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Math Help - power supplied to a resistive load cannot exeed (V\2)^2/R^2

  1. #1
    Super Member bigwave's Avatar
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    Cool power supplied to a resistive load cannot exeed (V\2)^2/R^2

    A battery has internal resistance R_i

    and open circuit terminal voltage V_b

    Show that the power supplied to a resistive load cannot exceed \frac{({\frac{V_b}{2})^2}}{R_i}

    well I only changed this to

    \frac{{\frac{V_b}{2}^2}}{R_i}<br />
\Rightarrow<br />
({\frac{V_b}{2})^2}\times\frac{1}{R_i}<br />
\Rightarrow<br />
\frac{V_b^2}{4R_i}<br />

    but not sure how you determine what cannot be exceeded

    any suggest...

    mod.... this should of been put in "other topics" its not an advanced topic.
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  2. #2
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    Internal resistance r.
    External R.

    <br />
 \displaystyle { I=\frac{V}{r+R} }<br />

    <br />
 \displaystyle {  P=I^2 \; R \;  = \;  \frac{V^2}{(r+R)^2}  \;  R<br />
}<br />

    Max power on R is

    <br />
  \displaystyle { \frac{dP}{dR} \; = \; 0<br />
}<br />

    we get

    <br />
R=r<br />

    and

    <br />
 \displaystyle {  P(max)=\frac{V^2}{4r}<br />
}<br />
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  3. #3
    A Plied Mathematician
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    I've attached a circuit model of the situation. Would you agree with this model?

    power supplied to a resistive load cannot exeed (V)^2/R^2-max-load-10-26-2010.jpg

    Let's say this model is accurate. Now what? Well, the one constant thing is the voltage of the battery. That doesn't change (at least, not theoretically). But as you change the load resistance, the relative voltage drops across each resistor changes. You want to maximize the power in the load. Now, Ohm's Law yields

    V_{\text{b}}=I(R_{\text{i}}+R_{\text{load}}).

    The power dissipated in the load resistor is P_{\text{load}}=I^{2}R_{\text{load}}=\left(\dfrac{  V_{\text{b}}}{R_{\text{i}}+R_{\text{load}}}\right)  ^{2}R_{\text{load}}.

    Now use the usual Calc I procedure of setting

    \dfrac{dP_{\text{load}}}{dR_{\text{load}}}=0, and solve for R_{\text{load}}.

    What do you get?
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  4. #4
    Super Member bigwave's Avatar
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    verify

    sure appreciate the help... now I see how this works...thanks...

    there was a follow up question on this.... not sure if input it right

    thus if V_b=96V and R_i = 50\Omega.

    Discrete loads of 150, 100, 50, 30, and 20m\Omega are connected, one at a time, across the battery. Plot the curve of power supplied versus the ohmic value of the load. I tried to use the Wolfram|Alpha to graph this but didn't get the expression right http://www.wolframalpha.com/input/?i=plot[P%3D%28%2896%29^2%29%2F%284\Omega%29%3B%28P%2C-50%2C200%29%2C%28\Omega%2C-50%2C200%29]

    Hence, verify that the maximum power transfer occurs when R_i = R_{load} = 50\Omega
    Last edited by bigwave; October 26th 2010 at 03:12 PM. Reason: added question
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  5. #5
    A Plied Mathematician
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    Well, here's a continuous plot. If you wanted to do a discrete plot, I'd use Excel.
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  6. #6
    Super Member bigwave's Avatar
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    appreciate the plot

    well. I need a lot of help with this EE subject.

    more ?? to post on the way
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  7. #7
    A Plied Mathematician
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    Did you mean you need more help with this problem, or just more help with problems in the same subject?
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  8. #8
    Super Member bigwave's Avatar
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    Im ok with this one

    but I will be posting more new problems

    its a new subject to me.
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  9. #9
    A Plied Mathematician
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    Ok. I'll be looking for more threads. Have a good one!
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