Electronic – Circuit with dependent source confusion

There are only two ways that the current through a resistor can be independent of the value of the resistor:

1. The current is zero. This means that the applied voltage must be zero as well. Your solution of making the capacitance infinite (effectively a short circuit) falls into this category.

2. The resistor is connected to a current source. Here's a hint: The circuit as given shows a voltage source connected in series with an inductor, a Thévenin source. It can be redrawn as its Norton equivalent: a current source in parallel with that same inductor. This effectively puts the inductor in parallel with the capacitor and the resistor, too. Is there a way to make sure that all the current from the current source flows through only the resistor?

Let me start with a direct answer. The existing current source, dependent or otherwise, can only drive through the existing \$2\:\textrm{k}\Omega\$ resistor and the voltage source. That's the only loop available to it. Let me redraw the schematic using this editor here:

^{simulate this circuit – Schematic created using CircuitLab}

Considering that current source, again, notice that I've added a sign around \$R_1\$ to indicate the polarity suggested by the current source direction and the loop it must go through. We'll use that as we create the KVL equation, shortly.

Let's walk through it, starting at the upper left corner where I wrote "starting point" on it. I'll follow around the loop shown by the green arrows:

$$\left(-4\:\textrm{V}\right) + \left(V_x\right) + \left(0\:\textrm{A}\cdot R_2\right) + \left(-R_1\cdot\frac{V_x}{4000}\right) = 0\:\textrm{V}$$

This is roughly speaking the same thing they wrote. Do you see how that works?

In any case, you can now easily solve for \$V_x=8\:\textrm{V}\$.

But you need to also understand that a current source is, effectively, infinite impedance. So the voltage source doesn't produce a separate current through it. There's no separate "I" there. The only current through \$I_1\$ is the dependent current. And that has nothing whatever to do with \$V_1\$, since \$V_1\$ can't affect it.

Does that help any?