Wire / cable voltage drop calculator and how to calculate.

* @ 68°F or 20°C

** Results may change with real wires: different resistivity of material and number of strands in wire

The voltage drop V in volts (V) is equal to the wire current I in amps (A) times 2 times one way wire length L in feet (ft) times the wire resistance per 1000 feet R in ohms (Ω/kft) divided by 1000:

*V*_{drop (V)} = *I*_{wire (A)} × *R*_{wire(Ω)}

= *I*_{wire (A)} × (2 × *L*_{(ft)} × *R*_{wire(Ω/kft)} / 1000_{(ft/kft)})

The voltage drop V in volts (V) is equal to the wire current I in amps (A) times 2 times one way wire length L in meters (m) times the wire resistance per 1000 meters R in ohms (Ω/km) divided by 1000:

*V*_{drop (V)} = *I*_{wire (A)} × *R*_{wire(Ω)}

= *I*_{wire (A)} × (2 × *L*_{(m)} × *R*_{wire (Ω/km)} / 1000_{(m/km)})

The line to line voltage drop V in volts (V) is equal to square root of 3 times the wire current I in amps (A) times one way wire length L in feet (ft) times the wire resistance per 1000 feet R in ohms (Ω/kft) divided by 1000:

*V*_{drop (V)} = √3 × *I*_{wire (A)} × *R*_{wire
(Ω)}

= 1.732 × *I*_{wire (A)}
× (*L*_{(ft)} × *R*_{wire
(Ω/kft)} / 1000_{(ft/kft)})

The line to line voltage drop V in volts (V) is equal to square root of 3 times the wire current I in amps (A) times one way wire length L in meters (m) times the wire resistance per 1000 meters R in ohms (Ω/km) divided by 1000:

*V*_{drop (V)} = √3 × *I*_{wire (A)} × *R*_{wire
(Ω)}

= 1.732 × *I*_{wire (A)}
× (*L*_{(m)} × *R*_{wire (Ω/km)} / 1000_{(m/km)})

The n gauge wire diameter d_{n} in inches (in) is equal to 0.005in times 92 raised to the power of 36 minus gauge number n, divided by 39:

*d _{n}*

The n gauge wire diameter d_{n} in millimeters (mm) is equal to 0.127mm times 92 raised to the power of 36 minus gauge number n, divided by 39:

*d _{n}*

The n gauge wire's cross sercional area A_{n} in kilo-circular mils (kcmil) is equal to 1000 times the square wire diameter d in inches (in):

*A _{n}*

The n gauge wire's cross sercional area A_{n} in square inches (in^{2})
is equal to pi divided by 4 times the square wire diameter d in inches (in):

*A _{n}*

The n gauge wire's cross sercional area A_{n}
in square millimeters (mm^{2}) is equal to pi divided by 4 times the square wire diameter d in millimeters (mm):

*A _{n}*

The n gauge wire resistance R in ohms per kilofeet (Ω/kft) is equal to 0.3048×1000000000 times the wire's resistivity *ρ* in
ohm-meters (Ω·m) divided by 25.4^{2} times the cross sectional area *A _{n}* in square inches (in

*R*_{n (Ω/kft)} = 0.3048 × 10^{9} × *ρ*_{(Ω·m)} / (25.4^{2}
× *A _{n}*

The n gauge wire resistance R in ohms per kilometer (Ω/km) is equal to 1000000000 times the wire's resistivity *ρ* in
ohm-meters (Ω·m) divided by the cross sectional area *A _{n}* in square millimeters (mm

*R*_{n (Ω/km)} = 10^{9}
× *ρ*_{(Ω·m)} / *A _{n}*