Nodal analysis three supernodes

Thread Starter

cheypr

Joined Oct 13, 2009
34
hi everyone, I'm new around here, trey nice forum, thanks I landed here. I'm trying to develop the equations to solve this problem. Sorry for tha image, i don have a scanner. Well, I'm all ears and eyes.
 

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Ratch

Joined Mar 20, 2007
1,070
cheypr,

hi everyone, I'm new around here, trey nice forum, thanks I landed here. I'm trying to develop the equations to solve this problem. Sorry for tha image, i don have a scanner. Well, I'm all ears and eyes.
There are 5 nodes plus ground, and neither of them is a supernode. Could you post your efforts to solve this problem?

Ratch
 

Thread Starter

cheypr

Joined Oct 13, 2009
34
cheypr,



There are 5 nodes plus ground, and neither of them is a supernode. Could you post your efforts to solve this problem?

Ratch
:confused:, so the 16v source between v1 and v2 is not a supernode? what about the 4ix dependent voltage source and the 8v source between v4 and v5?

o though an supernode was any dependent or independent voltage source that relates 2 nodes, none of them a reference node.

could you clarify this for me.

I was doing this:

16v=v1-v2
8v=v4-v5
4ix=v2-v4

at 16v: (v1-v3)/2 - (v2-v5)/2 =0

at 4ix: (v2-v5)/2 - (v4-v3)/3 = 1A

at 8v: (v4-v3)/3 - (v5-v2)/2 - (v5-0)/8 =1A
 

Ratch

Joined Mar 20, 2007
1,070
cheypr,

, so the 16v source between v1 and v2 is not a supernode? what about the 4ix dependent voltage source and the 8v source between v4 and v5?

o though an supernode was any dependent or independent voltage source that relates 2 nodes, none of them a reference node.

could you clarify this for me.
You are right. I misinterpreted what you wrote and misread the diagram. I read the 4Ix as a current source instead of a dependent voltage source. So at first, I did not see how you could get three supernodes. Now I see that you have one triple supernode.

The first thing I did was throw away the northeast 2 ohm resistor because it returns back to the supernode and cancels out.

Writing the first equation for the supernode we have

(V1-V3)*(1/2)+(V1-16-4*V3*(1/6)-V3)*(1/3)+(1/8)*(V1-16-4*V3*(1/6)-8)-1 = 0

and for the node V3

(V3-V1)*(1/2)+(1/3)*(V3-V1+16+4*V3*(1/6))+(1/6)*V3 = 0

Solving we get V1=24 volts and V3 = 12 volts. The other voltages are easily found from V1 and V3.

Ratch
 

Thread Starter

cheypr

Joined Oct 13, 2009
34
ok I'm ok with this part (V1-V3)*(1/2)+ could you explain the other part on red (V1-16-4*V3*(1/6)-V3)*(1/3)+(1/8)*(V1-16-4*V3*(1/6)-8)-1 = 0
I'm not seeing it or I'm getting dizzy jejeje:D hope you could!!!
 

Ratch

Joined Mar 20, 2007
1,070
cheypr,

ok I'm ok with this part (V1-V3)*(1/2)+ could you explain the other part on red (V1-16-4*V3*(1/6)-V3)*(1/3)+(1/8)*(V1-16-4*V3*(1/6)-8)-1 = 0
I'm not seeing it or I'm getting dizzy jejeje hope you could!!!
Sure. It is best to translate from type to hand written.

(V4-V3)/3 = (V1 - 16 - 4*V3/6 - V3)/3 where 4*V3/6 = 4*Ix .

V6/8 = (V1 - 16 - 4*V3/6 - 8)/8

-1 is the current source going into node V4

If you still cannot make sense of it, I can attach a PDF file.

Ratch
 

The Electrician

Joined Oct 9, 2007
2,847
I was doing this:

16v=v1-v2
8v=v4-v5
4ix=v2-v4

at 16v: (v1-v3)/2 - (v2-v5)/2 =0

at 4ix: (v2-v5)/2 - (v4-v3)/3 = 1A

at 8v: (v4-v3)/3 - (v5-v2)/2 - (v5-0)/8 =1A
Ratch has shown how to solve for V1 and V3 with two equations.

You can solve for all 5 node voltages at once like this:

Get rid of the northeast 2Ω resistor.

You have 3 voltage constraint equations:

V1 - V2 = 16
V4 - V5 = 8
V2 - V4 = 4*V3/6 (note that ix = V3/6)

You have one equation which sums currents at node V3, which is not part of the supernode:

(v3 - V1)/2 + (V3 - V4)/3 + V3/6 = 0

You need one more equation to deal with the supernode. Let's sum currents at V4; summing currents at any of the nodes which are part of the supernode should give this same equation:

(V1 - V3)/2 + (V4 - V3)/3 + V5/8 = 1

You now have 5 equations and 5 unknowns. Solve this system and you will get all 5 node voltages.
 

Thread Starter

cheypr

Joined Oct 13, 2009
34
wow:D, you guys are a good help. for now on i'll be around this forum i want to become more efficient analysing circuits. i'll do this forum my home:cool:.
 

stupid

Joined Oct 18, 2009
81
wow fantastic..of how the electrician's work fuses with u.

ratch,
could i have the PDF file as u had suggested...detailing the senses so that i could employ yur more direct approach ?

thanks in advance
stupid

cheypr,



Sure. It is best to translate from type to hand written.

(V4-V3)/3 = (V1 - 16 - 4*V3/6 - V3)/3 where 4*V3/6 = 4*Ix .

V6/8 = (V1 - 16 - 4*V3/6 - 8)/8

-1 is the current source going into node V4

If you still cannot make sense of it, I can attach a PDF file.

Ratch
 

stupid

Joined Oct 18, 2009
81
hi the electrician,
i dun quite grab how the node (V1 - V3)/2 works at the V4 summing..
why the 3ohm is missing from the formula, like (V1 - V3)/(2+3)?
could u explain?

thanks
stupid

Ratch has shown how to solve for V1 and V3 with two equations.

You need one more equation to deal with the supernode. Let's sum currents at V4; summing currents at any of the nodes which are part of the supernode should give this same equation:

(V1 - V3)/2 + (V4 - V3)/3 + V5/8 = 1

You now have 5 equations and 5 unknowns. Solve this system and you will get all 5 node voltages.
 
hi the electrician,
i dun quite grab how the node (V1 - V3)/2 works at the V4 summing..
why the 3ohm is missing from the formula, like (V1 - V3)/(2+3)?
could u explain?

thanks
stupid
The idea of a supernode is that if voltages sources connect two or more nodes together (V1, V2, V4 and V5 in this problem), those nodes behave, for the purpose of summing currents into the supernode, as if they were just one node. So summing currents into V4 is the same as summing currents into V1, or V2, or V5. But, to actually obtain the voltages at the nodes individually, when you sum currents you need to consider that the resistor currents are due to the voltages actually present at the nodes between which the resistors are connected.

The northwest 2Ω resistor is the only resistor connected between V1 and V3; the 3Ω resistor is not connected there. That's why we use (V1 - V3)/2 and not (V1 - V3)/(2+3). The 3Ω resistor is actually connected between V4 and V3, so we use (V4 - V3)/3 to determine the current in the 3Ω resistor.
 
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stupid

Joined Oct 18, 2009
81
thank u the electrician for clarfying my thought on that..
btw have u any online materials or readings dealing network analysis with dependent sources i can look up?

regards,
stupid
 
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