Basic Engineering Circuit Analysis 10th Ed., Problem 2.87

Discussion in 'Homework Help' started by jbates, Feb 2, 2011.

  1. jbates

    Thread Starter New Member

    Feb 2, 2011
    3
    0
    I've been playing around with this problem for a couple hours. I think it might be impossible, so asking here is my last resort.

    Code ( (Unknown Language)):
    1.             + 5V -
    2.  |-------|----R----|-----|
    3.  |       |         |     |
    4. 4kΩ   Is/|\       6kΩ   3kΩ
    5.  |       |         |     |
    6.  |-------|---------|-----|
    Find Is. Is is a current source, btw, in-case my text-art isn't clear enough.

    First, I did the obvious and combined the two rightmost resistors into a 2kΩ equivalent resistor.
    Code ( (Unknown Language)):
    1.             + 5V -
    2.  |-------|----R----|
    3.  |       |         |
    4. 4kΩ   Is/|\       2kΩ
    5.  |       |         |
    6.  |-------|---------|
    From the perspective of the node above Is, that leaves two paths to the node below Is where the voltage drops should be the same. Calling the current counter-clockwise through the left loop I1 and clockwise through the right loop I2, Is = I1+I2 and

    5V + I2*2kΩ = I1*4kΩ

    I've tried a million different things from this point, but I just can't solve for Is. I don't think it's possible. This is a new edition of the book and I think this problem must have a typo somewhere.
     
    Last edited: Feb 2, 2011
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Is = I1 + I2

    I2 = 5V/R

    I1 = ( 5V + ((5V/R)*2K) ) / 4K
     
  3. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    A numerical solution will be determined by R.
    JP
     
  4. jbates

    Thread Starter New Member

    Feb 2, 2011
    3
    0
    It's a resistor but the resistance isn't given, only the voltage across it.

    So it really can't be solved with the given information?
     
  5. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    I think so too. Is or R should be given. You can always find a solution as a function of one of them though.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    If R is unknown, the only think you can do is this


    Is = I1 + I2

    I2 = 5V/R

    I1 = ( 5V + ((5V/R)*2K) ) / 4K

    Is = 5V/R + ( 5V + ((5V/R)*2K) ) / 4K = 5V/R1 + (5V + 10K/R1) / 4k =

    = (6000 + R1) /(800 * R1)
     
  7. jbates

    Thread Starter New Member

    Feb 2, 2011
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    Okay, I thought so.

    Well, there goes two hours of my life.

    Thanks everyone.
     
  8. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I agree. You need to find Is as a function of R.

    It's obvious that the value of R is needed because if R=0, there is no solution (Is goes to infinity), and if R is very large then current through R is negligible and the current source has 5V across it resulting in 5V across the 4K resistor and Is is 1.25 mA in the limit as R goes to infinity.
     
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