What does "sum of voltage x1 due to v1" mean?

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Homework Help Overview

The discussion revolves around the interpretation of the phrase "sum of voltage x1 due to v1" in the context of circuit analysis, specifically involving voltage sources and their effects on circuit components.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore interpretations of the voltage x1 in relation to v1, questioning whether x1 can be expressed as a product of a1 and v3. There is uncertainty about the implications of treating v2 as zero and how that affects the calculations of voltages across components.

Discussion Status

Some participants have provided insights into how to approach the problem by considering the effects of individual voltage sources on the circuit. There is ongoing exploration of different interpretations and calculations, but no consensus has been reached regarding the correct approach or final values.

Contextual Notes

Participants note potential misunderstandings in their interpretations and calculations, particularly regarding the values of v3 and the conditions under which different voltage sources are considered. There is mention of previous problems that may influence current reasoning.

LongApple
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Homework Statement


upload_2015-1-19_17-33-18.png


Homework Equations


V=IR

The Attempt at a Solution



My first guess is the quote in the title of this thread meant the a1 except now a1 * v3 instead of a1 in the below image, which we already have from a previous problem. I think my interpretation may be wrong

upload_2015-1-19_17-31-57.png


I think that my interpretation may be wrong. I try to solvee for a1 and get 0.21. So I thought x1= a1 * v3 = 0.21 * v3

but v3 is not about 2 Volts so I have made a mistake

upload_2015-1-19_17-32-54.png
 

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LongApple said:
I thought x1= a1 * v3
No. "V1 acting alone" means x1 is the value v3 would have if v2 were 0.
 
LongApple said:

Homework Statement


View attachment 77960

Homework Equations


V=IR

The Attempt at a Solution



My first guess is the quote in the title of this thread meant the a1 except now a1 * v3 instead of a1 in the below image, which we already have from a previous problem. I think my interpretation may be wrong

View attachment 77958

I think that my interpretation may be wrong. I try to solvee for a1 and get 0.21. So I thought x1= a1 * v3 = 0.21 * v3

but v3 is not about 2 Volts so I have made a mistake

View attachment 77959

As you apply the typical loop rule sum, you can treat V3 as a sum of I1 * R3 and I2 * R3 ,
where the rest of the loop voltages come from V2 - I2 * R2 , or else V1 - I1 * R1 .
that treats the node almost like a switch, choosing the left branch or the right branch.

Here, they want one Voltage source, or the other ... that is,
if you short out V2 (=0), so that V1 supplies R1 in series with R3 || R2 ;
this would make the voltage across the parallel pair = V1 (R2||R3)/(R1+(R2||R3))
... so I agree that , a1=0.211
if you short out V1 , then V2 supplies R2 in series with R3 || R1,
this would make V3 = 0.493 V2 ... so in the end, V3 = 4.366 Volt
 
The phrase you should be querying is "the voltage x1 due to V1 acting alone."

It means the voltage of interest is the sum of a number of independent components. One component we shall call x1 and another component shall be called x2. These can be calculated separately, then added to find the actual voltage of interest.
 

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