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I'm very confused here! Most of the times I'm getting wrong answers or impossible equations because of this - even if I know the procedures for the analysis. It has to do with setting up nodal analysis and current flow direction.

In this example (image below), the tutor told to "consider" the current always going outwards from node 1 and node 2 - even if that's not the true behaviour for the current - because the sign will be taken care by the numerical value. Fair enough.

  1. First he considers it for node 1 as he just explained. Every current is going "outwards" from the node: including the current that goes directly from node 1 to node 2.
  2. But by the node 2 he ALSO considers the same current going "outwards". Meaning, from node 2 to node 1. How can the set of equations be consistent if the current cannot be both going from 1 to 2 and from 2 to 1? How is this system of equations possible?
  3. Then he considers the current imposed by the current source to flow into node 2. Now the direction matters? Wouldn't just the sign be taken cared of?

I'm very confused about all of this sign. Do we have to be consistent or not? Why is he able to do that? How can I always get away with the signs when doing nodal analysis reducing mistakes?

Thank you in advance!

enter image description here

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  • \$\begingroup\$ Ask your tutor is my advice. \$\endgroup\$
    – Andy aka
    Jul 19, 2022 at 12:40
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    \$\begingroup\$ Notice that +0 = -0. If you reverse the direction in each part of the equation, the equation will still have the same solutions. It doesn't matter whether you write (V2-V1)/2 + V2/10 - 3 = 0, or (V1-V2)/2 - V2/10 + 3 = 0 \$\endgroup\$
    – user253751
    Jul 19, 2022 at 13:01

2 Answers 2

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In nodal analysis, we are doing KCL for each node. And we can assume arbitrary currents directions. But to simplify things we assume that for a given unknown node the voltage at this node is the highest in the circuit. And this implies that every current in the node is leaving the node (going "outwards" from the node).

The next thing that you should do before starting the analysis is to decide for what current direction you will assign the "plus" sign and the "minus" sign. For example, I will give a minus sign if the current enters the node (coming into the node). And the plus sign if the current is leaving the node.

So for the first node, we can write:

Outflowing current (leaving the node) - Inflowinf current (entering the node) = 0

\$ \frac{V_1 - 10V}{1\Omega} + \frac{V_1}{5\Omega} + \frac{V_1 - V_2}{2\Omega} = 0\$

And the same thing for the second node.

\$\frac{V_1 - V_2}{2\Omega}+ \frac{V_2}{5\Omega} - 3A = 0\$

We give a minus sign for 3A current source because this current is entering the node based on our initial assumption (I will give a minus sign if the current enters the node (coming into the node). And the plus sign if the current is leaving the node).

Of course, we could have made different assumptions about the current direction and the sign we used.

For example, we could draw the current direction for each node like this:

enter image description here

And write KCL for node one like this assuming this

Inflowinf current = Outflowing current

\$\large I_1 = I_2 + I_3 \$

\$\large \frac{10V - V_1}{1\Omega}=\frac{V_1}{5\Omega}+\frac{V_1 - V_2}{2\Omega}\$

And for the secend node:

Inflowinf current = Outflowing current

\$\large I_3 + I_5 = I_4\$

\$\large \frac{V_1 - V_2}{2\Omega} +3A = \frac{V_2}{10\Omega} \$

And the solution will be exactly the same as before:

\$\large V_1 = \frac{750}{77} \approx 9.74V\$

\$\large V_2 = \frac{1010}{77} \approx 13.117V \$

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1)Remember that current flows from high to low voltage

2)We cant know if a current is entering a node we set them arbitarily but it doesnt matter.The equations which describe the current/voltage relationships are linear so the order doesnt matter.

But be careful if you set a current Ix entering a node then the voltage of the node is at lower potential than the other end of the resistor whose current is Ix.If you a current Iy leaving a node then the voltage of the node is at higher potential than the other end of the resistor whose current is Iy.

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