HOW TO UNDERSTAND DIFFERENT IMPEDANCE RATES?!!!
god at almighty.com
Thu Jun 12 16:37:20 EST 1997
>I believe the original poster asked whether impedance had anything to do
>with loudness. It surely does, as proven by my argument.
>Your problems with the use of the written language seem to have caused
>you to misunderstand my post, so let's see if I can make it easier for you.
>Let's assume you have a nine volt battery, a nine ohm ten watt resistor
>and an eighteen ohm ten watt resistor. Let's also say that the
>efficiency of the resistors is 100%, just to level the playing field.
>Next, connect the battery across the nine ohm resistor. The resistor
>will dissipate nine watts, will exhibit a rise in temperature, and will
>eventually come to thermal equilibrium at some temperature.
>Now, disconnect the battery from the nine ohm resistor and connect it
>across the sixteen ohm resistor. The resistor will dissipate 4.5 watts,
>will exhibit a rise in temperature, and will eventually come to thermal
>equilibrium at a temperature _lower_ than the nine ohm resistor used in
>the previous example.
>The point is that if you have a constant voltage source, the terminal
>temperature (which is analogous to sound pressure level) will vary as a
>function of the load impedance, which is the correct answer to the
>question originally posed. By the use of DC In the cases above, the
>imaginary portion of the impedance has been reduced to zero in order to
>allow you to, perhaps, grasp the concept.
John, I never disagreed with you. Your original conclusion correct,
however it was based on fallacious logic. (i.e. your conclusion,
albeit correct, did not follow from the premises provided.) Mr
Woodgate was able to point out non sequitur in your post as well, so
it was not a mere English misunderstanding on my part. I directly
address the fallacy in a previous post of this thread.
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