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Resistor & power dissipated question-please help

  1. Nov 14, 2008 #1
    Resistor & power dissipated question--please help

    1. The problem statement, all variables and given/known data
    Load resistor R is attached to a battery with EMF E and internal resistance r. For what value of the resistance R in terms of E and r will the power dissipated by the load resistor be a maximum?

    2. Relevant equations

    3. The attempt at a solution
    So, first utilizing Kirchoff's Loop Law:

    E-Ir-IR = 0
    E - I(r+R) = 0
    I(r+R) = E
    r + R = E/I
    R= E/I - r

    the answer is r, but I'm not understanding why the resistor has to have a maximum value, or how to get r.
  2. jcsd
  3. Nov 14, 2008 #2


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    Re: Resistor & power dissipated question--please help

    To see why there is a specific R value that corresponds to a maximum value, you can just think of the limits: if R=0, then the resistor R dissipates no power at all (because I2R=0). If R goes to infinity, the power dissipated again goes to zero, because the very large R causes the current to go to zero, so again I2R goes to zero.

    So for very large and very small R values, the power goes to zero, so the maximum power corresponds to some intermediate resistance.

    To actually calculate it, write an expression for the power dissipated by R. You can then take the derivative and set it equal to zero to find the maximum.
  4. Nov 14, 2008 #3
    Re: Resistor & power dissipated question--please help

    Ok, I just got it. I tried first substituting into P = IV and then P=V^2/R, and finally
    P=I^2R worked for me to get R=r.
  5. Nov 14, 2008 #4


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    Re: Resistor & power dissipated question--please help

    Right, I2R is the easiest way (I think) to get it.

    You can get it with the others, but for V you have to use the potential difference across just the resistor R. For example:

    I V = \left(\frac{E}{R+r}\right)\ \left( E \frac{R}{R+r}\right)
    which of course ends up the same as I2R which you used.
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