How to find a current is this specific circuit

[itzernie]

Homework Statement

Find the voltage V_o for this circuit.
Given: i_s = 36 A

Link to circuit: https://session.masteringengineering.com/problemAsset/1824621/4/pr_2-32.jpg

Homework Equations

V = IR
Kirchhoff's laws for current and voltage.

The Attempt at a Solution

I first noticed that the voltage source is labeled i_x , so I knew I needed to find that first. I tried using the Kirchhoff current law at the top middle junction, but realized that is no help since there are two currents at the junction I do not know. Then I tried using a loop law around the right and outer loops. When doing this I can simplify equations to V_o= 2i_x and V_o= 8i_x which does not make any sense, nor does that help me find i_x.

Once I find i_x, I can use the junction law to solve for the current into the 2 ohm resistor and then solve for V_o.

 

[TomHart]

If you use the current law, what if you write the current through the 2 ohm resistor in terms of the voltage across it.

 

[itzernie]

Thanks for answering! Ahh ok. I believe I see what you are saying. To be clear, by using the Current junction law on the top I would get the equation:

36 + (V_o/2) = I_x

then using the loop on the right I get the equation:

2I_x + 6I_x + V_o = 0 and then sub in the above equation and do algebra until

288 + 5V_o = 0V_o = 57.6 V ?

Is my logic correct in any ways?

 

[TomHart]

I'm getting a little confused on the signs.

288 + 5vo = 0

Solving this equation results in v_o = -57.6 V, but the answer should be +57.6 V, according to the way v_o is defined in the diagram.
You should go back and look at the signs of the terms in your equations to make sure they are correct.

 

[itzernie]

You are correct. I did not believe it should be negative either so I tried 57.6 and that was incorrect (no hint saying wrong sign came up so I know it isn't -57.6 as well).

Starting in the top left corner of the circuit and going clockwise, I don't believe my loop equation could be wrong ( enters + terminals and against current so all factors would be positive)

I think what I am most confused about is the direction of the current on the 2 ohm resistor.
In the above calculations I said it was going left (I assumed since I_x ) was going down that to complete a current loop it must be going left.

I just tried 36 - (Vo/2) = Ix to get an answer of 96 (from 288 - 3vo = 0) but that was incorrect as well.

I'm not sure where else I am going wrong. hmmm.

 

[TomHart]

Sorry, but I made a mistake when I worked out the problem.

Because of the way vo is defined in the picture, it implies that the current through the 2 ohm resistor is from left to right. So I would sum the currents into that node as follows:
36 = I_x + v_o/2
Now if I did a voltage loop on the loop on the right, I would get, going CCW:
2i_x + v_o - 6i_x = 0

As I'm looking at components when I do my voltage loop, if I come to a - sign first, that will be a positive voltage (a voltage increase). If I come to a + sign first, that will be a voltage drop. So because Ix is defined downward, the top side of the 6 ohm resistor will be the + side and the bottom of the 6 ohm resistor will be a - sign. That is why it is considered to be a voltage drop as I loop in the CCW direction. Does that make sense?

Edit: Just to try to clarify a little more, it is okay if you assume that the current through the 2 ohm resistor is flowing from right to left. But if you do that, you have to define the voltage across the 2 ohm to agree with that convention. In other words, when you do your voltage loop, the voltage drop of the resistor has to be in agreement with the way the definition of the current direction. So for this problem, it is just easier to assume the current through the 2 ohm flows from left to right because the definition of v_o is already set up for that direction. I hope that helps.

 

[itzernie]

It's ok without you I would not even be this far. But Ohhh Ok I believe I fully see what you are saying now! Now I get an answer of Vo = 48 V where Ix ends up being 12. The voltage source is 24 and now 24-72+48= 0 which does indeed fully cancel. Did you get 48 as well?

 

[NascentOxygen]

+48V is my answer

If v_o is not the polarity you'd choose, you can leave it off the diagram and solve for however you prefer the voltage polarity across the 2Ω resistor, V_R. Then right at the end, determine v_o by saying that v_o=‑V_R.

Just the same, it is good practice to go with the designated polarity and mark the resistor's current direction in accord with the designated polarity of v_o.

Try it both ways. You should get the same result. [emoji1]

 

[TomHart]

If vo is not the polarity you'd choose, you can leave it off the diagram and solve for however you prefer the voltage polarity across the 2Ω resistor, VR. Then right at the end, determine vo by saying that vo=‑VR.

That is what I wanted to say, but I couldn't seem to figure out how to say it.

 

[itzernie]

If vo is not the polarity you'd choose, you can leave it off the diagram and solve for however you prefer the voltage polarity across the 2Ω resistor, VR. Then right at the end, determine vo by saying that vo=‑VR.

Ahhh that makes sense. Thank you both for all of the help!

 

[NascentOxygen]

Just the same, it is good practice to go with the designated polarity and mark the resistor's current direction in accord with the designated polarity of v_o.

What I meant is "... it would be a useful exercise for you to try solving this using the designated polarity ..."

Try it both ways. You should get the same result.