On Mar 31, 2008, at 7:39 AM, William Jackson wrote: > 1. all 4 pairs of cables going back to a single circuit breaker of > ( 6000/48 = 125Amp ) > 2. each pair of cables going back to a separate circuit breaker of > ( 6000/4 = 1500/48 = 31.25 Amp ) > > > > I am not an electrical guy but I would have thought that the idea is > that the breaker trips before the cable burns, so I would assume > option > 2?
I just looked at a 7609 we have colo'd here in our DC room and it has THREE 40amp DC fuses on EACH power supply. So.. three 40A from feed A and three 40A from feed B, six feeds total all on 4AWG. Hope that helps. -- Robert Blayzor INOC [EMAIL PROTECTED] http://www.inoc.net/~rblayzor/ Mac OS X. Because making Unix user-friendly is easier than debugging Windows. _______________________________________________ cisco-nsp mailing list [email protected] https://puck.nether.net/mailman/listinfo/cisco-nsp archive at http://puck.nether.net/pipermail/cisco-nsp/
