quote:
Originally posted by justin_tx_16:
my first choice was the 570. I think I will go with my gut instinct on this. 64 ohm resistance means that Here is what the V=IR equation gets me... V=32I would mean to get 64 volts of power, I would need 2 amps, and to get 64 volts of power with 64 ohms, I would need 1 amp... haha, what is up with that? I must have something messed up here. How it should work is the higher the resistance, the more amps needed, right?
So, lets turn it to I=V/R, I=V/32, vs I=V/64, so yes, half the power. If only I could find DanF's info on the control pod... But we want watts right? actually milli watts. So we turn to this equation, P=I^2R, which we got from P=IV and V=IR, substitute the V in P=IV for IR and you get P=I^2R, so we want 100mW, or .1 watts, so .1=I^2(64), divide by 64, you get 0.0015625, square root it and you get 0.03952 amps are required. So, just gotta remember the watts to amps equation.
Well, lets look at a different equation. P=VI If I need 100mW's of power, .1 watts, that is .1=VI, .1=(j/c)(x c/s)........ i think that is the equation i need to work haha. Coulombs, the definition of fun!
edit, i see some typos here, just don't grade my physics
Justin:
You also need to take into account the efficency of the headphones.
In the end, its best to read online reviews (www.audioreview.com) and see what headphones are reccomended for portable/computer use. You could also alleviate the problem and buy a headphone amp, and achieve pure audio bliss ;-).
Again, I've got HD-535's. Very nuetral and revealing headphones. They are very true to the source, so poorly recorded MP3s will sound horrible. On the plus side, its great to hear nuances in the music, especially live music.
Jonathan