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AC Circuit Theory Maximum Power Transfer

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Discussion in 'Electronics Homework Help' started by serpico4321, Feb 23, 2013.

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  1. Feb 23, 2013 #1
    serpico4321

    serpico4321

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    Feb 23, 2013
    Hello All,

    I am currently hitting a barrier with regards to the basics of this question.

    A 50HZ supply is connected to various given impedance's connected in series and parallel. Calculate the value of the load impedance and the maximum power transfer.

    Maximum power transfer I understand is basically the Thevenin/Norton equivalent of Rth. But to acquire Rth surely you nee two values of ohms law?

    Though THERE IS NO EMF indicated in the question. Is it possible to simply use the frequency given 50Hz and the impedances to calculate the maximum power?

    Any help in the right direction would be vastly appreciated
     

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  2. Feb 23, 2013 #2
    The Electrician

    The Electrician

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    Jul 6, 2012
    When there are reactances involved, the load for maximum power transfer is not just Rth; tt's the complex conjugate of Rth.

    Replace the 50 Hz source with a short and calculate the impedance looking back into the output terminals. The load for maximum power transfer will be the complex conjugate of that impedance. Be sure to calculate the reactance of the inductor at 50Hz.
     
  3. Feb 24, 2013 #3
    serpico4321

    serpico4321

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    Feb 23, 2013
    Thanks for reply.

    With the Rth could you not use the modulus of impedance? That way removing the complex conjugate?

    To calculate the power, one must surely have two of three unknowns either voltage or current combined with the impedance to calculate the power?
     
  4. Mar 1, 2013 #4
    The Electrician

    The Electrician

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    Jul 6, 2012
    You cannot escape using the complex conjugate if you really want the maximum power transfer..

    Furthermore, power is only dissipated in the real part of the load impedance.. If the load is represented as an impedance, then the load power is given by the square of the load current times the real part of the load impedance.

    If the load is represented as an admittance, then the load power is given by the square of the load voltage divided by the real part of the load admittance.
     
  5. Mar 1, 2013 #5
    serpico4321

    serpico4321

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    Feb 23, 2013
    Thanks,

    Though without a value for voltage and only a value for the supply frequency is it possible to calculate power I'm assuming the answer for this is asking for a specific amount of power, in watts?

    Do you think it's the way it's worded?
     
  6. Mar 1, 2013 #6
    duke37

    duke37

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    Jan 9, 2011
    You must calculate the impedance of the source. This will be a resistor in series with an inductor.

    The load must be the complex conjugate and will be a resistor in series with a capacitor.
    The two resistors will be the same for maximum power transmission.

    The inductance and capacitance form a series resonant circuit so drop out of the calculations.
     
  7. Mar 1, 2013 #7
    john monks

    john monks

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    Mar 9, 2012
    You calculate the maximum power transfer in terms of your source voltage.
     
  8. Mar 1, 2013 #8
    serpico4321

    serpico4321

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    Feb 23, 2013
    Nice one, though I must be failing to get this.

    With the 5 values in the OP circuit how would the voltage be obtained?

    Or is it, as you say the thevenin equivalent which is the complex conjugate of the series and parallel elements of the circuit. Then if Rth = Rload the maximum power transfer would be 50% of the source current?
     
  9. Mar 2, 2013 #9
    john monks

    john monks

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    Mar 9, 2012
    Your on the right path except that current is not power. I think you meant source power.
     
  10. Mar 2, 2013 #10
    serpico4321

    serpico4321

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    Feb 23, 2013
    Cheers, I meant relation as P=I^2*R where p is watts as there is only resistance as in this case impedance then it can't be possible to calculate the amount of power (in watts) through the load resistor.

    The only value I can think to give is 50% of the total power available in the circuit provided by the source at 50Hz
     
  11. Mar 2, 2013 #11
    john monks

    john monks

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    Mar 9, 2012
    Your on the right path. I have to say though that I am a little disappointed with the colleges and universities in that they give you assignments, give you formulas like ohms law, work a few on the chalk board, and expect you to understand it just like that. But over and over again I work with technicians and engineers that hardly have a clue as to what is going on. As my physics professor, Dr James Conrad, put it you work only from the physics, not the math. And if you cannot prove a formula you should not use it. In this case you have an AC circuit with some resistance and inductance and you were rightfully advised to use the conjugate of the output impedance. My question to you is do you know why you use the conjugate. If you do not fully understand why I suggest you get some resistors, some inductors, capacitors, a function generator and find out why by building the circuit. It is not a waste of time.
     
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