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Mark Zarella
 
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Default Bridging an amp... Circuit theory question???

DUMB DUMB DUMB
Speakers don't produce any wattage. Ever. Better get a refund on your
education as well.

BTW - Power (wattage) is defined in many ways - and it isn't just "Current

&
Voltage" - ya got 3 variables...
Amongst MANY variations of Ohm's law a
P=VI
P=I^2 * R


Dumb dumb dumb. That's actually not ohm's law, but joule's law of heating.


P=V^2 / R = (Vmax - Vmin)/R


Forgot a squared.

==== BTW - this is the form of ohm's law that
applies to the bridging of amps, and the resultant non-linear function for
the performance of such


How is it nonlinear?

- the confusion comes because the current needed to
drive the same stereo load increases non-linearly.


Um...huh?

While you are correct
that the circuit does not "see" a 2-ohm load with a 4-ohm stereo load
(since, as I have also stated, a 4 ohm load is a 4 ohm load), the net
result in the current required is such that the circuit ACTS as if you had

a
single amp of twice the power with a 2-ohm load.


Depends what you mean. The "current required" for what?