Change Resistor to Deliver Maximum Power

Discussion in 'Homework Help' started by ElectronicGuru, Oct 30, 2014.

  1. ElectronicGuru

    Thread Starter New Member

    Sep 26, 2014
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    Here's the circuit below:
    Screen Shot 2014-10-30 at 7.35.57 PM.png
    I am tasked with adjusting the size of the 5 Ω resistor so that maximum power is delivered to the 9 Ω resistor. So I approach this as a maximum power transfer question. Turning off the ideal sources leaves me with a 5 Ω and 3 Ω resistance in series. So I would think that replacing the 5 Ω with a 6 Ω resistor will maximize the power transferred (to get a Thévenin resistance of 9 Ω—the same as the load).

    But I am wrong! The correct answer is that the wire with the 5 Ω resistor is open-circuited so as to produce a maximum current of 9 A going to through the 9 Ω resistor to deliver maximum power to that resistor.

    I was reading through my book and it gives some explanation on why maximum power transfer theorem failed me:

    I'm lost on the why this is. I've tried to think about the last sentence but I just don't understand. Could someone give an intuitive explanation as to why maximum power transfer works one way (finding the best load) but not the other (load is already specified)?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Have you verified the two cases to confirm which situation gives the greater power in the 9 ohm resistor?

    Is it when the 5 ohm is replaced with a 6 ohm resistor or with an open circuit?

    Once you establish the truth of the matter then you may be able to resolve in your mind as to why this is so?
     
  3. crutschow

    Expert

    Mar 14, 2008
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    It's quite simple.

    When the source impedance is fixed, then the maximum power theorem tells you that the maximum load power occurs when the power lost in the source impedance equals that dissipated in the load impedance.
    That occurs when both impedances are equal.

    When the load impedance is fixed, then to get the maximum power in the load, you want the minimum (ideally zero) power dissipated in the source. That, of course, occurs when the source impedance is zero.

    A good example, is a solid-state speaker power amplifier. Its output impedance is very low so as to maximize the power to the load (the speaker which has a fixed impedance).
    But interestingly, the tubes in an old tube amp had a high output impedance so an output transformer was used to match the tube impedance to the speaker impedance for maximum power to the speaker. Those amps typically had transformer taps for 4, 8, or 16 ohms to give the best match for the particular speaker used.
     
  4. WBahn

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    Mar 31, 2012
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    I'll take issue with this claim because maximizing power in the load doesn't necessarily equate with minimizing power dissipated in the source. Even if that were the case, it doesn't necessarily happen when the source impedance is zero.

    To justify my first claim, consider a source consisting of an ideal voltage source in parallel with a (finite) resistor. The Thevenin equivalent resistance of this source is 0 Ω, but power is most definitely being dissipated in the source.

    To justify my second claim, consider a source consisting of an current source. Here you do NOT want to minimize the source impedance, but rather to mazimize it (ideally infinite).

    The bottom line is that as soon as you start talking about a Thevenin or Norton equivalent circuit, you lose entirely the ability to talking about anything internal to the circuit that is being modeled because the equivalent circuits ONLY model the voltage/current relationships at the terminals. Power calculations for the equivalent source circuit are meaningless.
     
  5. WBahn

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    Mar 31, 2012
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    The last sentence gives you a hint to think of a case that provides a counter example to the notion that you want to choose the source resistance to equal the fixed load resistance. But it is a sloppy way to go about it because power lost in the equivalent impedance of the source is meaningless.

    But think of an actual source circuit composed of an ideal battery with a series source resistor (and thus directly comparable to a Thevenin equivalent circuit for that source circuit). Now put a load resistor into the circuit and ask yourself what value of series source resistor you want if the goal is to get the maximum possible voltage to appear across the load resistor or, equivalently, get the maximum possible current through the load resistor.

    For your problem, look at where the current from the 9 A source has to go. To get the maximum power into the 9 Ω resistor, how much of that 9 A do you want to flow through the 9 Ω resistor and how much do you want to flow through the 5 Ω resistor? If you can change the value of the 5 Ω resistor to something else, what would you need to change it to in order to get the amount of current that you want to flow in the 9 Ω resistor to actually flow in that resistor?
     
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  6. crutschow

    Expert

    Mar 14, 2008
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    I agree my statement wasn't precise. I meant there is no power dissipated in the equivalent source impedance. :rolleyes:

    I'm not sure what a current source has to do with the maximum power theorem but if you want to maximize the load power for an ideal current source than the load impedance should be ideally infinite, leading to infinite power dissipation. :eek:
     
    Last edited: Oct 31, 2014
  7. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    An ideal voltage source does not dissipate power when it is connected to a resistive load. If we have a 10v DC source and 10 ohm resistor, the resistor dissipates 10 watts and the source delivers this power to the resistor. If the source dissipated power too, the total power dissipated in that circuit would be more than 10 watts, but it's not. If the source dissipated 10 watts too then the total would be 20 watts and this would create more heat than just the resistor would create. Since it is only 10 watts that's all that is dissipated.

    Add an internal series resistance to the source so that it is then non ideal, and then it dissipates some power, but it's really the internal resistance dissipating the power not the internal ideal source part of it.
    So a 10v source with 1 ohm in series powering a 9 ohm resistor, the 1 ohm dissipates 1 watt and the 9 ohm dissipates 9 watts and the ideal source delivers 10 watts total.
    If the source was accepting power then that's a different story but that would take a higher voltage than 10v to force current through the first source. That is like battery charging. If the second source powered the first source with a current of 1 amp, then the first source (10v) would be dissipating 10 watts.

    To the OP:
    Do a circuit analysis keeping the 5 ohm resistor a variable like Rx. Then graph the power in the 9 ohm resistor as Rx varies from 0.001 to 100 ohms. Look at the graph, think about it. Do another graph if you like with different values for Rx.
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    What book are you reading?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    The maximum power theorem applies equally well to circuits with current sources as well as voltage sources.

    I don't know why you are saying the load impedance should be ideally infinite, as we have been talking about source impedances. Yes, if you have an ideal current source with an infinite load impedance, then the power goes to infinity. But if you have an ideal voltage source with a zero load impedance, then the power goes to infinity. Same difference.

    A current source with an infinite source impedance dissipates no power in the source impedance because no current flows through it and all of the current goes through the load. A voltage source with a zero source impedance dissipates no power in the source impedance because no voltage is dropped across it and all of the voltage appears across the load. Same difference.
     
  10. WBahn

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    Mar 31, 2012
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    No one has been talking about ideal sources dissipating power in the sources (I don't think so, anyway), so I'm not sure the point being made.
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Funny, i could have sworn at least one person did :)

     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Where am I ever claiming that the ideal voltage source is dissipating power?

    I described a circuit (i.e., the "source") that is constructed (i.e., "consists of") the combination of an ideal voltage source in parallel with a resistor. Regardless of whether a load is connected or not, power is being dissipated in the source circuit. None-the-less, the Thevenin equivalent circuit for this source has zero output impedance because the ideal source inside it masks the resistance from the load.

    In case I need to draw a picture:

    Source.png
     
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