Hi,
I think you'll find the difference if any is negligable. Be careful with your
measurements some of the straight translations may not be as simple as it sounds. 120
* 4.2 looks a lot less than 1800 but what was the power factor in your calculation.
What was your ammeter measuring (peak, average, rms). The average voltage of a sine
wave is "0 volts" does that mean the device consumes no power. Most AC ammeters use a
basic diode type bridge to convert AC - DC and tend to do a little float and you will
get less than .637 (off memory from 30 years back) of what it might infact be.
The greatest use of power is probably the fans. All the chips in the boards are live
anyway. A little extra would be consumed by each device plugged in but not much.
Just some thoughts as to why the power ratings are not what they seem. They usually
also overstate the ratings so as not to create unnecessary problems.
Teunis,
Hobart, Tasmania
Austalia.
On Monday, January 22, 2001 at 11:44:25 AM, Frankie wrote:
> Kinda interesting question:
>
> Does a catalyst 6500 (or any equipment) consume more power during periods of
> heavy traffic processing? If so, any rule of thumb as to how much more?
>
> I measured about 4.2A @120V on a 6506 w/ dual AC power supply, 1 supervisor,
> 4 flex-wan blades, and 7 OC3-MM port adapters. This is far less than 1800W
> max as listed in the product literature. In fact, anyone know the typical
> percentage difference between the LISTED max load vs the ACTUAL max load
> (fully loaded)? You'll find that even on a PC, there's a big difference
> between max and the actual load.
>
> Thanks
>
>
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