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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?


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.


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}\$

  • \$\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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