/\ Thanks for the explanation Tang. Why 2.5A though?
Also, what makes an amp not ideal, ie 28V for 50W? How do you get to 35V from 28V?
V=IR
you have an 8ohm resistor and you are putting a 20V battery across it
20 =I x 8
20/8 = I
I = 2.5A
I just just that first equation to show how DC current and differs from AC current calculations. For DC current you don't need to bother with the 1.414 division of the peak voltage and current to get correct power , you just use the dc voltage and current to get power dissipated.
The 35V is an example.
the lm3386 chip needs 35V rails to be able to drive 50W RMS wattage into a resistive 8 ohm load.
http://www.ti.com/lit/ds/symlink/lm3886.pdfcheck on page 3:
For 28V rails and 8ohm load the typical output power is 38W but National instruments promises the minimum a chip you buy from them will deliver is 30W.
(this is measured with the chip at 25C)
What makes an amp not ideal. Transistors are not 0ohm switches , they have internal resistance when ON , this resistance goes up when they are hot.
The input of the amplifier when powered from the same voltage rails as main output transistors(most amplifiers works like this) when rails dips for high volumes and low impedances might not be able to switch on the transistor fully.
look at the naim amplifier

note the two 0R22 resistors on the outputs of the transistors to keep things stable.
Technically you have a 0.22ohm resistor in series with your speaker.
lets say you have 28.28V , This is divided by this 0R22 and your 4ohm speaker.
This means that on your speaker you will only have 26.806V and 1.474V across the 0R22 resistor.
This calculates to 89.82W and not the full 100W you are hoping for. just over 10W is wasted in the 0R22 resistor