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I set up the equations $$ -\frac{V_a}{4\Omega} - 3A - \frac{V_a - V_b}{2\Omega} = 0 \\\frac{V_a - V_b}{2\Omega} + \frac{V_b}{3\Omega} + 4\Omega = 0$$ However when I row reduce them I get $$V_a = -12 \\V_b = -12 $$ but the solution is: enter image description here

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  • \$\begingroup\$ Your problem is with the second equation. You're assuming all the currents are going into the node b. But you are not consistent with the sign of \$\frac{V_b}{3}\$, it should be negative. Why? Because if you assume that current is going into the node, then you should have done \$\frac{0-V_b}{3}\$ (you are assuming current flowing from ground to node b.) \$\endgroup\$ – Big6 Sep 12 '16 at 4:57
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Just set it up and knock them down. The left side of the initial equation, for each node I set up, will be the currents leaving the node. The right side will be the currents entering the node.

Node 'a' first:

$$\begin{align*} \frac{V_a}{4\Omega}+\frac{V_a}{2\Omega}+3A&=\frac{0V}{4\Omega}+\frac{V_b}{2\Omega} \\ V_a\cdot\left(\frac{1}{4\Omega}+\frac{1}{2\Omega}\right)+V_b\cdot\left(\frac{-1}{2\Omega}\right)&=-3A \end{align*}$$

Node 'b' now: $$\begin{align*} \frac{V_b}{2\Omega}+\frac{V_b}{3\Omega}&=4A+\frac{V_a}{2\Omega}+\frac{0V}{3\Omega} \\ V_a\cdot\left(\frac{-1}{2\Omega}\right)+V_b\cdot\left(\frac{1}{2\Omega}+\frac{1}{3\Omega}\right)&=4A \end{align*}$$

Those are your two equations in two unknowns. In this case, \$V_a=-1\tfrac{1}{3}V\$ and \$V_b=4V\$, if I did things right.

I can reconstitute my equation form into yours, more or less, as:

$$\begin{align*} \frac{V_a}{4\Omega}+\frac{V_a-V_b}{2\Omega}+3A&=0 \\ \frac{V_b-V_a}{2\Omega}+\frac{V_b}{3\Omega}-4A&=0 \end{align*}$$

Or, in negated form:

$$\begin{align*} -\frac{V_a}{4\Omega}-\frac{V_a-V_b}{2\Omega}-3A&=0 \\ \frac{V_a-V_b}{2\Omega}-\frac{V_b}{3\Omega}+4A&=0 \end{align*}$$

Perhaps you can see the differences here? This last equation pair looks like yours except for the sign of \$\tfrac{V_b}{3\Omega}\$. Probably, because you failed to realize it was \$\tfrac{0V-V_b}{3\Omega}\$, as a guess.

Let me know if you need me to walk slowly through how I set up the initial equations.

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  • \$\begingroup\$ Thank you very much. I realized I didn't remain consistent with the KCL in the second equation and had a sign error. Your answer clearly spells it out. \$\endgroup\$ – Drew Sep 12 '16 at 14:47
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When you apply nodal analysis to a circuit, it's useful to assume that all currents are going out or in the node, unless a current source tells you otherwise. That's the approach you took. To analyze node a, you assumed all currents were going out the node and it's correct, but in node b, when you assume that all currents were going out, you included the current give by the 4 A source, when it clearly shows it's not going out the node, it's going in. In your second equation you should've written the 4 amp current on the other side of the equal sign, because it's going in.

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