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How is the following derived: Please help derive the following: dPh/dR

  1. Aug 16, 2014 #1
    Here is the working leading up to this:
    The circuit is a series circuit, with Ro being the circuit's internal resistance and Rh being the heater's resistance. We're only concerned about the power transfer to Rh. We have an input with a constant voltage V. Ohm's law: V = IR, where R is the total resistance of the circuit. In a series circuit this total resistance is the sum of the resistances in the circuit.

    I = V / (Ro + Rh)

    The voltage developed across the heating resistance will again be dictated by ohm's law:
    Vh = I * Rh
    Substitute: Vh = V * Rh / (Ro + Rh) P = I Vh
    so Ph = [V / (Ro + Rh)] * [V * Rh / (Ro + Rh)]
    Ph = V2 Rh / (Ro + Rh)2

    Rh is the only variable, since Ro and V are fixed in this context. In order to find the maximum power transfer, we optimize this equation. Optimization theorem: a functions optimum points occur at the function boundaries, and where the first derivative of the function is equal to zero. In this case we're not going to look at the boundaries, since they are at 0 and infinite.

    dPh/dRh = V2 (Ro - Rh) / (Ro + Rh)3.
  2. jcsd
  3. Aug 17, 2014 #2


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    What do you men on V2 and (Ro + Rh)2 ? Do those "2"-s mean square? Then use ^ as V^2 or (Ro+Rh)^2 or
    click to "Go Advanced" and you can find above the text the symbols X2 and X2, so you can write the square of something as something 2.

    So: I=V/(Ro+Rh)


    It is even better to use TeX
    [tex]P_h=I^2R_h= \frac{V^2R_h}{(R_0+R_h)^2}[/tex]

    You get the derivative of Ph with respect to Rh with the formula for the derivative of a fraction, and simplifying. If F(x) = g(x)/h(x) [tex]F'=\frac{g'h-gh'}{h^2}[/tex]

    [tex]\frac{d P_h}{dR_h}=V^2\left(\frac{(R_0+R_h)^2-2R_h(R_0+Rh)}{(R_0+R_h)^4}\right)[/tex]


  4. Aug 17, 2014 #3
    I am a highschool student, how would i simplify that, and in terms of maximum efficiency of a wire what does this mean? ie in terms of matching impedance
  5. Aug 17, 2014 #4


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    The power on Rh is maximum if its derivative dPh/dRh is zero. The derivative is shown in my previous post, but it can be simplified if you factorize the numerator. [tex]\frac{d P_h}{dR_h}=V^2\left(\frac{(R_0+R_h)^2-2R_h(R_0+Rh)}{(R_0+R_h)^4}\right)=V^2\left(\frac{(R_0+R_h)(r_0+R_h-2R_h)}{(R_0+R_h)^4}\right)=V^2\frac{r_0-R_h}{(R_0+R_h)^3}=0[/tex]

    What should be Rh to make the expression equal to zero?

    Last edited: Aug 17, 2014
  6. Aug 17, 2014 #5
    So Rh has to equal Ro??? wow thanks man
  7. Aug 18, 2014 #6


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    Yes, it is. You get the maximum power out of a voltage source on a load, equal to the internal resistance of the source.

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