Resistor Square Problem

Discussion in 'Homework Help' started by yan500, Jul 22, 2011.

  1. yan500

    Thread Starter Member

    Jul 12, 2011
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    Hey all, hope you're having a good Friday. I've got a question on Question 18 of this worksheet: http://www.allaboutcircuits.com/worksheets/dc_sp.html

    I can't understand why the answers are what they are. When they ask what is the resistance between A and B shouldn't it be 3k3 ohms?
     
  2. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    No because between points A and B, there is a 3k3 resistor in parallel with 4k7 + 1k5 + 2k7, which has to be accounted for.

    Rab = 3k3 || (4k7 + 1k5 + 2k7) = 2k41
     
  3. yan500

    Thread Starter Member

    Jul 12, 2011
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    Ohhhh, so you can essentially draw an imaginary line between the two leads then representing a wire?
     
  4. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    What do you mean by "imaginary line"?
     
  5. yan500

    Thread Starter Member

    Jul 12, 2011
    48
    0
    Sorry, I should have been clearer. What I mean is that point A essentially turns into a node that would split a current (if there was a current source) between AB and AC. And that would make it a parallel connection.
     
  6. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    Yes. If a current source were to be connected between nodes A and B. A current divider with two paths, A->B and A->C->D->B, would result.
     
    yan500 likes this.
  7. yan500

    Thread Starter Member

    Jul 12, 2011
    48
    0
    Thank you so much for your help.
     
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