100v line output

[PhilUK]

What is the 100v line output selection for?
I seem to remember this had something to do with installed PA or tannoy systems?
What is the effect on the amp and output transformer? Does the amp still see a resistance and could/should the option be used at all for speaker cabinets?
Thanks

[OldSchoolDave]

... could/should the option be used at all for speaker cabinets?

NO!

I can't imagine anyone using HIWATT power for their (institutional-style) Public Address System, but that's what the 100v setting was for:

http://en.wikipedia.org/wiki/100V_speaker_system

Dave

[PhilUK]

Thanks for that Dave, so if our OTs in that selection are giving out 100v, how much bigger does the OT have to be to provide it?
If it didnt have the 100v could it be physically smaller?

[OldSchoolDave]

If it didnt have the 100v could it be physically smaller?

WAY beyond my "expertise". Perhaps the good Professor could comment?

Anecdotally, HI-TONE OT's do not appear to be any smaller or lighter than Hylight HIWATT Partridges.

Dave

[PhilUK]

What I cant get my head around is that (as I understand it) its a 100v feed to speakers which then each have a step down transformer in them.
So what is the voltage range out of the back of a Hiwatt in say the 16 ohm setting?
In 100v line output, does the amp need to see a resistance or does a 100v system have no resistance and so it is provided by the transformer itself?
Sorry to be electronically inept, trying my best to understand!
(and appreciate your input)

[mhuss]

Ohm's Law will help here.

P = E^2/R, so E = sqrt(PR). For 16 ohms and 100 watts, that works out to 40 VAC.

Working the other way, R = E^2/P. For 100v and 100w that works out (by coincidence) to 100 ohms.

Why do 100V? The same reason mains are distributed via high tension instead of 120 or 240 volts. Again, Ohm's Law provides the answer. First, P = E x I -- if you have lower voltage you must have higher current to provide the same amount of power. Second, the more current you have passing through a wire, the more power you will lose heating the wires. Power dissipated in the wire increases as the square of the current (P = I^2 x R), so minimizing current by increasing voltage is a practical solution when transferring power over long wires.

--mark

[PhilUK]

Thanks Mark, over my head most of that but I get the idea.
Surprised the amp gives 40V into a 16 ohm load.
So why is the load not specified on the 100V setting? Is the amp running flat out with no resistance?

[mhuss]

"100v" (or "70v line" in the U.S.) is just the standard way of labeling such things. Everyone who might need t use such a setting knows what to do with it. As long as the impedances match, it will always be 100 watts,

--mark