“it is the overall lack of the voltage due to the internal impedance of the circuit”
\(V_{drop\left(V\right)} = I_{cable\left(A\right)} * R_{wire\left(ohms\right)}\) \(V_{drop\left(V\right)} = I_{wire\left(A\right)} * \left(2 * L_{\left(ft\right)} * \frac{R_{wire\left(\frac{Ω}{kft}\right)}}{1000_{\left(\frac{ft}{kft}\right)}}\right)\) \(V_{drop\left(V\right)} = I_{wire\left(A\right)} * R_{wire\left(Ω\right)}\) \(V_{drop\left(V\right)} = I_{wire\left(A\right)} * \left(2 * L_{\left(m\right)} * \frac{R_{cable\left(\frac{ohms}{km}\right)}}{1000_{\left(\frac{m}{km}\right)}}\right)\)
\(V_{drop\left(V\right)} = \sqrt{3} * I_{wire\left(A\right)} * R_{wire\left(Ω\right)}\) \(V_{drop\left(V\right)} = 1.732 * I_{wire\left(A\right)} * \left(L_{\left(ft\right)} * \frac{R_{wire\left(\frac{Ω}{kft}\right)}}{1000_{\left(\frac{ft}{kft}\right)}}\right)\) \(V_{drop\left(V\right)} = \sqrt{3} * I_{wire\left(A\right)} * R_{wire\left(Ω\right)}\) \(V_{drop\left(V\right)} = 1.732 * I_{wire\left(A\right)} * \left(L_{\left(m\right)} * \frac{R_{wire\left(\frac{ohms}{km}\right)}}{1000_{\left(\frac{m}{km}\right)}}\right)\)
Regardless of the section is, the calculator will take multiple seconds to show the actual loss in the voltage transmission.
For a cable having a diameter in inches and n gauges:
\(d_{n\left(in\right)} = 0.005 inches * 92^{\frac{\left(36-n\right)}{39}}\)
\(d_{n\left(mm\right)} = 0.127 mm * 92^{\frac{\left(36-n\right)}{39}}\)
\(A_{n\left(kcmil\right)} = 1000 * d_{n}^{2} = 0.025 in^{2} * 92^{\frac{\left(36-n\right)}{19.5}}\) \(A_{n\left(in^{2}\right)} = \left(\frac{\pi}{4}\right) * d_{n}^{2} = 0.000019635 in^{2} * 92^{\frac{\left(36-n\right)}{19.5}}\) \(A_{n\left(mm^{2}\right)} = \left(\frac{\pi}{4}\right) * d_{n}^{2} = 0.000019635 mm^{2} * 92^{\frac{\left(36-n\right)}{19.5}}\)
\(R_{n\left(\frac{Ω}{kft}\right)} = 0.3048 * 10^{9} * \frac{ρ\left(Ω.m\right)}{25.4^{2} * A_{n\left(in^{2}\right)}}\)
while the capacity at the end of the cord gets higher than the capability on the start, then it gives upward push to the capacity voltage drop.
The space is without delay proportional to the resistance and while the resistance will increase, the Voltage drop can even get most.
