# How to find a current is this specific circuit

• itzernie
In summary, the voltage source is labeled ix and the current through the 2 ohm resistor is from left to right. To solve for the voltage, Kirchhoff's current law and the loop on the right are used.
itzernie

## Homework Statement

Find the voltage Vo for this circuit.
Given: is = 36 A

## 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 ix , 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 Vo= 2ix and Vo= 8ix which does not make any sense, nor does that help me find ix.

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

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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 + (Vo/2) = Ix

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

2Ix + 6Ix + Vo = 0 and then sub in the above equation and do algebra until

288 + 5Vo = 0Vo = 57.6 V ?

Is my logic correct in any ways?

I'm getting a little confused on the signs.
itzernie said:
288 + 5vo = 0
Solving this equation results in vo = -57.6 V, but the answer should be +57.6 V, according to the way vo 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.

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 Ix ) 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.

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 = Ix + vo/2
Now if I did a voltage loop on the loop on the right, I would get, going CCW:
2ix + vo - 6ix = 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 vo is already set up for that direction. I hope that helps.

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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?

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.

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 vo.

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

itzernie and TomHart
NascentOxygen said:
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
NascentOxygen said:
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!

TomHart
Earlier I said:
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 vo.
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.

## 1. How do I determine the current in a specific circuit?

To find the current in a specific circuit, you can use Ohm's Law (I = V/R) where I is the current, V is the voltage, and R is the resistance. Alternatively, you can also use Kirchhoff's Current Law (KCL) which states that the sum of all currents entering and exiting a junction in a circuit must be equal to zero.

## 2. What is the difference between AC and DC current?

AC (alternating current) refers to a type of current that changes direction periodically, while DC (direct current) flows in only one direction. AC current is commonly used in household electricity, while DC current is used in electronics such as batteries and computers.

## 3. How does the placement of resistors affect the current in a circuit?

The placement of resistors in a circuit affects the current by altering the total resistance in the circuit. The higher the resistance, the lower the current will be. In series circuits, adding resistors will increase the total resistance and decrease the current, while in parallel circuits, adding resistors will decrease the total resistance and increase the current.

## 4. What is a short circuit and how does it affect current flow?

A short circuit occurs when there is a low resistance path in a circuit, causing a large current to flow. This can be dangerous as it can cause overheating or even fire. In a short circuit, the current bypasses the intended path and flows through the low resistance path instead.

## 5. Can current flow in an open circuit?

No, current cannot flow in an open circuit. An open circuit is a break in the circuit where there is no complete path for the current to flow. In order for current to flow, there needs to be a complete circuit with a source of voltage, a conductor, and a load. Without a complete path, there is no current flow.

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