ianganderton
Auckland, NZ
Step 3: Calculate Resistance for 166 amps
ΔV = I x R, so R = ΔV / I =
Does it matter which way this is calculated because
R = I / ΔV = 166 / 0.18V = 0.001084337 Ω which seems more viable. Thinking about it to rearrange the equation it would have needed to be divided by ΔV not by I
Then
Step 2: Calculate Cross Sectional Area for 1m distance
R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m
Solve for area: A = pL ÷ R;
A = 1.7 × 10-8 Ω m x 1 ÷ 0.001084337 = 1.56778E-05m2 = 15.67mm2 I think which makes more sense but is still very small compared to the 35mm2 / 2/0AWG recommended by other calculators
ΔV = I x R, so R = ΔV / I =
Does it matter which way this is calculated because
R = I / ΔV = 166 / 0.18V = 0.001084337 Ω which seems more viable. Thinking about it to rearrange the equation it would have needed to be divided by ΔV not by I
Then
Step 2: Calculate Cross Sectional Area for 1m distance
R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m
Solve for area: A = pL ÷ R;
A = 1.7 × 10-8 Ω m x 1 ÷ 0.001084337 = 1.56778E-05m2 = 15.67mm2 I think which makes more sense but is still very small compared to the 35mm2 / 2/0AWG recommended by other calculators