Relation between Sq.mm of wire and Its current carrying
capacity?
Answers were Sorted based on User's Feedback
Answer / y_suhas
you must know the current density then you find relation
between current & area
d=I/A,
d-current density
I- Current
A-conductor area in sq mm
Is This Answer Correct ? | 33 Yes | 6 No |
Answer / premnath
c = l/(r*a)
c = electrical conductivity
l = length of wire
r = resistance of wire
a = cross sectional area of wire
Is This Answer Correct ? | 11 Yes | 2 No |
Answer / k.a.sharma
Current carrying capacity is given by
I = (V*a)/(r*l)
I = Current carrying capacity
V = Max. allowable voltage drop in length 'l' by design.
a = cross sectional area of wire
r = specific resistance of material of wire
l = length of wire
Is This Answer Correct ? | 8 Yes | 0 No |
Answer / vinod
a=i/d
a=crosssectional area
i= current
d=current density
Is This Answer Correct ? | 13 Yes | 6 No |
Answer / víctor
As a thumbrule for home applications you can count 6 amp
per sq mm.
Large currents needs to take a look at the conductor
manufacturer.
Is This Answer Correct ? | 2 Yes | 4 No |
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