2012

### M. next

Re: Node-voltage Method, some misconception??

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

4. May 5, 2012

### Jony130

Re: Node-voltage Method, some misconception??

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

5. May 6, 2012

### M. next

Re: Node-voltage Method, some misconception??

What's nodal analysis? How does it differ from node-voltage method. Elaborate please.
[I only learnt 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.

Last edited: May 6, 2012
6. May 6, 2012

### M. next

Re: Node-voltage Method, some misconception??

I just checked, and they mean the same. Yet again, this is against the rule.

Last edited: May 6, 2012
7. May 6, 2012

### Jony130

Re: Node-voltage Method, some misconception??

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

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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8. May 6, 2012

### M. next

Re: Node-voltage Method, some misconception??

Lots of thanks Jhony, now I get it..

9. May 6, 2012

### Jony130

Re: Node-voltage Method, some misconception??

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

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10. May 6, 2012

### M. next

Re: Node-voltage Method, some misconception??

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