Solving for Unknown Currents in a Circuit: KCL and Homework Statement

[STEMucator]

Homework Statement

Find $I_x$ and $I_1$ in the following circuit:

Homework Equations

$\sum I = 0$

The Attempt at a Solution

I want to know if I'm actually envisioning these currents correctly.

Applying KCL to the upper left node:

$(1) \quad 6 mA - I_1 - I_x = 0$

Applying KCL to the bottom left node:

$(2) \quad I_x + 1 mA + 1.5 I_x - 6 mA = 0$

Right away now I see that $(2) \Rightarrow 2.5 I_x - 5 mA = 0 \Rightarrow I_x = 2 mA$.

Subbing into $(1)$ I see that $I_1 = 4 mA$.

Am I applying those properly ^? Why did I not need the other nodes at all? Is this because there are four equations and only two unknowns?

Would these be the correct equations for the upper right node and bottom right node respectively?

$(3) \quad I_1 - 1 mA - 1.5I_x = 0$ <- Using this with $(2)$ gives the same answer.

$(4) \quad 1 mA + 1.5I_x = 0$ <- This does not make any sense and I'm not quite sure why. Is there current flowing out of this node?

 

[berkeman]

Your first equation is fine, and the rest may be fine too, but I got vertigo when you chose the bottom right node for your second equation. A more typical approach would be to call the bottom node ground, and write your second KCL equation for the top right node. Do you get the same answer doing it that way?

 

[STEMucator]

Your first equation is fine, and the rest may be fine too, but I got vertigo when you chose the bottom right node for your second equation. A more typical approach would be to call the bottom node ground, and write your second KCL equation for the top right node. Do you get the same answer doing it that way?

Indeed $(3)$ and $(2)$ in combination give the same answer as $(1)$ and $(2)$.

So the bottom right node would be a reference to ground?

 

[berkeman]

Indeed $(3)$ and $(2)$ in combination give the same answer as $(1)$ and $(2)$.

So the bottom right node would be a reference to ground?

The whole bottom line is one node, and yes, normally I would label it as ground.