Engineering Superpostion circuit analysis question

AI Thread Summary
Determining series and parallel components in the circuit is crucial for analysis. By replacing I1 with an open circuit and V2 with a short circuit, it becomes clearer that R2 and R4 are in parallel, and this combination is in series with R3. Drawing the modified circuit helps visualize current flow and identify components that can be removed due to no current passing through them. The discussion emphasizes the importance of understanding current paths to simplify the circuit effectively. Clarifying these relationships aids in solving the circuit analysis problem.
fatmoe
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The problem I am having is determining what is in series and what is in parallel.

Now I am working with v1, so I1 has been removed and replaced with an open circuit and v2 has been replaced with a short.

So does that mean that r2 || r4? and if so, does that also mean that r2 || r4 which would then be in series with r3?



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I think you should actually draw the circuit diagram replacing I1 with an open circuit and V2 with a short circuit.

Then, you should see there are two more components you can remove, because they don't have any current flowing through them. That should answer your question

does that mean that r2 || r4? and if so, does that also mean that r2 || r4 which would then be in series with r3?
 
AlephZero said:
I think you should actually draw the circuit diagram replacing I1 with an open circuit and V2 with a short circuit.

Then, you should see there are two more components you can remove, because they don't have any current flowing through them. That should answer your question

Well I have redraw it but I don't see two other components that I can remove.. R6 is in the middle so I can't combine it with anything.. I'm confused.. Already spent 4 hours on this questions..
 
AlephZero said:
I think you should actually draw the circuit diagram replacing I1 with an open circuit and V2 with a short circuit.
K
Then, you should see there are two more components you can remove, because they don't have any current flowing through them. That should answer your question

I think I see now.. You mean r1 and r6? Then that would in series with the equivalent.
 
You have an open circuit for I1 and a short circuit for V2. The current will always flow the easiest route from + to -.

As Aleph said, draw your new circuit and look at how the current would flow through it. In your above post, R6 is correct, however, there will flow current through R1. What happens with the current on the node to the right of R2? Where will it flow from there?
 

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