Circuit analysis current source problem

AI Thread Summary
The discussion focuses on solving a circuit analysis problem to find V1 using the superposition theorem. Participants clarify the configuration of resistors, noting that the 80Ω and 120Ω resistors are in parallel, but not in series with the 8Ω resistor. They emphasize the need to calculate individual voltage drops before combining them for the final answer. Additionally, there is mention of a different approach needed for the 1A source, highlighting the importance of understanding series and parallel combinations. Accurate identification of resistor arrangements is crucial for solving the circuit correctly.
Jamessamuel
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Homework Statement


given the circuit shown, find V1.
12231610_1652042151703604_2077339994_n.jpg

Homework Equations

The Attempt at a Solution


see image:
am i right? Video says otherwise... I chose to use the superposition theorem here.
12200761_1652042148370271_1013343203_n.jpg


help appreciated,

regards,

james.
 
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Jamessamuel said:

Homework Statement


given the circuit shown, find V1.
[ ATTACH=full]91899[/ATTACH]

Homework Equations

The Attempt at a Solution


see image:
am i right? Video says otherwise... I chose to use the superposition theorem here.
[ ATTACH=full]91898[/ATTACH]

help appreciated,

regards,

james.
That second image is very difficult to read.

It appears that you took the 80Ω and 120Ω to be a parallel combination, which is correct. However, that combination is NOT in series with the 8Ω resistor.
 
SammyS said:
However, that combination is NOT in series with the 8Ω resistor.
I think they're considered in series for getting the current through them when the 1A supply is ignored, but must be separated to get individual resistor voltage drops. Once all individual voltage drops are gotten they are then combined (superpositioned?) for the answer.
 
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surely this is correct?
12242271_1652116441696175_986558241_n.jpg
 
Jamessamuel said:
surely this is correct?
[ ATTACH=full]91916[/ATTACH]
That's fine for the 6A source.

There is a different parallel/series combination you need to use for the 1A source. For this, there is no 56Ω combination.

(Thanks, insightful.)

Added in Edit:
The above refers to the following:
upload_2015-11-16_7-56-6.png


The 40Ω and 8Ω are in series.
That combination and the other two resistors, 80Ω and 120Ω, are all in parallel with the other.
 
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