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Maximum average power for a purely resistive load

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the circuit shown in the figure below. Suppose that R = 46Ω and Z=j14Ω. Determine the maximum average power that can be delivered to the load if the load is pure resistance. Note that the voltage source magnitude is given as Vmax, not VRMS

    Steif.ch06.p64_1.jpg

    2. Relevant equations

    j = [itex]\sqrt{-1}[/itex]
    V=IZ
    I1 = [itex]\frac{R_2*I_s}{R_1+R_2}[/itex]
    Zt = Thevenin equivalent impedance = VOC/ISC
    P = IRMS2*RL
    Power is max when RL = Rs, or ZL = Zs*
    Z* = complex conjugate of Z

    3. The attempt at a solution

    Shorting the voltage source to find the thevenin equivalents, I combine the capacitor and inductor in parallel then combine that value with the resistor:

    [itex]\frac{(-j10)(j14)}{-j10+j14}[/itex] = -j35
    -j35+46 = 46-j35 = Zt

    Going back to the original circuit and replacing the load resistance with a short circuit to find ISC:

    Current across capacitor = [itex]\frac{10}{-j10}[/itex] = j
    Using the current divider to find the current across the 46Ω resistor,
    ISC = [itex]\frac{j14*(j)}{j14+46}[/itex] = -0.278+j*0.8477

    Vt = ISC*Zt = (-0.278+j*0.0848)(46-j35) = -9.82+j13.63

    For max power, RL = |Zt| = |46-j35| = sqrt(46^2+35^2) = 57.8

    The current through the thevenin equivalent circuit with RL attached is:

    I = [itex]\frac{-9.82+j13.63}{46-j35+57.8}[/itex] = -0.125+j*0.0893
    IRMS = [itex]\sqrt{0.125^2+0.0893^2}[/itex]/[itex]\sqrt{2}[/itex] = 0.1086

    P = IRMS2*RL = 0.682 W which is incorrect. The correct answer is 2.95 W. Any ideas?
     
    Last edited: Apr 17, 2014
  2. jcsd
  3. Apr 17, 2014 #2

    gneill

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    Staff: Mentor

    Your Thevenin impedance looks good, but I'm not liking your Thevenin voltage or the short circuit current. How did you conclude that the voltage across the capacitor is j when ZL is replaced by a short? Shorting RL places the resistor R in parallel with the inductor, it doesn't ground the top of the R.
     
  4. Apr 17, 2014 #3
    Thanks for the reply.

    Sorry, I meant to say that was the *current* across the capacitor, not the voltage.

    So now repeating the same method starting from ISC while combining R and Z:

    R||Z = 46*j14/(46+j14) = 3.9+j12.8
    R||Z + C = 3.9+j12.8 -j10 = 3.9+j2.8
    ISC = V/R = 10/(3.9+j2.8) = 1.692-j1.215
    Vt = ISC*Zt = (1.692-j1.215)(46-35j) = 35.31-j115.1
    RL is still 57.8
    Current through Thevenin equiv. circuit with RL = I = (35.31-j115.1)/(46-j35+57.8) = 0.641-j*0.893
    IRMS = sqrt(0.641^2+0.893^2)/sqrt(2) = 0.777
    P = 0.777^2*57.8 = 34.9W which is still incorrect.
     
  5. Apr 17, 2014 #4

    gneill

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    Staff: Mentor

    Isc will be just the current through the resistor. You've taken it to be the total current produced by the source voltage driving the total impedance.

    If I may suggest, since you've found the Thevenin Impedance easily enough (and it is correct), why not find the Thevenin Voltage as the open circuit voltage rather than by way of the short circuit current and impedance? With the load removed, the open circuit voltage at the terminals is the result of the action of the voltage divider created by the capacitor and inductor. Easy!
     
  6. Apr 17, 2014 #5
    Isn't that what I did the first time, just with the mistake of me writing voltage instead of current?

    Just tried that, VOC = 35V which when plugged into the following equations gives me the right answer. However, I've noticed that this is very close to the real part of the Vt value I obtained in my first reply to you. Since it's asking for a purely resistive power, was I supposed to take just the real part of that voltage or is that just a coincidence that the values are similar? Nonetheless, thank you very much for your help.
     
  7. Apr 17, 2014 #6

    gneill

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    Staff: Mentor

    I don't think so. The source will see a total impedance of ZC + (Z||R), while you wrote that the capacitor current is 10/(-10j). That would only be true of the capacitor alone was across the source.

    I think that's a coincidence.
     
  8. Apr 17, 2014 #7
    Because it is [itex]35\,V[/itex]:
    [tex]V_t=V_{oc}=10\cdot\frac{14j}{14j-10j}=35[/tex]
     
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