Mistake in this node analysis work?

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influx
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Is the working out for node V2 wrong? In particular, I am referring to the (V1-V2)/2 term in equation 2?

Also, is the voltage drop always in the same direction as the current or is the voltage rise always in the same direction as the current?

Thanks
 
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Here is a good way to do nodal. have you ever heard of 'king of the hill' method?

You assume that your node is the point of highest potential.. that is, everything flows away from the node you're working on at that time. The signs will fall out correctly. Current sources are entered to the equations corresponding to their directions. Everything else is assumed leaving.

4Ix = V1/8 + (V1-V2)/2 (notice how everything was assumed to leave the node except current injections).

for node 2, you can do the same.

5-3Ix = V2/4 + (V2-V1)/2

Notice here in this equation we assumed V2 was at the higher potential. This is fine, as long as we assume it is the point of highest potential in all parts of the equation for that node.

I cannot solve the equations for you, but I sure hope this clarifies any confusion with Nodal analysis. ( I can say that the equations as I have written them are not quite where you want to start, you can entirely eliminate Ix from the equations).

Does it make sense? Any questions/corrections on what I have written are most welcome.
 
The trick is to leave the negative sign in front of the incoming currents, rather than expanding it out like you have done in the first term in equation 1. That way, you won't make common mistakes when identifying the correct direction of the current.
Also, if you move all your negative currents to the other side of the KCL equation, you should be able to see that the sum of all the incoming currents entering a node, equals the sum of all the outgoing currents leaving that node.
 
influx said:
Is the working out for node V2 wrong? In particular, I am referring to the (V1-V2)/2 term in equation 2?
I think it has been confirmed that that particular term doesn't belong in the equation. It's erroneous.

Also, is the voltage drop always in the same direction as the current or is the voltage rise always in the same direction as the current?
Current flows from higher potential towards lower potential.