Node-voltage Method, some misconception?

  • Thread starter M. next
  • Start date
  • Tags Tags
    Method
9 replies · 3K views · last reply
M. next
Messages
380
Reaction score
0
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.
 
Attachments
  • Hand-drawn circuit schematic with a voltage source on the left, two nodes labelled 1 and 2, three resistors, a ground symbol at the bottom, the label V thevenin, and a handwritten question about the reference node.
    Physics.png
    6.1 KB · Views: 550
Last edited:


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.
 
Last edited:


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


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

Circuit schematic with nodes 1 to 4: 9V battery E1 at node 1, resistors R1 9K and R3 9K in series, R4 9K across the top, R2 3K to ground at node 4, and Vth taken between node 3 and the bottom rail.


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)

[tex]\frac{V2}{R2} + \frac{V2 - V1}{R1} + \frac{V2 - V3}{R3} = 0[/tex]

And we notice that V1 = E1 = 10V

Now we write KCL for node 3.

[tex]\frac{V3 - V2}{R3} + \frac{V3 - V1}{R4} = 0[/tex]

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
 
Attachments
  • Circuit schematic with a 9V source E1 and resistors R1 9K, R2 3K, R3 9K and R4 9K, with nodes labelled 1 to 4 and Vth marked across the output terminals.
    bez tytułu.PNG
    1.6 KB · Views: 678


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

Four side-by-side diagrams of stacks of five 3V batteries in series, with node voltages marked: first with no GND shown, second with GND at the bottom reading +15V to 0V, third with GND at the middle reading +9V to -6V, fourth with GND at the top node reading +3V to -12V.


Three hand-drawn cartoon diagrams of a stick figure holding a battery labelled A and B with VAB = 1.5V and VAB = -1.5V; the figure's voltage reading changes with the chosen reference point GND.
 
Attachments
  • Four circuit diagrams of five 3V cells in series, each with a different ground reference: first shows the bare series stack, second labels nodes +15V, +12V, +9V, +6V, +3V, 0V with GND at the bottom, third labels +9V to -6V with GND at the middle, fourth labels +3V to -12V with GND at the top.
    0.1.PNG
    6.9 KB · Views: 1,003


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