DIng Ding Ding! I was just trying to keep things simple, e.g. "voltage"
It's an annoying notation but I've seen it in textbooks often. The first time I saw it in a supplementary reading, I was like "whaaaat? Why did they use E??" D:
Also, ravager is correct about all of the math, but headphones also have a sensitivity rating that affects how much power is needed to get a headphone to reach X dB SPL. More sensitive headphones (i.e. earphones) require less power to reach X dB SPL than less sensitive headphones (i.e. regular 'ol over-ear headphones), thus requiring less voltage (P = E^2 / R), and thus less "volume on the amplifier"; this is why headphones with a lower sensitivity rating require a "higher volume level on the amplifier" than a less sensitive headphone to get to the same volume level.
Yes, that is absolutely true, and something I did overlook. I remember the first time I discovered that with a decent pair of Klipsch speakers. Nearly blew out the windows in my German flat!
cool, that symbol makes more sense, E has always been reserved for Electric field, I thought ravager was going to go down to the nitty gritty part of physics lol oh well
Just to make my position clear:
I do not agree with Julian Hirsch.
I was rolling my eyes at Julian Hirsch!
Impedance acts like resistance in the analogue realm.
Therefore, using Ohms Law, we can generally say that Since E=I*R (E = voltage, I = current, and R = resistance), with E being constant, a higher resistance (or in our example, impedance) means less current drawn. Now POWER equation is P = I*E. So with that less current, means less power, for the same potential (voltage). The volume knob varies the voltage, just like Chris said, so in all cases, there will be more power to the headphones when the knob is turned towards max. (It is a little more complicated than this with the alternating current, but for illustration purposes, this will suffice).
So, using canned specs, and comparing 32 ohm to 300 ohm cans, we get something similar to this:
MyAmp puts out 0-11 Volts.
At 5 volts, we can plug in our 32 ohm cans to compute current and wattage (or by substitution, P=E^2/R)
But lets keep things simple. Calculate current, which is I = E/R = 5/32 = .156 amps or 156 milliamps (mA)
Then the power ( in watts) would be P = I*E = .156 * 5 = .78 watts or 780 milliwatts (mW)
Now do the same thing for the 300 ohm cans. I=E/R = 5/300 = .017 amps or 17 mA
Just by knowing what that equation is, we can see that will give much less power to the headphones
P = I * E = .017 * 5 = .085 watts or 85 mW. Even at 11 volts the power is P = E^2/R = 11*11/300 = 121/300 = 403 mW
In fact, you would need 15.3 volts to get the same "wattage" delivered to the 300 ohm headphones as compared to the 32 ohm at 5 volts.
Now take this information, and apply it what Chris said earlier about: "A low impedance Beyer DT990 needs just as much power as a high impedance Beyer DT990."
And you can see why the volume level will be lower, given the same output voltage on the high impedance cans.
There has to be a balance though, because if the impedance is too low, there is too much current drawn from the output and bad things can happen, with the least impactful being a blown fuse on the amp.
hope this helps.
(Geek hat off now)
Just wanted to add that the first sentence "Impedance acts like resistance in the analogue realm" is incorrect and rather confusing.
Digital signals AND Analog Signals must both deal with load impedance, source impedance, line impedance.
It would be more accurate to say that impedance does not come into play in steady state DC circuits.
Electronics can be confusing, which explains why I never leave the house without my tinfoil hat on!
i dono about any material but Celion Dion on a loop should encrypt all that goes in your head instantly :P