Nodal analysis with supernodes and voltage sources

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Cyb3rGlitch
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Homework Statement


Circuit schematic with nodes a, b and c and a reference node. A 24Ω resistor spans the top, a 10V source sits between a and b, a 40Ω resistor links b to the reference node, a 12Ω resistor hangs from a, a 6A current source points up from c to b, and a 12V source connects the reference node down to c.


Homework Equations


KCL

The Attempt at a Solution


Circuit schematic with a supernode containing a 10V voltage source between nodes a and b, a 6A current source from b to c, resistors 24Ω, 12Ω and 40Ω, a 12V source, and a labelled reference node at the right.


I recognise that the voltage sources form supernodes. This produces:

[tex]A - B = 10[/tex] -> [tex]A = 10 + B[/tex]

We also know that node C is 12v.

I then form KCL equations:
[tex]6 + \frac{A - C}{12} = \frac{B}{40} + \frac {A}{24}[/tex]

When I simplify this and sub in C and A, I get a ridiculous results. What am I missing here? Cheers!
 
Last edited:
I think Vb-Va=10 -> Va=Vb-10.
To see this, write a kcl around the loop starting from the reference node to Node A and then Node B:

-Va-10+Vb=0 ->Va=Vb-10.

I have explained this on one of my posts:
http://circuits.solved-problems.com/resistive-circuits/nodal-analysis-supernode/
Scroll down to section three of the solution. Read the paragraph starting with "The voltage of Node III can be written in terms of the voltage of Node I."
It is also explained in my Nodal Analysis ebook, which can be downloaded from my website for free (a free and simple registrating is required though):
http://circuits.solved-problems.com/nodal-analysis/

Hope this will be of help.
 
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