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!

Current and Electricity Question

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    i got it until ' R P(1) = 3 ohm '
    but shouldn't the effective resistance of arm ADE be 6+3=9
    and effect. resistance b/w A & E be (9*6) / 9+6 = 54/15 ?
     

    Attached Files:

  2. jcsd
  3. Apr 7, 2012 #2
    Why is it you say ADE, 6+3 not 3+3?

    Going from A->D->E, A->D, 2 parallel resistors. D->E only one resistor.
     
  4. Apr 7, 2012 #3
    please help me to understand how to calculate resistances in arms like the above e.g. .
     
  5. Apr 9, 2012 #4
    You have to simplify one branch at a time where resistors are clearly either in series or parallel. Starting at the left branch, it begins to simplify like so:
    Req2.png

    [1] Recognize that the two 3Ω resistors are in series, add them: 3Ω + 3Ω = 6Ω. Replace that resistor in next step.

    [2] With the new 6Ω resistor in place we now recognize that it is in parallel with the AD 6Ω resistor, so we do the math to simplify them: 6Ω || 6Ω = 3Ω

    [3] The 3Ω resistor replaced the previous parallel ones.

    With these simplifications we see that the "arm ADE" is actually the equivalent of two 3Ω resistors in series, thus 6Ω.

    Also, I don't like how your book uses phrases like "Therefore Effective Resistance between A and E is given by." This is confusing, as it is only one path between points A and E, NOT the equivalent resistance. The true effective resistance between A and E would be when all resistors in the diagram are simplified down to one resistance (with respect to A and E, e.g. when all that remains is one resistor in between point A and point E).

    Hope that helps.
     
    Last edited: Apr 9, 2012
  6. Apr 9, 2012 #5
    ok , i think i got it
    thank you very much sir...
    appreciate what you are doing !!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Current and Electricity Question
Loading...