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schematic

simulate this circuit – Schematic created using CircuitLab

I drew a simplified schematic of the circuit I was analyzing. I was writing the node equation for node B and came up with this:

\$\frac{V_a-V_b}{2}+2=0\$

I put zero because there is no resistance between \$V_b\$ and the ref. Then \$I_{R1}=-4\$, is that correct? Doesn't seem to make sense that the current will be that no matter what is to the rest of the circuit...

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  • \$\begingroup\$ Do the wires going off to the left connect to anything? The answers you've gotten so far assume they don't! \$\endgroup\$ – The Photon Oct 23 '16 at 0:49
  • \$\begingroup\$ Yes they do but I didn't drew then because it wouldn't affect the node equations (I guess), I had this confusion when writing the nodal equation for node B but is fine now. Thanks! \$\endgroup\$ – João Pedro Oct 23 '16 at 22:14
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The node equation for \$V_b\$, expressed as outgoing currents on the left and incoming currents on the right, is:

$$\frac{V_b}{R_1} = \frac{V_a}{R_1} + I_2 + I_{V_1}$$

But you know that \$I_{V_1}=0\$ as there is no galvanic path for any current through it. This results in:

$$ \frac{V_a-V_b}{R_1} + I_2 = 0$$

Which is exactly as you wrote. You didn't do anything wrong there, at all.

Just solve for \$V_a\$:

$$\begin{align*} \frac{V_a-V_b}{R_1} + I_2 &= 0 \\ \\ \frac{V_a-V_b}{R_1} &= -I_2 \\ \\ V_a-V_b &= -I_2\cdot R_1 \\ \\ V_a &= V_b-I_2\cdot R_1 \\ \\ V_a &= 4\:\textrm{V}-2\:\textrm{A}\cdot 2\:\Omega = 0\:\textrm{V} \end{align*}$$

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    \$\begingroup\$ @JoãoPedro I thought you did fine, but just didn't complete your thinking all the way through. From this, of course, you can work out the fact that the current in \$R_1\$ must be \$2\:\textrm{A}\$, of course. (Which is the only path for that current source.) \$\endgroup\$ – jonk Oct 22 '16 at 22:44
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I(R1) is 2A. That's set because it's the only path where the current from the current source can go through. That means |Va-Vb|==2A*2Ω==4V. With Vb=4V, that means Va=0V.

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  • \$\begingroup\$ The current couldn't go to the voltage source? What if the voltage source polarity were reversed (positive connected to ground)? \$\endgroup\$ – João Pedro Oct 22 '16 at 22:22
  • \$\begingroup\$ The voltage source has only one connection to the circuit, so there's no current going through it. It only lifts the potential of the upper circuit part on 4V above ground potential. If it was connected reversed, Vb would be -4V against ground and Va would be -8V against ground. \$\endgroup\$ – Janka Oct 22 '16 at 22:26
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I think there is some confusion here because not everyone caught that there is additional circuitry to the left. Your equation needs an addition +Iv1 term. Because there is no way to calculate this current it is an addition unknown in the analysis. That means you need one more equation to solve for the new unknown. The additional equation would be Vb=4V. Whenever you have voltage sources doing node analysis you will need this extra equation. So you add the new variable Iv? to the nodes it connects to. The additional equation will be Vx-Vy = Vsource if the source is connected between nodes x and y.

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