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Calculate New Power when adding a Resistor

  1. Sep 14, 2012 #1
    As this is my first post please feel free to correct me if I am posting this is the wrong section or on improper etiquette in general. I'm having a bit of trouble with what should be a pretty simple problem.

    1. The problem statement, all variables and given/known data
    A resistor R1 = 23Ω is connected to a battery that has negligible internal resistance and electrical energy is dissipated by at a rate of 44W.

    If a second resistor R2 = 20Ω is connected in series with R1 , what is the total rate at which electrical energy is dissipated by the two resistors?



    2. Relevant equations
    P=VI=I2R, V=IR


    3. The attempt at a solution

    I first solved for the current I by using the equation P=I2R using my R1. I found the current to be I=1.38A.

    From there I found the equivalent resistance by adding my resistances as they are in series to get an equivalent resistance of 43Ω.

    I then used the equation P=I2R plugging in the 1.38A and 43Ω.
    This gave me a new power of 82W but the site says this is not correct. Any suggestions?
     
  2. jcsd
  3. Sep 14, 2012 #2

    vk6kro

    User Avatar
    Science Advisor

    Try calculating the battery voltage in the first calculation.

    This will stay constant for the second calculation.
     
  4. Sep 14, 2012 #3
    Thanks! I got it! I had it in my head that the current would be what remained constant since it was in series but wasn't quite sure about that. Thanks again!
     
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