Sloving a depentant voltage source problem using node voltage

  • Thread starter cybhunter
  • Start date
  • #1
25
0
Hi everyone,

I currently stumped in the process of approaching a dependent voltage source problem using node voltages. Attached is the circuit.

What I know:

the Node north of the 60V supply with also be 60V

there are four nodes, two when supernodes are used


What I'm confused about:

with the super node being the 6iψ, KCL states the sums must be zero

when I try to solve, I end up getting the equation (using v2 as the junction of the dependent, and both 3Ω resistors):

[(60-v1)/2]+[(V2-V3)/3]+[V2/3]=0
I am not sure how to calculate the current across the 60Volt source

:confused:


In addition, I end up having a recursion problem trying to solve for iψ

Since V2 is 60+6iψ, and iψ is ([V2-V2]/3), I end up with

V2=60+6([V2-V3]/3)


any help would be greatly appreciated

Joe
 

Attachments

Answers and Replies

  • #2
625
138
Simply treat A and C as a one big (super) node.
So we only have one unknown node Vb

attachment.php?attachmentid=44355&stc=1&d=1330171555.jpg


Current leaving the node is equal current entering the node.

For node B

Vb/24 + (Vb - 60)/2 + (Vb - Vc)/3 = 0 (1)

And one additional equation

Vc = Va + 6*Iψ = Va + 6*(Vc - Vb)/3 = 2Vb-Va = 2Vb - 60 (2)

So we have

Vb/24 + (Vb - 60)/2 + (Vb - (2Vb - 60))/3 = 0

http://www.wolframalpha.com/input/?i=+b/24+++(b+-+60)/2+++(b+-+(2b+-+60))/3+=+0

But we can remove the supernode from the circuit. We you can transform this circuit, legally, and get a new circuit which is equivalent to the old one

attachment.php?attachmentid=44354&stc=1&d=1330171517.jpg
 

Attachments

Last edited:

Related Threads on Sloving a depentant voltage source problem using node voltage

Replies
4
Views
4K
Replies
15
Views
812
  • Last Post
Replies
2
Views
445
  • Last Post
Replies
1
Views
511
Replies
2
Views
652
Replies
5
Views
1K
Replies
3
Views
3K
Replies
9
Views
7K
  • Last Post
Replies
1
Views
1K
Top