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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There are 5 nodes plus ground, and neither of them is a supernode. Could you post your efforts to solve this problem?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.
, 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?cheypr,
There are 5 nodes plus ground, and neither of them is a supernode. Could you post your efforts to solve this problem?
Ratch
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., 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.
Sure. It is best to translate from type to hand written.ok I'm ok with this part (V1V3)*(1/2)+ could you explain the other part on red (V1164*V3*(1/6)V3)*(1/3)+(1/8)*(V1164*V3*(1/6)8)1 = 0
I'm not seeing it or I'm getting dizzy jejeje hope you could!!!
Ratch has shown how to solve for V1 and V3 with two equations.I was doing this:
16v=v1v2
8v=v4v5
4ix=v2v4
at 16v: (v1v3)/2  (v2v5)/2 =0
at 4ix: (v2v5)/2  (v4v3)/3 = 1A
at 8v: (v4v3)/3  (v5v2)/2  (v50)/8 =1A
You're quite welcome.thanks to you both for the help. I really apreciate it.
cheypr,
Sure. It is best to translate from type to hand written.
(V4V3)/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
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.
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.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
I haven't looked for any online help. I just use textbooks for reference.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