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The info in soundmind's response is incorrect

Mr. Curl's point was that when the load impedance is dropped to one third of nominal, the voltage drop ACROSS the output devices remains approximately the same, controlled by the voltage gain of the amp, but the current THROUGH the output devices is increased by a factor of 3.

Power dissipated in a device is equal to V x I.

In the schools that I attended: 1 x 3 = 3 , not 9

This has NOTHING to do with the power dissipated in the load, which has obviously increased by I2.

Anyone with an elementary knowledge of power amp design would have noticed this. It's exactly why a reacive load is the worst for an amp, since the voltage across the output devices and the current through them peak at the same time.

Peter

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