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Power dissipated in resistor

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    image.jpg


    2. Relevant equations
    P=V2/R
    P=I2R
    V=IR
    3. The attempt at a solution
    As the resistance of R increases, the potential difference across R increases, so the power dissipated in R increases too. But I can't seem to see how the power decreases afterwards. Is it because, as time passes, the e.m.f of the battery dies out eventually?
     
  2. jcsd
  3. Nov 2, 2015 #2

    gneill

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    Staff: Mentor

    Power is a product of current and potential difference: P = IV. What happens to the potential and current as the external resistance gets very large? Can the potential across R increase indefinitely?
     
  4. Nov 2, 2015 #3
    The current becomes very small and the potential gets larger. Hmm I'm confused on which quantity to look at to determine the power.. Because if current becomes too small, then according to P=IV, power will then decrease. But as resistance increase, potential difference increases too, and again according to P=IV , power will then increase..
     
  5. Nov 2, 2015 #4

    gneill

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    Staff: Mentor

    Yes, you need to consider both quantities if you want to make a choice by logical deduction. Consider V and I when R is at the extremes of its values, say 0 Ω and ∞ Ω.

    Alternatively, you can analyze the circuit and write an expression for the power as a function of R, then you can examine the curve mathematically.
     
  6. Nov 2, 2015 #5
    I see.. So if R -> ∞ Ω, power decreases, as current decreases right? Ok I got the picture, thanks !
     
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