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Maximum Power Transfer

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the value of RL that will draw the maximum power from
    the rest of the circuit shown below. Calculate the maximum power. (This problem was included in our exam this week, and I don't really recall the exact values of the parameters, but I'm pretty sure that the circuit looks more or less like this.)

    attachment.php?attachmentid=41273&stc=1&d=1322295774.jpg

    2. Relevant equations
    V=IR
    P=I2R
    KVL

    3. The attempt at a solution

    Computing for Vab:
    mesh 1:
    -10+5I1+15(I1-I3)=0
    but I3=0, so I1=2A

    mesh 2:
    I2=3Vx
    but Vx=-6I3, and I3=0, so I2=0

    Since no voltage exists in the 3 and 6 ohm resistors, I shorted them and Vab will now be equal to the voltage across the 15 ohm resistor. (Will it? I'm not really sure about this)

    Therefore,
    V5ohm=Vab=15I1=15(2)
    Vab=30V

    Computing for Rab:

    mesh 1:
    20I1-153=0 --> eqn.1

    mesh 2:
    I2=3Vx
    but Vx=-6I3
    I2=3(-6I3)
    I2+18I3=0 --> eqn.2

    mesh 3:
    -15I1-3I2+24I3=0 --> eqn. 3

    I1=-0.011A
    I2=0.270A
    I3=-0.015A

    I=-I3=-(-0.015)
    I=0.015A

    Rab=1/I=1/0.015
    Rab=66.667ohms

    Computing for the maximum power:

    Pmax=I2Rab=(30/(66.667+66.667))2(66.667)
    Pmax=3.375W

    That is what I did. Sirs, did I solve it the right way? I'm especially uncertain about how I solved Vab.
     

    Attached Files:

    Last edited: Nov 26, 2011
  2. jcsd
  3. Nov 26, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    The general method looks okay, but you've got a small problem with mesh1. The voltage is 10V and it's flowing through 5+15 = 20Ω. That should be a current of 0.5A, not 2A. As a result your Thevenin voltage does not turn out correctly.

    Also, if I might make a suggestion that could simplify things a bit, why not first convert the 10V supply and its 5Ω series resistor to a Norton equivalent current? The Norton resistance will be in parallel with the 15Ω resistor so you can combine them. Then convert back to Thevenin. You'll have eliminated one mesh.
     
  4. Nov 26, 2011 #3
    ah. thanks for pointing those out, sir. so it means that the major mistake is the 2A that's supposed to be 0.5A? nothing more?
     
  5. Nov 26, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    That's about it!
     
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