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Homework Help: 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:

    [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 !!
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