Nodal analysis of DC ladder network

Thread Starter

PK1248

Joined Aug 19, 2012
6
Just to add more info, for the voltage at node 2;

(V(2) - V(1)) / 120 + V(2) / 820 + (V(2)-V(3)) / 330 = 0

and for node 3;

(V(3)-V(2)) / 330 + V(3)/470 + (V(3)-V(4))/100

now i can see that V(4) is goint to be 10V but i have to prove it through nodal analysis eqns.
 

Thread Starter

PK1248

Joined Aug 19, 2012
6
well that's where im a bit confused. At node 1 for example, we have (V(1) - V(2)) / 120 leaving to the right but leaving downwards, we have no resistance between the Node 1 and the voltage source, therefore infinite current ( I = V/R).

so i am assuming that supernode is used to ''bypass'' the voltage source or something in that sense? but i really haven't used supernode and lecturer hasn't brought it up at all
 

Jony130

Joined Feb 17, 2009
5,186
In this case you can swap E1 with R6 and solve the circuit.
Or you can treat V1 and V5 as a one big (super) node. Simply short V1 with V5.

So we can write for node 1

(V1 - V2)/R1 + V5/R6 = 0

For node 2

V2/R2 + (V2 - V1)/R1 + (V2 - V3)/R3 = 0

And for node 3

(V3 - V2)/R3 + V3/R4 + (V3 - E2)/R5 = 0


And one additional equation for our supernode

V1 - V5 = E1 ----> V5 = V1 - 18V

And now we can solve this circuit

But you can also use this trick
http://forum.allaboutcircuits.com/showthread.php?p=446404#post446404
 

WBahn

Joined Mar 31, 2012
26,304
Keep in mind that nodal analysis is simply a formalized application of KCL. With that in mind, you can see that when you write the node equation for Node 5, you need the current flowing up to Node 1. But, that current IS the current flowing to the right from Node 1 to Node 2.

When you have a voltage source, there is nothing wrong with writing the node equations of one node relative to the other. In fact, there's pretty much no way around it. What you generally do is define a "supernode" in which the nodes within the supernode have relationships to each other that are defined by voltage sources or other things that make them hard/impossible to do with KCL. You then develop equations that relate the various node voltages of the nodes within the supernode separately but treat the supernode as a black box with regard to the nodes outside it.

So in this case you have a supernode involving Nodes 1&5 and the internal equation is simply V1 = V5 + 10V. Similarly, you have a supernode involving Nodes 0&4. So that gives you two of the five equations you need.

You know apply KCL at Nodes 2, 3, and 5 and wherever you need the current entering a supernode, you simply recognize that, by KCL, it is equal to the net current leaving the supernode from all of the other exit points.
 
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