Calculating Node Voltages with KCl"

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Discussion Overview

The discussion revolves around calculating node voltages using Kirchhoff's Current Law (KCL) in a circuit analysis context. Participants are attempting to derive equations for node voltages A and B, but are encountering issues with their equations and the signs of the terms involved.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents equations for node A and node B but questions their correctness.
  • Multiple participants identify potential sign errors in the equations, but do not agree on the specific nature of these errors.
  • There is a discussion about the relationship between node B and the voltage source, with suggestions that the term for I2 may be incorrect.
  • One participant suggests that the voltage drop from Vb to the 0.2 ohm resistor should be represented as (Vb - 10), indicating confusion over the correct sign for the voltage terms.
  • Another participant expresses uncertainty about the correct answers, stating their derived values for Va and Vb, which are echoed by another participant.
  • There is a suggestion that the provided answers in the original problem may be inconsistent with the circuit as drawn.

Areas of Agreement / Disagreement

Participants generally disagree on the correct formulation of the equations, particularly regarding the signs of the voltage terms. There is no consensus on the correctness of the original equations or the provided solutions.

Contextual Notes

Participants express confusion over the signs in their equations and the relationships between the voltages and the components in the circuit. The discussion highlights the dependency on how voltage drops are defined in relation to the circuit elements.

TheRedDevil18
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Homework Statement



2dw183.jpg


Homework Equations

The Attempt at a Solution



I'm not getting the correct answers, here are my equations

Node A:
80 = ix + i2
80 = Va/0.143 + (Va-(Vb+10))/0.2......1

Node B:
20 = iy-i2
20 = Vb/0.125 - (Va-(Vb+10))/0.2......2

Are these equations correct ?
 
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equation 1... i see a sign error
equation 2...same problem
 
donpacino said:
equation 1... i see a sign error
equation 2...same problem

I don't see where the sign error is
 
There is an error with your I2 term. Look at the relationship between VB and the voltage source
 
donpacino said:
There is an error with your I2 term. Look at the relationship between VB and the voltage source

I don't know, Is it -10V ?, I'm confused with the signs :frown:
 
TheRedDevil18 said:
80 = Va/0.143 + (Va-(Vb+10))/0.2......1
You drop down from Vb by 10v to get to the 0.2 ohm resistor. Should be ...(Vb-10)...

If you draw an arrow from (-) to (+) on the battery, which you should do and label it +10V, you can see the drop in potential in going from Vb towards the 0.2 ohm. The bottom of the battery is 10v less than the top.
 
NascentOxygen said:
You drop down from Vb by 10v to get to the 0.2 ohm resistor. Should be ...(Vb-10)...

If you draw an arrow from (-) to (+) on the battery, which you should do and label it +10V, you can see the drop in potential in going from Vb towards the 0.2 ohm. The bottom of the battery is 10v less than the top.

I still don't get the correct answers. I get Va = 5.65V and Vb = 7.56V
These are my equations
80 = Va/0.143 + (Va-(Vb-10))/0.2

20 = Vb/0.125 - (Va-(Vb+10))/0.2
 
TheRedDevil18 said:
I still don't get the correct answers. I get Va = 5.65V and Vb = 7.56V
These are my equations
80 = Va/0.143 + (Va-(Vb-10))/0.2

20 = Vb/0.125 - (Va-(Vb+10))/0.2

that is what I got too. either we both made the same mistake, or the 'solution' is incorrect.
 
The answers accompanying the question in post #1 are consistent with the 20A source being directed downwards, so are not appropriate for the circuit as drawn.
 
  • #10
Ok, maybe the solutions are incorrect. I'm a bit confused with the second equation though, why is it Vb+10, because the 0.2 ohm resistor is still connected to the negative end of the battery. Shouldn't it be Vb-10 ?
 
  • #11
It should be Vb - 10 for the reason I gave.
 
  • #12
NascentOxygen said:
It should be Vb - 10 for the reason I gave.

For both the equations ?
 
  • #13
TheRedDevil18 said:
For both the equations ?
Yes, it defines the voltage at one end of that 0.2 Ω resistor.
 
  • #14
Ok thanks guys
 

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