Help figuring out nodal analysis

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Discussion Overview

The discussion revolves around the application of nodal analysis in a circuit problem, focusing on identifying nodes, setting up equations, and calculating equivalent resistance. Participants explore different approaches to solving the problem, including the use of Kirchhoff's laws and the implications of current sources.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether the branch with the current source should be considered a node.
  • Another participant suggests that the total current in the circuit is 5 mA, while there is uncertainty about the meaning of "total current."
  • Some participants propose finding the equivalent resistance as an alternative approach, with one calculating it to be 3.6 kΩ.
  • There is a discussion about setting up simultaneous equations for nodal analysis, with specific attention to the signs in the equations and the inclusion of the current source.
  • Participants express confusion regarding the correct formulation of equations based on Kirchhoff's current law, particularly concerning the signs and terms related to the current source.
  • One participant confirms the need to apply KCL at a specific node and provides a formulation for the equation involving the current source.
  • There are corrections and refinements to the equations proposed by participants, with ongoing adjustments to signs and terms as they work through the problem.
  • At the end, one participant suggests a potential value for Vo, which is affirmed by another participant.

Areas of Agreement / Disagreement

Participants generally agree on the need for nodal analysis and the use of KCL, but there are multiple competing views regarding the correct formulation of equations and the treatment of the current source. The discussion remains unresolved in terms of confirming the correctness of the proposed equations.

Contextual Notes

Some participants express uncertainty about the signs in their equations and the implications of the current source, indicating potential limitations in their understanding of the circuit configuration and the application of nodal analysis.

Who May Find This Useful

Students or individuals seeking assistance with circuit analysis, particularly those learning about nodal analysis and Kirchhoff's laws in electrical engineering contexts.

naivy
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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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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.
 
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?
 
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.
 
Here's some work in trying to set up the simultaneous equations. Please show me where I messed up?

g1Hzs.jpg
 
Can someone please look at this for me? Thanks
 
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.
 
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
 
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! :)
 

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