Max Power Calculation

Discussion in 'Homework Help' started by Ryuk, Feb 13, 2014.

  1. Ryuk

    Thread Starter New Member

    Oct 9, 2012
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    [​IMG]

    In the above circuit, I am to calculate the max power that can be dissipated in RL.

    By killing the sources, the thevenin resistance is 2 kΩ.

    With an open at RL, the current flowing through the 2 kΩ resistor is 3 mA * ((4 / (4 + 2 + 3)) kΩ) which is 4/3 mA. Since we know the node to the left of the open is 6 V, the thevenin voltage will be 6 V minus whatever the voltage to the right of the open is. That voltage will be 4/3 mA * 3kΩ so that's 4 V. Hence, the thevenin voltage is 2 V.

    So shouldn't max power be Vth^2 / Rth = (2 V)^2 / 2 kΩ ???

    That's 2 mW but the solution I read says 0.5 mW. The solutions I'm reading don't show the steps clearly but it says that in the above formula, Rth should be multiplied by 4 in the denominator. Where would that come from?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Only half the Thevenin voltage appears across the load. That's just 1V.
     
  3. Ryuk

    Thread Starter New Member

    Oct 9, 2012
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    Could you clarify what that means? Why does only half the thevenin voltage appear across the load?




    Also, I have another question about another circuit. I couldn't make a separate thread for some reason (I kept getting a blank page when posting).

    [​IMG]

    In the above circuit, we can easily find the open circuit voltage using the voltage divider formula which is 1 V * (1 kΩ/(2 kΩ + 1 kΩ) which gives us 1/3 V.

    Another way is to write a KCL equation for the node to the right of the 2 kΩ resistor. Let's call that node Vx.

    So assuming all currents are going out and ground reference is the bottom of the circuit, (Vx - 1)/2 kΩ + (Vx - 0)/1 kΩ = 0. This makes Vx also 1/3 V as expected.

    Now why doesn't it work if we were to write a KCL equation for the node above the voltage supply? Vx is still the node to the right of the 2kΩ and all currents are still going out we assume.

    So (1 V - 0)/1 kΩ + (1 V - Vx)/2 kΩ = 0. This simplifies to 2 V + 1 V - Vx = 0 and so we get Vx = 3 V which is wrong.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    At MPT the load voltage is always half the effective source voltage.

    In your second example you equate the lhs (of your equation) to zero. This is not correct.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Take a step back and forget about max power and Thevenin circuits and such.

    Take a battery with Vdc volts and apply it to two series-connected resistors, R1 and R2, that are equal in value (call that value R).

    What is the power in R2?
     
  6. Ryuk

    Thread Starter New Member

    Oct 9, 2012
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    D'oh! I forgot about the current traveling on the pathway where the voltage supply is.

    So it wouldn't be possible to find Vx with nodal analysis from that node alone.
     
  7. Ryuk

    Thread Starter New Member

    Oct 9, 2012
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    (Half of Vdc)^2 / R.

    Oh...of course. So because thevenin resistance equals load resistance, the circuit behaves just like your example.

    Although, I still don't understand why in the solution they have it as (Vth)^2 / (4 * Rth)

    It's the same answer but it seems like a strange way to represent it.
     
    Last edited: Feb 13, 2014
  8. WBahn

    Moderator

    Mar 31, 2012
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    As you say, it's the same answer. Arguably, it is the "simplest" form of the equation, but it does obscure some of the underlying quantities.
     
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