stupid question about SPL vs power

Example:

I have two sets of speakers.

One is ~3R5 with 83db "typical in room" SPL (Micca RB42)
Two is ~6R with 95db "typical in room" SPL (Energy RC-70)

Assuming the same perceived volume, and the fact that I know it takes more power because I had to increase the volume to compensate, is that power just being wasted as heat in the crossover/drivers then? I assume yes, but I'm having a brain fart.
 
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R == ohms in technical nomenclature. Example 5R1 means 5.1 ohms. Like 2u2 means 2.2uF

Also I've corrected my post. 3.5R means nothing. 3R5 does.

R as opposed to k or M.... decimal points get lost in crappy printing... 5R1 vs 5.1 is more likely not to result in someone seeing 51...
 
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... is that power just being wasted as heat in the crossover/drivers then?
Yes, loudspeakers convert most electrical power to heat rather than acoustical watts.

This site gives an explanation of how to do the conversions:

Sound power level SWL and sound pressure level SPL distance compare acoustic power sound source noise Conversion of sound pressure to sound intensity conversion sound level energy level strength directivity factor coefficient sound intensity SIL - se
 
Power is square law. 10V into 10R load = 1x1x10= 10W
Add 10R, say to reduce speaker volts, load now 20R. 0.5x0.5x20 = 5W total.
I.e. the power into the speaker doesn't (necessarily) stay the same, but sensitivity is reduced. In this case the impedance has gone up. Hence the qualification.

In your speaker example the impedance has fallen. Maybe the idea is to increase power to the bass driver as well as padding mid/HF. Sensitivity is also confusing as some quote it relative to volts into 8 ohm load as opposed to per watt actually consumed. Electrical energy converts to heat mainly, whatever the situation.