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Circuit solution

Can anyone help me to solve this circuit using node voltage method? Because in the solution, I don't understand why there is just 3I1 only.

winny
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Jin En Ng
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  • 3I1 is not a solution, it is one of the parameters – jsotola Jun 30 '23 at 15:08
  • Where's I1 in your circuit? – Andy aka Jun 30 '23 at 15:20
  • Jin, I'm not sure what your question is. The solution is just what you wrote at the bottom. And it looks like you were starting out using KCL, which is the node method. So I'm not quite understanding what you feel you need to hear. Or is that a copy of notes you took, but don't follow? – periblepsis Jul 01 '23 at 00:02

1 Answers1

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Look, I think I completely agree with your process. But before I go there, let's take a look at the schematic again:

schematic

simulate this circuit – Schematic created using CircuitLab

I think you can see why I've set up the currents this way. In order for the currents at \$V_x\$ to sum to zero, the above must be true. The value of \$R_3\$ really doesn't matter much. All it does is change the voltage across \$I_1\$. But otherwise it has no impact on that branch.

Getting back to the KCL (nodal):

$$\begin{align*} \frac{V_x}{R_1}+\frac{V_x}{R_2}+3\cdot I_1&= \frac{V_{_\text{IN}}}{R_1} \\\\ \frac{V_x-V_{_\text{IN}}}{R_1}+\frac{V_x}{R_2}+3\cdot I_1&= 0\:\text{A} \end{align*}$$

And that's pretty much what you wrote, already. So I don't find any problems with your work product.

Since \$I_1=\frac{V_{_\text{IN}}-V_x}{R_1}\$:

$$\begin{align*} \frac{V_x-V_{_\text{IN}}}{R_1}+\frac{V_x}{R_2}+3\cdot \frac{V_{_\text{IN}}-V_x}{R_1}&= 0\:\text{A} \\\\ \frac{V_x}{R_1}+\frac{V_x}{R_2}+3\cdot \frac{V_{_\text{IN}}}{R_1}&= \frac{V_{_\text{IN}}}{R_1}+3\cdot \frac{V_x}{R_1} \\\\ V_{_\text{IN}}&=\left(1-\frac12\frac{R_1}{R_2}\right)V_x \end{align*}$$

From that, and given your resistor values, it's clear that \$V_x=2\cdot V_{_\text{IN}}\$.

So what's your difficulty here? I think you did just fine.

periblepsis
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