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Resistance in a series-parallel combination

  1. May 28, 2006 #1
    The circuit in question is this:

    There are two heating elements with a resistance R_0 = 20 Ohm. I have to find R so that the total power output of the heating elements stays the same no matter if I only use one heating element or both.

    I first thought of using [tex]R = \frac{R_0 * R_0}{R_0 + R_0}[/tex], but I think that would mean that the power output would be higher when both heating elements are in use. Am I right?

    Then I tried solving [tex]R = \frac{U^2}{P}[/tex] for P and doing [tex]U^2 / R_1 = U ^2 / R_2[/tex] for R_1 being the resistance when only one heating element is in use and R_2 being the resistance when both heating elements are in use. Obviously that just leaves me with [tex]R_1 = R_2[/tex], which doesn't help with this problem.

    Another approach I tried was just saying that R has to be equal to R_0, because then the maximum available power P would arrive at the resistor at R_0 when only one heating element is in use and P/2 would arrive at each of them when both of them are turned on. However I don't know how to prove this or even if this is the right idea at all.

    Does anyone have an idea as to how I could solve this problem?
  2. jcsd
  3. May 28, 2006 #2

    Andrew Mason

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    Homework Helper

    Just work out the equation for total power loss for each scenario, equate them and solve for R.

    [tex]P = I^2R_{eff}[/tex]

    [tex]I = U/(R+R_0)[/tex]
    (1)[tex]P = I^2R_0 = \frac{U^2}{(R+R_0)^2}R_0[/tex]

    [tex]I = U/(R + R_0/2)[/tex]
    (2)[tex]P = I^2R_0/2 = \frac{U^2}{(R+R_0/2)^2}R_0/2[/tex]

    Equate (1) and (2) and solve for R. You will need to solve a quadratic equation.

  4. May 28, 2006 #3
    You are the best! Thank you very much.
  5. May 29, 2006 #4
    If you don't mind my asking: How did you arrive at the conclusion that you need the equations for total power loss? I simply can't figure this out.
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