Partial ramblings on ohms law. To be continued.
*note: I started writing this up in response to a general question on the TT forum. Then I started writing. And I wrote more. And then I realized I was way off base with the guys original question. Since it would have sucked to trash an otherwise useful write-up, I'll dump it here.
Everything boils down to good 'ol Ohms law. Ohms law defines the relationship between resistance (ohms), voltage (volts), current (amperes), and power (watts):
(power = volts * amps)
Ohms law can also figure in the resistance:
R = E / I
(ohms = volts / amps)
With these two equations (and a little rememberance of our 9th grade algebra) we can find the third value if we know the other two. For example, if we know the watts (P) and the voltage (E), we can figure out how many amps something draws. Take a hair dryer with "1800W" silk-screened on it. How many amps is that pulling on a US 120V circuit?
1800 = 120 * I
-------- = 15 amps
...which is coincidentally the max draw for a standard home circuit. If we all ran 20amp breakers, I have no doubt they'd be selling us 2400W hair dryers.
Now, back to audio. Amplifiers put out voltage and current. How much of each (volts * current) tells you how many watts. This power is is fed into a speaker, which is a resistive circuit (put simply, the ohms rating). We know that speakers move back and forth based on the amount of voltage applied to them. And lets say we have a 100W, 8ohm speaker. Driven with 100W, how much voltage is required by the amp?
E = sqrt(P * R)
E = sqrt(100 * 8)
The amplifier must put out 28.28 volts. So how many amps is that?
I = P / E
Run the numbers, and the amp has to put out about 3.5 amps of current. So what happens if we want to run that same 28.28V of signal through a 4ohm speaker? How many watts are we dissipating? How many amps?
P = E^2 / R
-------- = 200W
I = P/E
----- = 7 amps
Whoa! Now we're asking our amplifier for twice as much current.