KCL Problem with Dependant sources

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Discussion Overview

The discussion revolves around a circuit analysis problem involving Kirchhoff's Current Law (KCL) and dependent sources. Participants are attempting to determine the voltage drop V1 in a circuit with given resistances and an input voltage of 1V, while exploring various approaches to formulating the KCL equations.

Discussion Character

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

Main Points Raised

  • One participant attempts to apply KCL at a node but questions the correctness of their derived equation, suggesting that their logic may be flawed.
  • Another participant challenges the initial node equation, asking for clarification on how certain terms were derived, particularly regarding the use of Vin in the equations.
  • Some participants suggest combining resistors in parallel to simplify the analysis and recommend re-evaluating the KCL equation.
  • There is a discussion about the potential difference across the resistors and the correct identification of node voltages, with some participants expressing uncertainty about how to find the node voltage.
  • One participant proposes using a reference node for simplification and questions the necessity of introducing additional variables for node voltages.
  • Eventually, a participant arrives at a value for V1 and calculates the power delivered by the dependent source, which is acknowledged as correct by others.

Areas of Agreement / Disagreement

Participants express various viewpoints on the correct application of KCL and the identification of node voltages. While some progress is made towards a solution, there is no clear consensus on the initial steps or the accuracy of the earlier equations.

Contextual Notes

Participants note limitations in their understanding of dependent sources and the potential differences across resistors, indicating that assumptions about node voltages and circuit configurations may not be fully resolved.

bran_1
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Homework Statement


I'm trying to find the voltage drop V1 of the following circuit, assuming Vin=1v, and for the life of me can't seem to get the right answer.
QWnesTY.jpg

Homework Equations


KCL, Ohm's law

The Attempt at a Solution


I basically tried to use KCL at the top node,
Code: 
Vin/2+2V1 = V1/3 + Vin/6, so 1/2=-2V1+V1/3+V1/6, (1/2)/(1.5)=V1=1/3, but apparently that's the incorrect answer

Thanks for any help
 
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Hi bran_1, Welcome to Physics Forums.

I don't understand the terms of your node equation. For example, the first term of the LHS (left hand side) is Vin /2. How do you arrive at that? And on the RHS you have Vin/6, but Vin is not the node voltage. Can you explain your reasoning?
 
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gneill said:
Hi bran_1, Welcome to Physics Forums.

I don't understand the terms of your node equation. For example, the first term of the LHS (left hand side) is Vin /2. How do you arrive at that? And on the RHS you have Vin/6, but Vin is not the node voltage. Can you explain your reasoning?

Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node. As for the Vin over 6, since the 3ohm and 6ohm resistors are in parallel, they share the same voltage, do they not? This is just ohm's law again, except for the current leaving the node through this branch.Edit: I didn't get it correct though, so my logic is flawed somewhere...
 
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bran_1 said:
Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node. As for the Vin over 6, since the 3ohm and 6ohm resistors are in parallel, they share the same voltage, do they not? This is just ohm's law again, except for the current leaving the node through this branch.Edit: I didn't get it correct though, so my logic is flawed somewhere...
You need not assume any value for Vin.
You should combine the 3 ohm and 6 ohm resistors in parallel and use their equivalent resistance (for simplicity). You'll still have the two nodes, and three currents instead of four.

Write the KCL (node voltage) equation again and simplify.
 
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cnh1995 said:
You need not assume any value for Vin.
You should combine the 3 ohm and 6 ohm resistors in parallel and use their equivalent resistance (for simplicity). You'll still have the two nodes, and three currents instead of four.

Write the KCL (node voltage) equation again with and simplify.

I tried that also, it didn't seem to make a difference, as I got the same answer.
Vin/2 + 2V1 = V1/2
Vin = -3V1
V1=-(1/3)V, which is incorrect.

Is there some rule with dependent sources that I'm unaware of?
 
bran_1 said:
Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node.
Vin is not the potential difference between the two ends of the 2 Ω resistor. Only the left side of that resistor is at potential Vin. What is the potential at its other end? So then what is the potential difference?
 
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Wouldn't it be VA (of the node) on the other side? I'm not sure how to find the node voltage though
 
bran_1 said:
Wouldn't it be VA (of the node) on the other side? I'm not sure how to find the node voltage though
It's an unknown at this point. You'll solve for it after writing your node equation.
 
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gneill said:
It's an unknown at this point. You'll solve for it after writing your node equation.

Alright, so I'd have (Vin-Va)/2+2V1=V1/2, right? Could I call the bottom node node B, and say it's V=0? So then V1=Va-Vb? -> V1=Va?
 
  • #10
bran_1 said:
Alright, so I'd have (Vin-Va)/2+2V1=V1/2, right? Could I call the bottom node node B, and say it's V=0? So then V1=Va-Vb? -> V1=Va?
Better.

Note that you can see that Va = V1 just by inspecting the circuit diagram. You don't need to introduce a different variable. Why not just use V1?

Yes, the bottom node makes an excellent choice for the reference node.

upload_2017-1-25_9-13-34.png
 
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  • #11
gneill said:
Better.

Note that you can see that Va = V1 just by inspecting the circuit diagram. You don't need to introduce a different variable. Why not just use V1?

Yes, the bottom node makes an excellent choice for the reference node.

View attachment 112093

So using that, I have
Vin-V1=-3V1+V1 -> V1=-0.5V
 
  • #12
Looks good.
 
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  • #13
gneill said:
Looks good.

So then the power delivered by the dependent source would be:

P=VI , -0.5 * (2*-0.5) =1/2 watts
 
  • #14
bran_1 said:
So then the power delivered by the dependent source would be:

P=VI , -0.5 * (2*-0.5) =1/2 watts
Yes, that looks right.
 
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  • #15
gneill said:
Yes, that looks right.
Thanks for all your help!
 

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