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Hello, I am stumbling across an issue here. So in Nodal Analysis, we identify our reference point and label the voltage nodes. Then we use KCL to find out equations and solve. My book is providing an example in the picture I've provided. In the highlighted red, why did they decide to multiply it by a negative? -3A is coming out of Node 2. When do we need to multiply it by a negative? And also, how do I make my direction currents consistence so I can get the correct answer? I know we have to choose our current direction but I keep getting the signs wrong. Much help is appreciated.

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Preface note: I originally used \$3\:\text{A}\$ as the current source. But I mis-read the problem, as stated. They specify, on the schematic, that the out-flowing current source value is \$-3\:\text{A}\$. The original error is mine. It is fixed, now, below.

The node is (by super-position):

$$\overbrace{\frac{V_2}{3\:\Omega}+\frac{V_2}{1\:\Omega}+\frac{V_2}{7\:\Omega}+\left(-3\:\text{A}\right)}^{\text{Outward-flowing currents}}=\overbrace{\frac{V_1}{3\:\Omega}+\frac{0\:\text{V}}{1\:\Omega}+\frac{V_3}{7\:\Omega}}^{\text{Inward-flowing currents}}$$

Moving everything from the right side over to the left side gives:

$$\frac{V_2-V_1}{3\:\Omega}+\frac{V_2}{1\:\Omega}+\frac{V_2-V_3}{7\:\Omega}+\left(-3\:\text{A}\right)=0\:\text{A}$$

Moving the constant to the right and you have:

$$\frac{V_2-V_1}{3\:\Omega}+\frac{V_2}{1\:\Omega}+\frac{V_2-V_3}{7\:\Omega}=3\:\text{A}$$

Does that help at all?

Note

Voltage spills away from a node via available paths and voltage spills into a node from nearby nodes via the same available paths. The only caveat is that a current source can only flow into, or flow out of, a node. But not both directions. (Current sources only work one way. So it will either appear on the outward-flowing side or on the inward-flowing side -- but not both sides.)

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  • \$\begingroup\$ I like your explanation on which currents flow inwards and outwards which helps me understand. In the final simplified equation I have gotten that exactly but I'm not sure why the author converts that -3A into a positive. \$\endgroup\$ Sep 22 at 16:40
  • \$\begingroup\$ @GreenLeaf21 That's my mistake. I misread the problem's schematic. I'll fix my answer. Thanks for catching that!! \$\endgroup\$
    – jonk
    Sep 22 at 18:35
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    \$\begingroup\$ thanks for the explanation \$\endgroup\$ Sep 23 at 19:14
  • \$\begingroup\$ @GreenLeaf21 If you find the answer sufficient, feel free to select it. This saves the time of others who may misunderstand and imagine you want another answer or two to help out. Doing so respects the time of others. On the other hand, if you really do want to see some different answers to this question, then do NOT select my answer. \$\endgroup\$
    – jonk
    Sep 24 at 8:36

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