Quick Nodal Anaylsis Question

Thread Starter

Digit0001

Joined Mar 28, 2010
100
Hi
DSC00069.JPG
i have another quick question, i am finding the voltage nodes in this circuit and wondering are my equations correct?

KCL SuperNode
V1/2+V2/4+V3/3=0

KVL(V1-V2)
10+V2=V1

KVL(V2-V3)
V3+5i=V2

KVL(V1-V3)
V1+6+V3=0

P.S
 

jegues

Joined Sep 13, 2010
733
Hi
View attachment 31159
i have another quick question, i am finding the voltage nodes in this circuit and wondering are my equations correct?

KCL SuperNode
V1/2+V2/4+V3/3=0

KVL(V1-V2)
10+V2=V1

KVL(V2-V3)
V3+5i=V2

KVL(V1-V3)
V1+6+V3=0

P.S
I'm not 100% certain, but I think you should do this in 2 super nodes rather than 1 big one encompassing the 2 sources. (If that's not at all what you've done then I can't follow your work)

Also, in your KCL for the supernode equation there is no mention of any current through the 6Ω resistor.

I've managed to write a set of equations for two seperate supernodes around each source.

Here's the supernode equation I've come up with around the independent source.

\(\frac{V_{1} - V_{3}}{6} + \frac{V_{1} - 0}{2} + \frac{V_{2} - 0}{4} = 0 \)

Give the rest a shot.
 
Last edited:

Thread Starter

Digit0001

Joined Mar 28, 2010
100
yes i was trying to use one big supernode. This is what i got for the other three equations

SuperNode between V2 and V3
V2/4+V3/3=V1-V3/6

KVL between V1 and V2
10+V2=V1

KVL between V2 and V3
V2+5i=V3
 

jegues

Joined Sep 13, 2010
733
yes i was trying to use one big supernode. This is what i got for the other three equations

SuperNode between V2 and V3
V2/4+V3/3=V1-V3/6

KVL between V1 and V2
10+V2=V1

KVL between V2 and V3
V2+5i=V3
Why are you writing KVL's?

From each of the two supernode equations you wrote, you should be able to obtain another equation.

For example, using the independent source, one can note that,

\(V_{1} - V_{2} = 10\)

Now do the same for the dependent source, and express i in terms of your node voltages.

Your 2 supernodes give you a total of 4 equations, more than enough to solve for the 3 unknown node voltages.
 
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