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Unknown resistor in circuit - find current through it (tricky)

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data
    http://imageshack.us/a/img210/3330/homeworkprob15.jpg [Broken]

    Find I0 in the network.

    (The middle resistor value was not given)


    2. Relevant equations
    V = IR

    Voltage Division:
    (Voltage across series resistor) = [(resistance) / total series resistance)](total input V)

    Current Division (for 2 parallel resistors):
    (current across parallel resistor) = [(other resistor) / (sum of parallel resistors)](total incoming current)]

    Parallel resistors = (1/R1 + 1/R2)-1
    Series Resistors = R1 + R2

    Delta Y conversion and back for resistors

    3. The attempt at a solution

    I really do not know where I should start.

    Other than all I can do being to simplify the 8 and 4 ohm resistors to 12 ohm. But then there's still that unknown resistor in the middle.

    I don't see how it would help to use y to T or the other way around to help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 20, 2013 #2

    NascentOxygen

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    Staff: Mentor

    If you redraw this, so that the unknown resistor is "vertical" you can see that it is parallel to some other/s.

    Are you sure you aren't given the value of another parameter (e.g., voltage) of the circuit? Otherwise, you can only determine the current as an algebraic expression (in terms of R).
     
  4. Jan 20, 2013 #3

    rude man

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    Homework Helper
    Gold Member

    Obviously has no solution if unknown resistor remains unknown.
     
  5. Jan 20, 2013 #4

    gneill

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    Staff: Mentor

    ...Or if the total current or some other voltage drop is not provided.
     
  6. Jan 20, 2013 #5
    Yep, this is exactly all that was given.

    Darn... looks like my professor really dropped the ball then (he wrote this problem up himself).


    Thanks anyway guys.

    (By the way, the given answer by him is I knot = 1A)
     
    Last edited: Jan 20, 2013
  7. Jan 20, 2013 #6
    Well like NascentOxygen says, you could try to work out the algebraic expression for it... All you'd have to do at a later stage would be to insert the numbers...?
     
  8. Jan 20, 2013 #7

    rude man

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    I say, good point there, gneill! :smile:
     
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