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Find Maximum Power to Resistor R

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Byjn4TH.png

    2. Relevant equations
    Pmax = ((VTh)2)/(4RTh)

    Rparallel = (R1R2)/(R1 + R2)

    KCL ΣI = 0
    KVL ΣV = 0

    3. The attempt at a solution
    Okay, so to find RTh, I remove all independent sources and set the "terminals" at R to have a 1A current running through them. Since there is a 30kΩ resistor in parallel with the 1A current that I added, I assumed that the voltage across the 1A that I added was 30kV and that RTh = 30kV/1A = 30kΩ.

    Then I returned to the initial circuit to find VTh. I applied nodal analysis such that v1 = v0 and v2 = VTh.

    Applying KCL @ v1:
    (v1 - 100)/10 + v1/40 + (v1 - v2)/22 = 0

    Applying KCL @ v2:
    (v1 - v2)/22 - .003v1 + v2/30 = 0

    After simplifying and solving the system, I got v1 = 70.1V and v2 = 42.7V. Then I plugged v2 into the equation for pmax and I calculated .0152W. However, the solution in the back of the textbook says that the power should be infinity. What have I missed?
     
  2. jcsd
  3. Feb 21, 2015 #2

    gneill

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    Staff: Mentor

    Why did you ignore the rest of the circuit? The dependent current source may well influence the Thevenin resistance (hint: It does!).

    Stick on the 1A external current source and determine the voltage at the output with the rest of the circuit intact (including the 100 V source). Find your Rth from there by the ratio of I/V.
    Your first node equation looks fine, but the second seems to be mixing incoming and outgoing current terms in the sum. Check it.
    You have a couple of glitches to iron out in your derivations, but I must admit that I don't understand their claim to infinite power. Now, the Thevenin resistance may turn out to be a bit "different", but in operation with a load resistor attached the voltage and current should both be finite according to my reckoning.

    Let's see how you get on after addressing the items I pointed out above.

    EDIT: I just realized why they claim infinite power for the maximum delivered to the load. Let's see if you can spot the reason :smile:
     
    Last edited: Feb 21, 2015
  4. Feb 21, 2015 #3
    I ignored most of the circuit while calculating RTh because that had worked a few times in the past when there was a resistor in parallel with the 1A of added current, but I guess that was just a coincidence.

    I've never calculated an RTh before while leaving an independent source in tact. Is the 100V supposed to stay in tact while calculating RTh because of the dependent voltage source? How do I know when the independent sources should or shouldn't be removed when finding a Thevenin equivalent?

    Attempting to calculate RTh without ignoring most of the circuit:
    I added the 1A of current, and then since the whole circuit is there, I just adjusted my nodal analysis from earlier, remembering that v2 will instead equal the voltage that has a relationship with RTh (v2 ≠ vTh ... v2/1A = RTh).

    I believe I have found the error at v2...
    KCL @ v2:
    (v2 - v1)/22 - .003v1 + v2/30 - 1 = 0

    KCL @ v1:
    (v1 - 100)/10 + v1/40 + (v1 - v2)/22 = 0

    Simplifying and solving the system, I got that v1 = 67.9V and v2 = 34.5V, which would mean that RTh = v2/1 = 34.5kΩ...

    When I remove the -1 from the equation for KCL @ v2 and resolve the matrix, I get similar values for v1 and v2, which I believe would equal VTh...I still don't see how the power would equal infinity.
     
  5. Feb 21, 2015 #4

    gneill

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    Okay, that was not the best advice on my part. You should suppress the independent sources when you drive the circuit with an external stimulation. Sorry about that. o:)
    You need to be a bit careful when you write the resistance values in k Ohms in your equations. You are effectively dividing through the denominators by 1000, or equivalently, multiplying the whole equation by 1000. This will affect the sizes of current supplies too! So instead of a 1 amp source stimulating your circuit it's 1000 mA. And the dependent source needs to be multiplied by 1000 as well.

    I think you'll find different results when you do that.
     
  6. Feb 21, 2015 #5
    Okay, so if I remove the independent voltage source and send a 1mA current through the terminals such that v2/1mA = RTh, would these equations be correct?

    KCL @ v1:
    (v2 - v1)/22 - 3v1 + v2/30 - 1 = 0

    KCL @ v2:
    v1/10 + v1/40 + (v1 - v2)/22 = 0
     
  7. Feb 21, 2015 #6

    gneill

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    Yes they should work.
     
  8. Feb 21, 2015 #7
    Okay, so when I simplified and solved the above equations, I got v1 = 4.33E-4 and v2 = .0016, which means .0016V/1mA = 1.6Ω, I think. Then I adjusted my nodal analysis equations by removing the 1mA and returning the 100V, and these are the equations that I got:

    KCL @ v2:
    (v2 - v1)/22 - 3v1 + v2/30 = 0

    KCL @ v1:
    (v1 - 100)/10 + v1/40 + (v1 - v2)/22 = 0

    Simplifying and solving these, I got v1 = 125.7V and v2 = 251.4V.

    When I square 251.4 and divide by 4RTh, I'm going to have a very large power, but it won't be infinite. What am I still missing?
     
  9. Feb 21, 2015 #8

    gneill

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    I think you need to show step by step how you solved the previous two node equations and found Rth. Something's amiss.
     
  10. Feb 22, 2015 #9
    For RTh:

    KCL @ v1:
    (v2 - v1)/22 - 3v1 + v2/30 - 1 = 0

    KCL @ v2:
    v1/10 + v1/40 + (v1 - v2)/22 = 0

    For the first equation, I used 660 as a common denominator...
    (30/30)((v2 - v1)/22) - ((30*22)/(30*22))3v1 + (22/22)(v2/30) - 1((30*22)/(30*22)) = 0

    30v2 - 30v1 - 1980v1 + 22v2 = 660
    -2010v1 + 52v2 = 660 ← Final Equation 1

    For the second equation, I used 880 as my common denominator...
    ((4*22)/(4*22))(v1/10) + (22/22)(v1/40) + (40/40)((v1 - v2)/22) = 0

    88v1 + 22v1 + 401 - 40v2 = 0

    150v1 - 40v2 = 0 ← Final Equation 2

    Then I put these final two equations into my calculator as matrices, this time getting v1 = -.364V and v2 = -1.363V.

    To find VTh:
    First, I adjusted Final Equation 1 by removing the term caused by the 1mA current:
    -2010v1 + 52v2 = 0

    Then I adjusted Final Equation 2 by adding a term for the 100V source:
    150v1 - 40v2 = 8800

    Then I put this into my calculator as matricies to get v1 = -6.3V and v2 = VTh = -243.6V
     
  11. Feb 22, 2015 #10

    gneill

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    So you're pushing 1 mA into the circuit across a voltage of v2 = -1.363 V. What does that make Rth?

    Your Vth looks good.
     
  12. Feb 22, 2015 #11
    Doesn't that make RTh = -1.363kΩ?
     
  13. Feb 22, 2015 #12

    gneill

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    Yes it does!
     
  14. Feb 22, 2015 #13
    Okay so...

    Pmax = VTh2/(4RTh)

    P = (-243.6)2/(4*(-1363)) = -10.8W

    Are you allowed to have negative power?
     
  15. Feb 22, 2015 #14

    gneill

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    Negative power implies that the circuit is absorbing power from the load, which can happen. But there's more to it here.

    Draw the circuit as a load resistor connected to your Thevenin equivalent model. You have a voltage divider consisting of the Thevenin resistance and the load resistor. The load resistor is a standard one with positive resistance. Can you find a value that might lead to "interesting" behavior?
     
  16. Feb 22, 2015 #15
    If RL = -RTh, then when I apply KVL to the circuit you told me to draw...

    VTh + RThi + RLi = 0
    243.6 - 1363i + 1363i = 0
    243.6 = 0
    Which is wrong, which might have something to do with the power being infinity?
     
  17. Feb 22, 2015 #16

    gneill

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    Indeed. Leave the load resistance as a variable R and write an expression for the power. Take a limit as R approaches Rth.
     
  18. Feb 22, 2015 #17
    I think I'm confused or maybe using the wrong formula... regular power P = V2/R

    P = VTh2/RL
    lim RL → RTh VTh2/RL = constant, at least in this case where VTh and RTh are constants...

    Or should I be using this formula:

    P = ((VTh)/(RTh + RL))2RL

    In which case, I can take the limit as RL → -RTh and get infinity.
     
  19. Feb 22, 2015 #18

    gneill

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    Staff: Mentor

    The latter is closer. The voltage should be squared.
    The first one ignores the Thevenin resistance which is in series with the load.
     
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