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Right, first the diagram, I tried using the schematic function,but it never actually added the schematic.

Schematic

I picked this because it illustrates my confusion in the simplest schematic I could find.

We've got three nodes here, with the bottom (N3) being the ground node, current directions are provided for N1, so writing up KCL for the nodes give the following.

$$N1 :2mA - \frac{V1}{3000}\ -\frac{V1-V2}{6000}\ = 0$$ (current enters from the 2mA supply, leaves trough the rest of the legs)

$$N2 : \frac{V1-V2}{6000} - 4mA - \frac{V2}{12000} = 0$$ (current enters from the 6k resistor, leaves trough the rest of the legs)

This in turn gives out the correct node voltages for V1 and V2. (\$\frac{-12}{7}\$ for V1 and \$\frac{-120}{7}\$ for V2)

However, I am supposed to be able to arbitrarily define the direction of the current over the 12k ohm resistor, so if I decide that current flows into node 2 and instead and use the following N2 equation

$$N2 : \frac{V1-V2}{6000}\ - 4mA + \frac{V2}{12000}\ = 0$$

I get the wrong result for both V1 and V2, so clearly the direction of the current across the 12k ohm resistor matters, what am I doing wrong ?

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    \$\begingroup\$ If you follow the convention where +current enters the node and -current leaves the node, then stick with it all the way through. Your 2nd N2 equation is not following the convention. \$\endgroup\$ – Mattman944 Jul 23 '19 at 13:53
  • \$\begingroup\$ Please use proper English capitalisation and puctuation in your writing. It is site policy, makes your post much easier to read and adds to your credibility. \$\endgroup\$ – Transistor Jul 23 '19 at 14:16
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Indeed you can define the current direction in which ever way you want. If you want to get the current going into the node N2, it should be \$\frac{0-v_2}{12k} = -\frac{v_2}{12k}\$. If you add this current, then you get the same equation: $$\frac{(v1-v2)}{6000}\ - 4mA + (\frac{-v2}{12000})\ = 0$$

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    \$\begingroup\$ of course! because the current flows from node 3 to node 2, the actual voltage i should be calculating is v3-v2/12000, but as node 3 is reference, it's voltage is automatically 0, thus it becomes 0-v2/12000 and the math works out. thanks to both you and Elliot Alderson, i knew there was something i didn't quite grasp. :) \$\endgroup\$ – Draken Jul 23 '19 at 14:31
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Your problem is that you are not following the passive sign convention when using Ohm's Law. You can define the direction of current and voltage however you like, but if you don't follow the passive sign convention then a minus sign is introduced into Ohm's Law.

The passive sign convention, when applied to Ohm's Law, says that current is assumed to enter the end of the resistor that is assumed to have a higher voltage. If you follow this convention, then \$I_R=\frac{V_R}{R}\$. So, if you assume that current is flowing down through the 12k resistor it is true that \$i = \frac{V_2}{12\mathrm{k}\Omega}\$.

However, if you assume that current flows up through the 12k resistor then you are not following the passive sign convention and \$I_R = \frac{-V_R}{R}\$. Your last equation would then be

$$\frac{V_1-V_2}{6000} - 0.004 + \frac{-V_2}{12000}= 0$$

You can see that this is algebraically identical to the first (correct) equation for node 2, and, sure enough, it also gives the correct answer. Note that I have converted all of the quantities in your equation to the base units: volts, amperes, and ohms. Using mixed units is asking for trouble.

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