Calculating Node Voltages with KCl"

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The discussion focuses on calculating node voltages using Kirchhoff's Current Law (KCL) and highlights errors in the initial equations provided. Participants identify sign errors in the equations related to the voltage drop across resistors and the influence of a voltage source. The correct formulation should reflect that the voltage at Node B is Vb - 10V due to the battery configuration. There is consensus that the original answers provided in the homework may be incorrect, as they conflict with the circuit's layout. The conversation emphasizes the importance of accurately representing voltage relationships in circuit analysis.
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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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