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schematic

simulate this circuit – Schematic created using CircuitLab

The problem is to find the node voltage at e2 (I added the current lables and directions) After getting it wrong, and seeing the correct answer I know that the correct equation must be

$$ \frac{e2-5}{3} + \frac{e2}{5} - 3 = 0 $$

However, I can't see how I would arrive at the above without having known the answer (i.e I'm bound to get the next one wrong too)

So this is how I tried to work it out. I defined the currents into the node as positive, giving

$$ i1 - i2 + i3 = 0 $$ Where $$ i1 = \frac{e2-e1}{R1} = \frac{e2-5}{3} $$ $$ i2 = \frac{e2-e3}{R2} = \frac{e2}{5} $$ $$ i3 = I = 3 $$ So $$ \frac{e2-5}{3} - \frac{e2}{5} + 3 = 0 $$

Which, with a couple of incorrect signs, gives the wrong answer.

What is wrong with my logic?

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What is wrong with my logic?

Your 3rd equation is not consistent with the passive sign convention.

Since you've chosen \$i_1\$ to enter the \$e_1\$ terminal of \$R_1\$, the correct equation is, by the passive sign convention,

$$i_1 = \frac{e_1 - e_2}{R_1}$$

Note that one does not need to know in advance if \$e_1\$ is a higher potential than \$e_2\$. It may be that, when the equations are solved, \$e_1 \lt e_2\$.

But that's irrelevant since, in that case, \$i_1\$ will be negative and thus, as desired, the current will be entering the more positive terminal of the resistor (a negative current to the right is a positive current to the left).

Given the reference polarities and reference directions chosen, the correct KCL equation for node 2 is

$$i_1 - i_2 + i_3 = \frac{5 - e_2}{3} - \frac{e_2}{5} + 3 = 0$$

which is equivalent to your 1st equation.

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Current goes from higher potential to lower. So as you assume \$i_1\$ goes from \$e_1\$ to \$e_2\$ then you actually assume \$e_1\$ is higher then \$e_2\$ .So current equation is:

\$i_1 = \frac{e_2-e_1}{R_1}\$

\$i_1 = \frac{5-e_1}{3\Omega}\$

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  • \$\begingroup\$ It's not a matter of which way the current flows, but of using consistent conventions throughout the computation. \$\endgroup\$ – microtherion Aug 31 '14 at 3:42
  • \$\begingroup\$ I'm not telling 'it's matter which way the current flow' . I just try to explain why he have to write equation this way or you can say 'why the conventions is this way ' \$\endgroup\$ – Anklon Aug 31 '14 at 5:23

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