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Homework Help: Seemingly difficult problem

  1. Mar 5, 2005 #1
    oK here it is!

    Three identical resistors are connected in parallel. The equivalent resistance increases by 700 ohms when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor.

    Ok my thoughts:

    Ok in situation one, we have:


    In situation two we have:

    1/Req +700 ohm=(R1+R2)+1/R3, R1+R2=Reff.

    So 1/Req +700 ohm=1/Reff+1/R3.

    Ok, now I'm stuck...Any tips welcome!
  2. jcsd
  3. Mar 5, 2005 #2
    Let the resistance of the resistors be R.

    For the first situation, we have [itex]R_{eq1} = (\frac{1}{R} + \frac{1}{R} + \frac{1}{R})^{-1}[/itex].

    For the second situation, we have [itex]R_{eq2} = R + (\frac{1}{R} + \frac{1}{R})^{-1}[/itex].

    And we know that [itex]R_{eq1} + 700\Omega = R_{eq2}[/itex]. Can you take it from here?

  4. Mar 5, 2005 #3
    Ok lets try:

    So I'm assuing I can sub Req2 in for the bottom equation.

    Req1+700ohm=R+ (2/R)^-1

    From Req1=(3/R)^-1.

    Ok so (3/R)^-1 +700 ohm=R +(2/R)^-1

    Solve for R.

  5. Mar 6, 2005 #4
    You made an error solving for R. Try it again. :smile:

    You're good once you get R = 600 ohms.

  6. Mar 6, 2005 #5
    Ah HA!! I got it now. Ok so R/3+700ohm=R+R/2.
    Ok now 700=R+R/2-R/3
    Then 700=7R/6
    And R=600!

    Thanks a bunch :)
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