I've edited the opening paragraph of the section on Horn loudspeakers so as to give a better background of their use. (More correctly, a brief history of the use of horns to direct and amplify sound, not just their use with drivers.)
I don't think this edit would count as "original research" as it's a bit of a link-and-cut-and-paste from other Wikipedia articles, but others might disagree. Also, I might not have conformed with proper Wikipedia style and I'd be happier if someone more familiar with Wikipedia's style standards would double-check what I've done. Thanks.
Based on my reading of a commercial website that compares various headphones, my impression is that headphones with a higher impedance will require a higher voltage to drive them. On the other hand, in Wikipedia's treatment of the history of headphones there is a mention that some headphones “... used with early wireless radio had to be more sensitive and were made with more turns of finer wire; impedance of 1,000 to 2,000 ohms was common”, this is as opposed to the statement that “headphones used in telegraph and telephone work had an impedance of 75 ohms”.
Presumably what is true of headphones is also true of loudspeakers. Assuming that there's any validity to my impression that impedance and sensitivity are related then I'd appreciate some discussion of that. (Perhaps a discussion would be appropriate if I'm mistaken in my impression, if it's a common impression.)
I suppose that the same discussion would be appropriate in the Wikipedia articles for Headphones and Electrical impedance.
As far as the relationship between impedance and sensitivity goes for loudspeakers (and indeed the relationship between resistance/impedance and efficiency for most electrical devices) a higher number of turns of wire in a coil of a given size generally translates to higher efficiency (as field strength = amps * turns) and also higher impedance. Because the coil has a greater impedance it is necessary to power it with a higher voltage source to achieve the same power input (power = voltage * current, (power ≠ voltage), voltage = current * impedance), but (again because of its higher impedance) it will not draw nearly as much current at that given power level (current = voltage/impedance).