First of all, since you are talking like someone knowledgeable in EE, I would add that for ac circuits,
P = V*I*cos(Φ)*Distortion factor, that's just for tease, practically speaking we can mostly ignore the distortion factor that people rarely mentioned anyway.
So you have the formula but there is no simple answer because impedance (a phasor that has magnitude and phase angle) varies with frequency. Take a look of the graphs below and you can see how you can determine the current and power (in watts) on a point by point (frequency) basis.
For current, at a given voltage you can simply pick the lowest impedance magnitude point.
If the BMR is used as an example, it appears to be about 5.5 ohm at around 1,600 Hz. No idea why it would be lower than the woofer's minimum impedance, perhaps from cemf? (would need to do some digging..)
So you can find the current using just Ohm's law:
At 28.3 V (just an example), I = 28.3/5.5 = 5.15 A that's rms, peak would be 2X5.15 = 10.3 A.
The phase angle at that point was near 0 degrees so totally negligible. Suffice to say, the BMR can be considered easy to drive. It's sensitivity is on the low side, but still, something like the 3000 series Denon AVR should be able to do a good job for a lot of people.
If it were say 60 degrees, then it would be bad, because cos (60) = 0.5. It however, does not mean the calculated current drawn by speaker will be higher because of the 60 degrees phase angle. What it does mean, is that the "real power" (watts) consumed/dissipated by the speaker would be half of that of if the angle was 0 degree, and that means the power amplifiers output transistors would have to dissipate much more heat at high phase angles.
For a worked example, it is best you read it yourself in the article linked below:
Phase Angle Vs. Transistor Dissipation
sound-au.com
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www.audioholics.com