Help figuring out nodal analysis

In summary, the conversation revolved around solving a circuit problem using nodal analysis and finding the equivalent resistance. The participants discussed the correct labeling of nodes and equations, as well as the inclusion of the current source in the equations. The final answer was determined to be Vo=4.
  • #1
naivy
40
0

Homework Statement



Lpa6j.png


Homework Equations



Equivalent resistance?
Nodal analysis?
V=IR

The Attempt at a Solution



2V0lf.png

I was thinking of using nodal analysis at the points designated by the arrows. The bottom one would be connected to ground. My question is if I picked out the nodes correctly. Is the branch with the current source supposed to be a node?

Also, is the total current of the circuit .4444444 ?
 
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  • #2
Welcome to PF, naivy! :smile:

You're nodes look about right.
But your leftmost 2 arrows are connected with a wire without resistor, so their voltages must be the same.

I'm not sure what you mean by total current though.
What do you mean by total current?

As far as I'm concerned, as far as you can talk about a total current that would just be 5 mA.
 
  • #3
Could I also just do something like finding the equivalent resistance?

I calculated it to be 3.6 kΩ. Then I multiplied this by 5 mA to get Vo to be 18V.

Is this answer correct? Thanks. If not, could you please help me some more with nodal analysis?
 
  • #4
Yes, you can also find the equivalent resistance.
But I don't think that makes your problem easier.

And no, your calculated values are not correct.

If you try for the equivalent resistance, you should rearrange the circuit in such a way that the top of the current source is on one side, and the bottom of the current source is on the other side.
From there you can calculate the equivalent resistance, after which you can calculate the Vo voltage drop.

For nodal analysis you should label your top 3 (not 4!) arrows as, say V1, V2 and V0.
The bottom arrow would be 0 V.
From there you can set up a set of 3 equations applying Kirchhoff's current law, (combined with Ohm's law).
Solve the set of equations and you have Vo.
 
  • #5
Here's some work in trying to set up the simultaneous equations. Please show me where I messed up?

g1Hzs.jpg
 
  • #6
Can someone please look at this for me? Thanks
 
  • #7
Just got up. :zzz:

You're on the right track! :smile:

But what did you do with the current source?
The 5 mA is supposed to be in there somewhere.
You should add it in your 3rd equation.

And in your 2nd equation the sign of (V2-V1)/4k appears to be wrong.
 
  • #8
Is it supposed to be (V1 - V2)/4k.

For the third equation, how am I supposed to add the current source?
Does the third equation look like this?

For V': (V1-V')/6k + 5mA = 0

I really appreciate your help! Thanks :D
 
  • #9
naivy said:
Is it supposed to be (V1 - V2)/4k.

Yep! :smile:


naivy said:
For the third equation, how am I supposed to add the current source?
Does the third equation look like this?

For V': (V1-V')/6k + 5mA = 0

I really appreciate your help! Thanks :D

If you look at the diagram in your problem statement at the point you have effectively labeled V', you need to apply KCL there.

Current in is 5 mA, current out is V'/4k + (V'-V1)/6k.
 
  • #10
So then according to KCL, it should be something like V'/4k + (V'-V1)/6k = 5mA ?
 
  • #11
Yep! :wink:
 
  • #12
Wait, was there a negative sign missed there?

is it supposed to be (-)V'/4k + (V'-V1)/6k = 5mA ?? If so, would you mind checking the equations that I have so far to make sure it's right?

(V1-V2)/4k - V2/6k = 0 or (V1-V2)/4k + Vo/6k = 0

(V'-V1)/6k -V1/10k + (V1-V2)/4k = 0

(V'-V1)/6k - V'/4k = 5mA

Thank you!
 
  • #13
naivy said:
Wait, was there a negative sign missed there?

is it supposed to be (-)V'/4k + (V'-V1)/6k = 5mA ?? If so, would you mind checking the equations that I have so far to make sure it's right?

No.
Following what you did earlier, it would be:
-V'/4k + (V1-V')/6k + 5mA = 0


naivy said:
(V1-V2)/4k - V2/6k = 0 or (V1-V2)/4k + Vo/6k = 0

(V'-V1)/6k -V1/10k + (V1-V2)/4k = 0

(V'-V1)/6k - V'/4k = 5mA

Thank you!

eq1: Since you chose V2 equal to Vo, Vo should have the same sign.

eq2: Your sign of (V1-V2)/4k is still wrong.

eq3: The sign of V'/4k is wrong.


Aaaaaaand... I'm off to work. ;)
 
  • #14
Does that mean that Vo=4?
 
  • #15
Yup.
That's what I got too! :)
 

1. What is nodal analysis and why is it important?

Nodal analysis is a method used in circuit analysis to determine the voltage at each node (junction) in a circuit. It is important because it allows scientists and engineers to analyze and design complex electrical circuits by breaking them down into smaller, more manageable parts.

2. How do I perform nodal analysis?

To perform nodal analysis, you will need to follow these steps:

  • Identify all the nodes in the circuit.
  • Select a reference node (usually the one with the most connections) and assign it a ground potential of 0V.
  • Write down the Kirchhoff's Current Law (KCL) equation for each node, using the currents entering and leaving the node.
  • Solve the resulting system of equations for the unknown node voltages.

There are also online tools and software that can help with nodal analysis calculations.

3. What is the difference between nodal analysis and mesh analysis?

Nodal analysis and mesh analysis are two different methods used to analyze circuits. Nodal analysis focuses on determining the voltage at each node in a circuit, while mesh analysis focuses on determining the current through each loop (or mesh) in a circuit. They both use Kirchhoff's Laws, but mesh analysis is typically used for more complex circuits with multiple voltage sources.

4. Can nodal analysis be used for AC circuits?

Yes, nodal analysis can be used for both DC and AC circuits. However, for AC circuits, the equations used will involve complex numbers and phasors, and the analysis will need to be done in the frequency domain.

5. What are some common mistakes to avoid when using nodal analysis?

Some common mistakes to avoid when using nodal analysis include:

  • Incorrectly identifying nodes or reference node.
  • Missing or incorrectly applying Kirchhoff's Laws.
  • Not considering the direction of currents in the equations.
  • Forgetting to include all components in the circuit, such as voltage sources or dependent sources.
  • Using incorrect units for voltage or current.

It is important to double check all steps and calculations to avoid these mistakes and ensure accurate results.

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