1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Superpostion circuit analysis question

  1. Feb 6, 2014 #1
    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?



    nqVKsGY.jpg
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2014 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    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

     
  4. Feb 6, 2014 #3
    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..
     
  5. Feb 6, 2014 #4
    I think I see now.. You mean r1 and r6? Then that would in series with the equivalent.
     
  6. Feb 7, 2014 #5
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Superpostion circuit analysis question
Loading...