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maximum power to load?

Discussion in 'Misc Electronics' started by zooeb, Feb 9, 2005.

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  1. zooeb

    zooeb Guest

    in a simple electric circuit, in which you have a DC voltage
    generator, an internal resistence and the resistence of the load, why
    the maximum power that the generator is able to tranfer to load is
    when load resistence is equal to internal resistence? I try to think
    about it: if internal resistence is zero, power developed by load is
    V*I, where V is the voltage generator and I=V/RL; then if the internal
    resistence goes up, the power developed by load is V*I, where V minor
    than the voltage generator and I=V/(RI+RL), which is less then the
    previous current. So, why in the first case power isn't max?
     
  2. R.Lewis

    R.Lewis Guest

    If the internal resistance of the source is zero ohms and the load resitance
    is equal to the internal resistance then the power is infinite.
    Difficult to get any higher than that.

    Otherwise do some sums.
    If I have a 12v battery with an internal resistance of 6ohms and I connect a
    6ohm load resistor to it the power dissipated in the load is 6 watts.
    Just try any other value of load resistance and see if you can get its
    dissipation above 6 watts.
     
  3. :in a simple electric circuit, in which you have a DC voltage
    :generator, an internal resistence and the resistence of the load, why
    :the maximum power that the generator is able to tranfer to load is
    :when load resistence is equal to internal resistence? I try to think
    :about it: if internal resistence is zero, power developed by load is
    :V*I, where V is the voltage generator and I=V/RL; then if the internal
    :resistence goes up, the power developed by load is V*I, where V minor
    :than the voltage generator and I=V/(RI+RL), which is less then the
    :previous current. So, why in the first case power isn't max?

    Matching the load resistance to the generator resistance for maximum
    power transfer applies when you have a generator with a known, fixed
    internal resistance. If you change the characteristics of the generator
    by lowering its internal resistance, of course you can deliver more
    power to the load.
     
  4. DM

    DM Guest

    But still, if you change the internal resistance of the generator you're
    going to decrease the power at the load, until the load resistance
    changes to match the internal resistance of the source.

    You can increase the voltage to the load by making its resistance larger
    than the resistance of the source but you will only get max power
    transfer when they are equal.
     
  5. John Fields

    John Fields Guest

    ---
    That's not true. Consider the generator to be a voltage source with a
    resistance in series with it and the load to be a resistance to
    ground, and you'll have this:

    E1
    |
    R1
    |
    +----E2
    |
    R2
    |
    GND

    A simple voltage divider, where:

    E1 is the generator's voltage source
    R1 is the generator resistance
    E2 is the generator's output voltage, and
    R2 is the load resistance

    Now, if E1 = 2V
    R1 = 1R, and
    R2 = 1R

    then

    E1R2 2V * 1R
    E2 = ------- = --------- =1V
    R1+R2 1r + 1R

    and the power being dissipated in R2 will be:


    E² 1
    P = ---- = --- = 1W
    R2 1


    Now, if we change the generator's internal resistance by _lowering_ it
    to 0.5 ohm, the voltage across the load will _increase_ to:



    2V * 1R
    E2 = ----------- = 1.333V
    0.5R + 1R


    and the power the load will dissipate will be


    1.333²V
    P = --------- = 1.333W
    1R

    Which is _higher_ than the dissipation with the generator resistance
    at 1 ohm, but lower than it would be if the load resistance was equal
    to the generator resistance.

    Since we know that if R1 and R2 are equal E2 has to be 1/2 of E1, then
    with R1 and R2 bothe equal to 0.5 ohms, R2 will dissipate E2²/0.5R = 2
    watts, and so will R1.
    ---
     
  6. Fred Abse

    Fred Abse Guest

    No. The current is infinite. The power delivered to the load is zero,
    since power is I^2R and R is zero. Anything multiplied by zero is zero.

    A zero ohm load is called a short circuit.
     
  7. DM

    DM Guest

    You are correct. I did all my calculations based on a constant source
    resistance and varying the load resistance, if I'd reversed my values I
    would have seen that.
     
  8. R.Lewis

    R.Lewis Guest

    Yep. You are certainly corrent that the power dissipated in the resistor
    would be zero.

    A zero ohm load is just that.
    An unwanted zero ohm load is a f*****g short.
     
  9. redbelly

    redbelly Guest

    No Fred. The power is infinite since power is V^2 / R and R is zero.
    Anything divided by zero is infinite.
     
  10. John Fields

    John Fields Guest

     
  11. Fred Abse

    Fred Abse Guest

    V=IR, hence if R is zero, V is zero.
    Except zero.
     
  12. Fred Abse

    Fred Abse Guest

    But not the converse. Sometimes we *want* a short.
     
  13. R.Lewis

    R.Lewis Guest

    V=I*R
    if R=0 then V=0, i.e. V=R
    the power deveoped in the resistor is V^2/R but since V=R this becomes
    (V^2/V=) V or conversely (R^2/R=) R. The watts developed is thus V volts.
    This conclusively proves that a resistor of zero ohms has a resistance of N
    volts, where N is the source voltage.

    The power developed in the resistor may also be expressed as I^2R. Now since
    R=V this may be written as I^2V.
    Since the power is also expressed as V^2/V, equating these two, V^2/V=I^2V
    which may be re-aranged as I^2=V^2/(V*V).
    I is thus SQRT(1) which is one.
    i.e. the current flowing through the resistor equals1 volt independant of
    the source voltage (provided the source resistance is equal to the supply
    volts in volts).

    I hope this explains it fully for you.
     
  14. Fred Abse

    Fred Abse Guest


    ROFL!
     
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