Problem with Node Voltage Method

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Discussion Overview

The discussion revolves around the application of the Node Voltage Method to analyze a circuit involving a 5V voltage source and various currents. Participants are working through the equations derived from Kirchhoff's Current Law (KCL) at different nodes and expressing uncertainty regarding sign conventions and the relationships between currents and voltages in the circuit.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The original poster presents KCL equations for nodes 1 and 2 but expresses uncertainty about the sign convention related to the voltage source and the current source with resistors.
  • One participant suggests that at node 2, KCL indicates the current in the remaining branch can be determined based on the specified currents in the other branches.
  • Another participant confirms the KCL equation for node 1 and asks for validation of their expression.
  • There is a suggestion that knowing all but one current at a node allows for the determination of the unknown current.
  • One participant proposes a value for i2 based on their understanding of the currents at node 1.
  • Another participant confirms the calculation of i1 and expresses confidence in their understanding of the currents at node 2.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of the sign conventions and the relationships between the currents and voltages. There are multiple interpretations of the KCL equations and the values of the currents involved, indicating ongoing uncertainty and exploration of the problem.

Contextual Notes

Participants express confusion regarding the sign conventions and the impact of the voltage source on the KCL equations. There is also a lack of clarity on how to proceed with the analysis when current sources are involved.

challarao
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1. Find the power delivered by the 5V Voltage source in the network of Fig.
node.jpg

This is my attept at solution:

At node 1 V1...assuming current leaving V1:
KCL equation: -1A+1A+[(V1-5)/3]+[(V1-(V2-10))/2]=0

At node 2 V2...assuming current leaving V2:
KCL equation: -2A-1A+[V2-(V1+10)]=0

But, I'm not certain about the sign convention when there is a voltage source with a resistor between V1 and V2...
I'm also confused that how to proceed when there is a current source and resistor between two nodes...
Where did I go wrong...?
Please help me...
 
Last edited:
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If you look at the node labelled 2, you'll note that three branches come together and two of them have specified currents. What does KCL tell you about the current in the remaining branch? Can you now repeat this feat at node 1?
 
gneill said:
If you look at the node labelled 2, you'll note that three branches come together and two of them have specified currents. What does KCL tell you about the current in the remaining branch? Can you now repeat this feat at node 1?
Thank you
Yes, there are four branches coming together at node 1.Currents are specified for two of the branches.

At node 1 KCL: -1+1+[(v1-(-5))/3]+[v1-(v2+10)]/2=0

Did I write it correctly?
Thanks...
 
challarao said:
Thank you
Yes, there are four branches coming together at node 1.Currents are specified for two of the branches.

At node 1 KCL: -1+1+[(v1-(-5))/3]+[v1-(v2+10)]/2=0

Did I write it correctly?
Thanks...

You won't even need the node voltage... If you know all but one current
flowing through the node, the last one is thus determined y the others.
So what must i2 be?
 
I think i2=(v1+5)/2, am I correct?
 
Look again at node 1. You've determined what i1 must be, coming from node 2, right? So all branches leading to node 1 have known currents except for the branch with i2. So what's i2?
 
As 2A and 1A are reaching node 2...3A must leave it..

So, i1=3A, am I correct..?
and wow i2=3A...I got it now ...I think I'm correct...amn't I?
Thank you very very much...
 
Last edited:
challarao said:
As 2A and 1A are reaching node 2...3A must leave it..

So, i1=3A, am I correct..?
and wow i2=3A...I got it now ...I think I'm correct...amn't I?
Thank you very very much...

Yup. :smile:

Now you can figure out the power delivered by the 5V supply.
 

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