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.

 

[rezeew]

I can confirm that you are applying KCL (Kirchhoff's Current Law) correctly in this problem. KCL states that the algebraic sum of currents entering and leaving a node must equal zero. In this case, you correctly identified the two nodes where KCL can be applied (upper left and bottom left). Since there are only two unknown currents ($I_x$ and $I_1$), you only need two equations to solve for them. This is because the other two nodes (upper right and bottom right) are already connected to the nodes where you applied KCL, so the currents entering and leaving those nodes are already accounted for in your equations.

Your equations for the upper right and bottom right nodes are correct. For the upper right node, you correctly identified that the current leaving the node is equal to the current entering the node ($I_1$). For the bottom right node, you correctly identified that the current leaving the node is equal to the current entering the node ($I_x$).

As for your question about the fourth equation, you are correct that it does not make sense. This is because the bottom right node is not a separate node from the bottom left node. They are connected by a wire, so they are essentially the same node. Therefore, you do not need to apply KCL to the bottom right node separately.

Overall, you have correctly applied KCL to solve for the unknown currents in this circuit. Keep in mind that KCL can only be applied to nodes in a circuit, and not to individual components. Also, you may not always need to use all of the nodes in a circuit to solve for unknown currents. It depends on the specific circuit and the number of unknowns.