Superposition, circuit analysis.

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Discussion Overview

The discussion revolves around circuit analysis using superposition, specifically focusing on the behavior of resistors and current sources in a given circuit. Participants are analyzing voltage and current values across various resistors based on their calculations and interpretations of the circuit configuration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents calculated values for currents and voltage across resistors, indicating a potential confusion regarding the presence of a 50Ω resistor.
  • Multiple participants assert that there is no 50Ω resistor in the circuit, suggesting a need for clearer labeling of components.
  • Another participant proposes a voltage divider approach to analyze the circuit, referencing the formula for voltage dividers.
  • Further calculations are provided for current through a 2Ω resistor, with a participant expressing uncertainty about the current through the 10Ω resistor.
  • One participant questions the simplicity of the circuit divider approach, indicating that it may not be straightforward.
  • A later reply discusses the process of visualizing the equivalent circuit and emphasizes the importance of experience in circuit analysis.

Areas of Agreement / Disagreement

Participants generally disagree on the presence of certain resistors and the clarity of the circuit configuration. There is no consensus on the correct approach or final values, as multiple interpretations and calculations are presented.

Contextual Notes

Some calculations appear to depend on assumptions about the circuit configuration, and there are unresolved questions regarding the correct identification of components and their relationships.

lam58
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Hi, could check if I have the right answer please. The question (Q 4b) is in the attached image below.

My answer is:First for analysis the DC voltage source.

i2 = 0.82A, i1 = 1.97A

therefore V across 10Ω resistor = 11.5V.

For the analysis of the current source I got a bit confused, however, I think I might have got it right. I shorted the voltage source and found the current across the 50Ω resistor by doing:

Rt= (10//2) + 12 = (41/3)

i50Ω = (Rt*is)/(Rt+50Ω) = 0.86A

Then to find the current across the 10Ω resistor:

Rt = 2Ω hence i10Ω = (2*0.86)/(2+10) = 0.14A.

Thus V10Ω = 1.4V.

Then adding the two separate source responses across the 10Ω resistor the total voltage V10Ω= 1.4 + 11.5 = 12.9V.

Is this correct? :confused:
 

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There is no 50 ohm resistor.

First thing is to label the resistors and sources. Put in numbers only at the very end.
 
rude man said:
There is no 50 ohm resistor.

First thing is to label the resistors and sources. Put in numbers only at the very end.

Woops lol.I'll try again.
 
First if we remove current source we have a simply voltage divider:
Vx1 = 110V * (10||14)/(5 + 10||14)
Next we connect a current source but we short a 110 voltage source we have a very simple current divider.
http://en.wikipedia.org/wiki/Current_divider
 
Jony130 said:
First if we remove current source we have a simply voltage divider:
Vx1 = 110V * (10||14)/(5 + 10||14)
Next we connect a current source but we short a 110 voltage source we have a very simple current divider.
http://en.wikipedia.org/wiki/Current_divider

The circuit divider doesn't seem that simple to me. ?
 
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Jony130 said:
First we have a voltage divider
http://en.wikipedia.org/wiki/Voltage_divider#Resistive_divider

Vout = Vin * R2/(R1 + R2)

So in our case
R2 = 10Ω||14Ω; And R1 = 5Ω; Vin = 110V;

attachment.php?attachmentid=65541&stc=1&d=1389556772.png


Next we have a two current dividers
http://en.wikipedia.org/wiki/Current_divider
attachment.php?attachmentid=65542&stc=1&d=1389556823.png


The current that is flow through 2Ω resistor is equal to:

I = -4A * 12Ω/(12Ω + (2Ω + 5Ω||10Ω) = -2.76923076A

And 10Ω resistor current is equal to ??

Ix = -2.77A * ??/( ?? + ?? ) = ??

Thanks that's excellent, but how do you know to make the equivalent circuit like that? I was trying to draw it with the 4Amp going into the circuit splitting between the 5ohm resistor and the other branch.
 
lam58 said:
Thanks that's excellent, but how do you know to make the equivalent circuit like that?
Hey what can I say, I simply look at the circuit and I see equivalent circuit after some thought.
So all you need is some experience, that's all.
 

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