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As far as I know, current flows from the high potential to the low potential of the voltage. In that case, they should follow situation 1, but I have seen lots of solution and problems where the author used situation B for nodal analysis or so on. Why?

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2 Answers 2

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It doesn't matter so long as you are consistent in direction across an edge at the nodes at both ends. If you chose the wrong direction because you got the prediction wrong, or chose a direction for simpler equations, then you will end up with negative currents in the final solution. This means that the current is in the opposite direction to what you expected.

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  • \$\begingroup\$ You've pointed out a major problem with KCL - establishing a consistent method of dealing with direction causes more grief than anything else. Fixing in your mind (with confidence) a way to assign signs (+/-) to currents is key. The last step in a solution reverses signs that you assigned incorrectly. \$\endgroup\$
    – glen_geek
    Commented Jun 13, 2023 at 12:16
  • \$\begingroup\$ So you think it is always better to go with situation A? \$\endgroup\$
    – Obiick
    Commented Jun 13, 2023 at 12:22
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    \$\begingroup\$ @Obiick, no, this answer is saying it doesn't matter what you choose. The only difference is in set-up B you will get a negative value for I1. Which will have the exact same physical meaning as a positive value for I1 with set-up A. \$\endgroup\$
    – The Photon
    Commented Jun 13, 2023 at 14:39
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If you express KCL with a formula you will say that in a node \$\sum_{1}^{M}I_j = 0\$, or that \$I_1 + I_2 ... + I_M = 0\$; then in the solution, you will solve for each of them, with their sign (or direction). This will always be correct, no matter their direction.

To me, it looks like simple algebra, more than circuit analysis.

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  • \$\begingroup\$ Right.. But is it better to make the equation considering their direction to find the accurate sign and values? \$\endgroup\$
    – Obiick
    Commented Jun 13, 2023 at 13:27
  • \$\begingroup\$ Is it better to say \$I_1 + I_2 + I_3 = 0\$ or \$I_1 - I_2 + I_3 = 0\$? You have to assume that the circuit might be so complex that you could not know in advance how the current flows - as it is with multiple voltage/current sources. So the general form of adding all the variables is preferable IMHO, as it doesn't force any prior assumption, that could even mislead to any wrong solution. \$\endgroup\$
    – LuC
    Commented Jun 13, 2023 at 14:25

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