When selecting wire diameter, we usually look for cross-reference tables to find the recommended wire diameter for the maximum current drive. But sometimes maybe more useful to calculate by formula than look into the tables. This way, you can have more accurate results. There is nothing new, just simple physics.

Wire resistance (Ω)is calculated as follows:

R = ρ·l/S or R = (1.27·ρ·l )/d^{2};

Where ρ – resistivity of material (Ω·m), found in tables, l – wire length (m), S – cross selection area(m^{2}), d – wire diameter (m). According to this, we can calculate wire length when other parameters are known:

l = R·S/ρ or l = (0.785·R·d^{2})/ρ

Cross selection area can by found as follows: S=0.785·d^{2.}

Wire resistance depends on temperature. Lets say R_{2} is resistance for temperature t_{2} and R_{1} is for t_{1}(usually t1 = 18°C), then:

R_{2}=R_{1}·(1+α·(t_{2}-t_{1}))

Where α – temperature coefficient (K^{-1}).

Let’s say that the maximum current for cross selection area is marked as Δ (A/mm2); then maximal current can be found I=0.785·Δ·d^{2}. Needed diameter can be found by formula(√ – square root symbol):

d=√((1.27·I)/Δ)

So if allowed load is Δ=2A/mm^{2}, then d=0.8√I.

If the diameter is less than 0.2mm, then melting current can be calculated by the formula:

I_{m}=(d-0.005)/k

where k – coefficient (k_{copper}=0.034; k_{nickel}=0.07; k_{iron}=0.127). Then diameter would be d=k·I_{m}+0.005. This formula may be used when calculating fuses.

For example, if we need a fuse that melts at 5A, then copper wire diameter would be:

d_{5A}=0.034·5+0.005=0.175mm

Why the hell would you use so confusing and complicated symbols…

Agreed… use Latex next time

i want to know which copper cable of 4 core is suitable for me for my ac’s of 50kv