More resistance is less load??

I thought it was E/IR   :-)

Good responses by the others above, except that the reference to E/IR should be E = IR (meaning E equals I multiplied by R). (E is commonly used in the context of Ohm’s Law to denote volts, and means the same thing in that context as V).

So when Ghosthouse referred to V/I = R, he was correct. By simple algebra V/I = R is equivalent to V (or E) = I x R = IR.

Regarding the voltages that are typically provided to speakers, for a resistive load:

Power = (I squared) x R = (E squared)/R
where, for example, P is expressed in watts, I in amps, R in ohms, and E in volts.

Therefore E = (Square root (P x R)).

So for example 100 watts into an 8 ohm resistive load corresponds to:

Square root (100 x 8) = 28.28 volts.

Regards,
-- Al

In other words, think of the flow of electrons like a water hose as someone already mentioned. The less resistance, the more flow and therefore the more draw on the amplifier to push those electrons through the wires. The more draw, the harder the amplifier has to work. It's rather counter intuitive unless you think about it in those terms. 

Thanks this is very helpful. I agree falconquest it is counter-intuitive, but the equations and explanations are clear & helpful. I have some clarity now.

almarg has brought up the notion of "power".

Used to be, when I paid attention to these things, 40 watts per channel was 40 watts "RMS".

How is RMS (root mean square I believe) to be understood?