Max power transfer

Discussion in 'Homework Help' started by minduka, Mar 8, 2009.

  1. minduka

    Thread Starter New Member

    Mar 5, 2009

    I need some help on two max power transfer questions.

    This is my approach to the first one.

    R_{L}= 8 \Omega - because max power occurs when R_{TH} = R_{L}

    Then I found the max power transfered in (b) by using

    P_{L,Max} = \frac{v_{L}^{2}}{4R_{TH}}

    V_{L} = IR = 2A*8\Omega = 16v

    R_{TH} = 16//8//16 = 4\Omega

    P_{L,Max} = \frac{16^{2}}{16} = 16

    I'm confused when they give you R_{L} and need to find the max power. I'm assuming you have to use R_{L} to find the max power.


    For the second problem:

    R_{TH} = 10//10+10 = 15\Omega

    V = 20v

    P = \frac{20^{2}}{4*15} = 6.67 W

    But that is not the answer according answer sheet.
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  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Hello minduka,

    In the first question you recognized that RL = 8Ω for the left hand circuit.

    The questioner is not asking for the maximum power transfer case for the right hand circuit - they are asking you to place the load RL = 8Ω into the right hand circuit and then solve for the power dissipated in the load, which with RL = 8Ω would give 32W - as indicated by the correct answer.

    In the second question, you are nearly there, but just need to realize that the voltage you should use is not 20V. Suppose you reduce the circuit to its Thevenin equivalent. It would be a 10V source in series with a 15Ω resistance. So the maximum power case is ((10/2)λ2)/15 = 1.67W.

    Hope this helps.
  3. minduka

    Thread Starter New Member

    Mar 5, 2009
    That was so helpful, thanks!

    I think the first problem is a little decieving :p
  4. nirosha

    New Member

    Nov 6, 2008
    In the first problem,by using voltage division rule we get voltage across RL.use this voltage and given 2watts to find RL then place this RL value in the next ckt and solve the rest.
  5. Ratch

    New Member

    Mar 20, 2007

    Not necessily so. If the source impedance and voltage cannot be changed, then yes, max power occurs when the load impedance equals the source impedance. But if you can make the source impedance smaller while keeping the source voltage the same, then you will transfer more power with the smaller source impedance.