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RE: Techie Question (old style)
The rule of thumb is that a three gauge difference doubles or halves the
current carrying capacity (or resistance) of any wire.
Bill
ARINC
-----Original Message-----
From: NIXON, Grant (Kanata) [mailto:GNIXON@MDS.Nordion.com]
Sent: Tuesday, August 05, 2003 1:36 PM
To: 'Bruce Bugg'; RADSAFE (E-mail)
Subject: RE: Techie Question (old style)
The most important factor is cross-sectional area, not gauge.
10 AWG dia = 2.989 mm
5 AWG dia =5.338 mm
For stranded wire, we have:
10 AWG: resistance is about 0.00334 Ohm/m [area ~5.261 mm^2].
5 AWG: resistance is about 0.00105 Ohm/m [area ~16.773 mm^2].
We see that the 5 AWG wire has more than 3 times the carrying capacity of 10
AWG wire.
If we calculate resistance in parallel for 2 of the 10 AWG wires:
L/R=1/0.00334+1/0.00334 -> R/L=0.00167 Ohm/m, which is the same as 7 AWG
wire.
See, e.g., http://www.mogami.com/e/cad/wire-gauge.html
Best regards,
Grant
-----Original Message-----
From: Bruce Bugg [mailto:obbugg@dmvs.ga.gov]
Sent: Friday, August 01, 2003 5:11 PM
To: RADSAFE (E-mail)
Subject: Techie Question (old style)
Not really a Rad question, per se, but I figure someone on this list will
have a resource or the answer.
I have 2 parallel wires of a given gauge. Is there a formula (or is it
straight math) to tell me the effective gauge?
In other words, are two 10-gauge stranded wires in parallel equivalent to
one 5-gauge wire?
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Capt. Bruce Bugg
Special Projects Coordinator
Law Enforcement Division
Georgia Department of Motor Vehicle Safety
P.O. Box 80447
Conyers, GA 30013-8047
voice: 678.413.8825
fax: 678.413.8832
e-mail: obbugg@dmvs.ga.gov
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