You are wrong. Take a look:

According to NAD, the watts when driving a 4 ohm or an 8 ohm load are constant, so we have

x^2 * (4 Ohms) = y^2 * (8 Ohms) -- divide by 4 Ohms
x^2 = y^2 * (2 Ohms) -- take a square route
x = y * sqrt(2 Ohms)

x is the amp's current when driving a 4 Ohm load
y is the current when driving an 8 Ohm load

As you can see, x is greater than y.

This is what we expected, since more current flows into a less resistant load.

Last edited by pmbuko; 04/29/05 08:47 PM.