Question: [![enter image description here](https://i.stack.imgur.com/kR0Y7.png)](https://i.stack.imgur.com/kR0Y7.png)

This is just a solved problem from a book on circuit analysis (Hayt), but I got a little confused on the voltage drop across the 2K resistor. This confusion might be about dependent sources. Why did he write the current as a result of only the current source and ignored the voltage source? Why not R\*(Vx/4000 - I) where I is the current resulted from the voltage source? This question could be generalized to: dependent sources takes account of the whole circuit? What I mean by that is, this function (Vx/4000) takes account of the whole network, so the current in that wire will always be that of the function regardless of the rest of the circuit? Appreciate any insight.

Answer: 
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:

![schematic](https://i.stack.imgur.com/v1IYu.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fv1IYu.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

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?
An ideal current source will "overpower" and ideal voltage source. 

In this case, because the output Vx is open there is no way current can flow that way. The only place the current has to go is through the 2k resistor and into the voltage source. 

An ideal voltage source can both sink and source current. In this case the voltage source is sinking current.