1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Resistance in a series-parallel combination

  1. May 28, 2006 #1
    The circuit in question is this:
    [​IMG]

    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

    User Avatar
    Science Advisor
    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.

    AM
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Resistance in a series-parallel combination
Loading...