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Finding Resistance from internal resistance and Power in a circuit

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A resistor with resistance R is connected to a battery that has emf 15.0[V] and internal resistance r = 0.38[Ω].

    For what two values of R will the power dissipated in the resistor be 78.0 W?

    2. Relevant equations

    P=I2R=(ε2/(R+r)2)R

    3. The attempt at a solution

    ε2R=(R2+2Rr+r2)P

    R2+(2r-(ε2/P))R+r2=0

    R=1/2[((ε2/P)-2r)±√(((ε2/P)-2r)2)-4r2


    At this point I should be able to plug in what I know, and I get roughly
    R=1.06±0.77

    However that is not right, I have tried again making sure I kept my sig figs in order and all that good stuff, I get a couple similar but different answers, still all wrong though.

    I have only one attempt left and it's worth 27% of the grade on this HW so any help would be appreciated greatly.

    This is an image of my attempts, the problem, and how the solutions manual did the quadratic.
    M7Wgi.jpg
     
  2. jcsd
  3. Oct 11, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Your method and manipulations look okay up to this point. But I think perhaps the algebra got away from you doing the quadratic formula. Why not rewrite it at this point, substituting in numerical values for the known values and show us what that resulting quadratic looks like?
     
  4. Oct 11, 2012 #3
    I've tried this it kinda looks like
    R=1/2[(2.885)-0.76]±√((2.885-0.76)^2)-0.5776

    What messes with me is that the root term should just cancel most itself and become the first number again.

    My professor just got back to me saying that his answer for the root term was 0.9895. which if this is true then working it out would be

    R=1.0625±(0.9895-0.5776)
    so R= 1.47,0.65

    Which maybe, I'm honestly afraid to try. Can anyone tell me how he got that for his root figure?
     
  5. Oct 11, 2012 #4
    Am I just being an idiot? is the -4r2 supposed to be under the root?
    I mean that makes sense to me in terms of a quadratic formula -4AC being under the root.
     
  6. Oct 11, 2012 #5
    Just dawned on me what you were saying
    R2-2.125R+0.1444=0 is simple haha

    Thank you so much.
     
    Last edited: Oct 11, 2012
  7. Oct 11, 2012 #6

    gneill

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    Staff: Mentor

    Your equation resolved to:
    $$R^2 + \left( 2 r - \frac{E^2}{P}\right) R + r^2 = 0$$
    This is of the form:
    $$X^2 + B X + C = 0$$
    with roots:
    $$X = \frac{-B \pm \sqrt{B^2 - 4C}}{2}$$
    So what are numerical values of your constants B and C?

    EDIT; I was a bit slow typing up this response, so I missed your last post. Yes, once you boil it down to the fundamental parameters of the quadratic it is easier (less error prone) to find the roots with so much less algebra in the way :smile:
     
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