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I have done the nodal analysis for the following question, and everything appears to have been done correctly as far as I can see. However, when I simulate the results I get values for the circuit which are wildly different.

Where would the error have been introduced? All sign conventions are followed and I've doubled checked the working out.

Edit : Ix = (v1-v2)/2 | This was not included in the diagram enter image description here

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    \$\begingroup\$ Is there a multiplication by 2 missing in middle equation ? The source is 2 ix. (V1 - V2)/2 is only 1ix ? Also mark the reference direction of ix in the diagram. It seems to be marked 2 different ways in the current diagram. \$\endgroup\$
    – AJN
    May 9, 2021 at 8:05
  • \$\begingroup\$ Hi @AJN, I've edited Ix, it should have been (V1-V2)/2. And the dependent current should be 2Ix. I thought with regards to nodal analysis, we could specify the direction of the current (by default) to point inwards into a node, with the exception being currents that outright were pointing away from the node. \$\endgroup\$
    – user36278
    May 9, 2021 at 8:10
  • \$\begingroup\$ Does the original question have ix reference direction clearly marked ? \$\endgroup\$
    – AJN
    May 9, 2021 at 8:16
  • \$\begingroup\$ Yes, it does, it has it marked from left to right at the V1 node. The current is marked as moving from V1 to V2. But I don't think that would make a difference as I have remained consistent with the current directions. Furthermore, I tested the equation and removed the dependent current source and replaced it with an independent current source (9.6A) and it provides the correct answer. I find that very strange as no other change was made to the equation. \$\endgroup\$
    – user36278
    May 9, 2021 at 8:26
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    \$\begingroup\$ This is what I get. The numbers for the voltages don't seem to match yours. \$\endgroup\$ May 9, 2021 at 9:48

1 Answer 1

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Your mistake is again the current direction.

For \$V_1\$ node you have wrote: $$+12A + \frac{V_1 - V_2}{2\Omega}....= 0$$

As we can see from a circuit diagram the \$12A\$ current is entering the node(incoming current). But we have a problem here because you decided to give a "+" for 2 ohms resistor current as well.

But this notation \$\frac{V_1 - V_2}{2\Omega}\$ imposes the current direction. And this current direction is from \$V_1\$ to \$V_3\$ (the current is flowing out of the node).

And here we have an error in your current direction assumption. And this is why you are getting the wrong answer.

The correct equation should look like this:

$$-12A + \frac{V_1 - V_2}{2\Omega} + \frac{V_1 - V_3}{4\Omega} = 0$$ $$\frac{V_2}{4\Omega} + \frac{V_2 - V_1}{2\Omega} + \frac{V_2 - V_3}{8\Omega} = 0$$ $$ 2\times\frac{V_1 - V_2}{2\Omega} + \frac{V_3 - V_1}{4\Omega} + \frac{V_3 - V_2}{8\Omega} = 0$$

Or like this (not recommended way):

$$+12A + \frac{V_2 - V_1}{2\Omega} + \frac{V_3 - V_1}{4\Omega} = 0$$ $$\frac{V_2}{4\Omega} + \frac{V_2 - V_1}{2\Omega} +\frac{V_2 - V_3}{8\Omega} = 0$$ $$ 2\times\frac{V_1 - V_2}{2\Omega} +\frac{V_3 - V_1}{4\Omega}+\frac{V_3 - V_2}{8\Omega} = 0$$

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  • \$\begingroup\$ Wow, thank you. I confused myself, but you've explained it soo well. I now get it. \$\endgroup\$
    – user36278
    May 9, 2021 at 10:19

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