Engineering Circuit problem - need a sanity check

[DrummingAtom]

Homework Statement

Find V_x. With R₁, R₂ and i₃ given.

*In my attachment, I renamed the given values.*

Homework Equations

KCL, KVL, Ohm's law

The Attempt at a Solution

For KVL, I solved for V₃ in L₂ then subbed it in L₁ to give:

3V_x - V_x + V₁ = 0

From here, I tried to get my KCL equations to have i₁ in terms of i₃ so I could replace V₁ with R₁ and i₁. But the KCL won't let me do that.. in fact, I can't get any of those KCL equations in different terms they just cancel each other out in every case.

I'm lost if this is an error in the problem or I'm doing something wrong. Thanks for any help.

 

[gneill]

I think you're making your life more difficult than need be; the two resistors are in series and thus form a single branch and have a single current. So you really only need one node (where R2 and the two current supplies join) in order to write your KCL equation.

The two current supplies feed into that node, and one path leads out (through R2). Call the latter current i2. What's Vx in terms of this current i2?

 

[DrummingAtom]

I think you're making your life more difficult than need be; the two resistors are in series and thus form a single branch and have a single current. So you really only need one node (where R2 and the two current supplies join) in order to write your KCL equation.

The two current supplies feed into that node, and one path leads out (through R2). Call the latter current i2. What's Vx in terms of this current i2?

V_x = i₂R₂ or = 5i₂ with the given value.

 

[gneill]

V_x = i₂R₂ or = 5i₂ with the given value.

Okay, that makes one equation. What's the KCL equation for the node?

 

[DrummingAtom]

Node = i₃+i₄-i₂ = 0

Then plugging that into V_x = 5(i₃ + i₄).

 

[DrummingAtom]

The main thing I'm still confused about was my original KCL equations. They showed no solutions, does that mean it's in series?

 

[gneill]

Node = i₃+i₄-i₂ = 0

Then plugging that into V_xi₂ = 5(i₃ + i₄).

Careful, you've made an assumption about the direction of current i₂ when you wrote the expression for Vx. Better make sure that you remain consistent!

 

[DrummingAtom]

Hmm, I'm still confused about when I can make assumptions on the direction.

So, it would be:

V_x = 5(-i₃ - i₄)

 

[gneill]

The main thing I'm still confused about was my original KCL equations. They showed no solutions, does that mean it's in series?

A few things spring to mind when I look at your original KCL equations. First, note that the two resistors are in series and thus form a single branch. This would eliminate your node "A" from consideration. Second, you don't really need to write the equation "C", since if you select it as the choice for the reference node then its equation will be redundant, duplicating information in the others. Third, you've introduced two currents (i₁ and i₂) in a single branch; that's bound to lead to either redundancy or unsolvable equations.

To write KCL equations to solve a circuit, start by selecting a reference node (or "ground" node as it's sometimes called). Then identify the other independent nodes (In this circuit there's only one). Write the KCL equation for each independent node. If there's more than one independent node you'll want to assign variables for node voltages and use them along with KVL on the branches between nodes to replace the unknown currents with voltage/resistance expressions. Then solve for the node voltages.

 

[gneill]

Hmm, I'm still confused about when I can make assumptions on the direction.

So, it would be:

V_x = 5(-i₃ - i₄)

When you wrote the equation V_x = i₂*R₂ you tacitly made the assumption that i₂ is flowing from left to right through R₂ and thus INTO your node "B". Thus all the currents are taken as flowing into node B, and the KCL for that node is:

i₂ + i₃ + i₄ = 0

Now you can replace i₂ and i₄ with expression involving V_x (which is, after all, the variable you're hoping to solve for!). You've got your V_x = i₂*R₂ (so solve for i₂) and your i₄ = 3*V_x.

 

[DrummingAtom]

Ok, I was going that route but wouldn't it be 3V_x = i₄R₄? That was the only reason I chose not to do that because I thought I would be bringing in another unknown variable..

 

[DrummingAtom]

Thanks for your help by the way, I really appreciate it.

 

[gneill]

Ok, I was going that route but wouldn't it be 3V_x = i₄R₄? That was the only reason I chose not to do that because I thought I would be bringing in another unknown variable..

I only see resistors R1 and R2 in that circuit. i₄ = 3V_x.