The superposition theorem with dependent sources

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Discussion Overview

The discussion revolves around applying the superposition theorem to determine the current Ix in a circuit with dependent sources. Participants explore various methods of analysis, including nodal analysis and Kirchhoff's laws, while addressing specific calculations and interpretations of circuit behavior.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant describes their approach to finding Ix' using Kirchhoff's Voltage Law (KVL) and arrives at a value of 2A, but expresses confusion regarding the nodal analysis for V''.
  • Another participant corrects their previous answer from 2A to 1.4A, indicating uncertainty in their calculations.
  • Questions arise about the interpretation of V'' in relation to the current through resistors and the dependent source, with participants discussing whether their reasoning aligns with KVL.
  • One participant references a similar problem from another thread, suggesting it may provide clarity.
  • There is a mention of a calculated value for Ix'' as -1.75A, with participants questioning if this value is correct and how it relates to KVL equations.
  • Another participant emphasizes the importance of confirming calculated currents using Kirchhoff's laws to ensure consistency.

Areas of Agreement / Disagreement

Participants express varying degrees of confidence in their calculations and interpretations, with no consensus reached on the correctness of specific values or approaches. Multiple competing views remain regarding the application of KVL and the relationships between circuit elements.

Contextual Notes

Participants note potential confusion regarding the application of KVL and nodal analysis, as well as the interpretation of dependent sources in the circuit. There are unresolved mathematical steps and assumptions that could affect the outcomes.

DODGEVIPER13
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Homework Statement


Use the superposition principle to determine the value of Ix.


Homework Equations


I1+I2+I3=0 KCL
V1+V2+V3=0 KVL
Ix'+Ix''= answer

The Attempt at a Solution


So I understand how to get Ix' its just standard KVL it is -10+2Ix'+Ix'+2Ix'=0 which solving for Ix' gives 2A. This is the part I don't get so using nodal analysis (V''/2)+((V''-2Ix'')/1)=3 which when I simplify I get 6=3V''-4Ix''. So staying on this part why V''/2 is it this because that gives the current that goes across the 2 ohm resistor and V'' is the voltage on the wire? Also the second part ((V''-2Ix'')/1) is it this because your getting the current going from left to right over the 1 ohm resistor and thus by using the potential over 1 ohm which gives the current flowing from the left side of the resistor to the right. You would subtract off the dependent source part because it flows against the other current, so you get the current that was being solved for which is from left to right? essentially you are doing this I1=I2+I3 where I1=3A. So am I right there? On the next part I am confused Vx''+2Ix''=0 which when solved gives Vx''=-2Ix'' is this because the 3A source is in parallel with the dependent source and so the voltage is the same across both and thus V''=-2Ix''. Or is it because of KVl and so starting from the far left of the circuit I get 2Ix''+Vx''=0. I guess what I am asking is am I doing so form of KVL here or is it just simply relating the dependent source quantity. Anyways when this is all done I get the answer which is 2A.
 

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Sorry I meant 1.4 not 2A on the last part
 
Sorry if it is really confusing
 
Oh also is my picture readable I am sorry if it is too small let me know
 
I can re upload or clarify anything just let me know please
 
Ok thanks on his problem I get Ix''=-1.75 A is that correct
 
So 3-1.75=1.25
 
Where v''=-3Ix'' but why is this is it because of KVL equations giving me 3Ix''+v''=0 or is it just simply a relation
 
  • #10
DODGEVIPER13 said:
Ok thanks on his problem I get Ix''=-1.75 A is that correct
Once you have worked out a value for the current, use Kirchoff's Laws to confirm that that value satisfies ∑ currents into a node = 0. If it does, it must be right. Or check your current together with the knowledge that the voltage at the centre node is 20¼ volts.
 

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