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

## Homework Statement 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 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 #### Attachments

haruspex
Homework Helper
Gold Member
I thought x1= a1 * v3
No. "V1 acting alone" means x1 is the value v3 would have if v2 were 0.

lightgrav
Homework Helper

## Homework Statement

View attachment 77960

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

NascentOxygen
Staff Emeritus