Voltage drop is the decrease of electric potential along the path of a current flowing in an electrical circuit. Voltage drops in the internal resistance of the source, across conductors, across contacts, and across connectors are undesirable because some of the energy supplied is dissipated.

120-volt Single Phase Voltage Drop Table, Wire & Cable Chart

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*Calculated at 3% Maximum Voltage Drop

240-volt Single Phase Voltage Drop Table, Wire & Cable Chart

Voltage Drop Calculator

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Voltage Drop Formulas & Calculations

DC / single-phase voltage drop calculation   (dc voltage drop formula)

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))

3 phase voltage drop calculation

The line to line voltage drop V in volts (V) is equal to the 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 the 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))

Wire diameter calculations

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 (in) = 0.005 in × 92^(36-n)/39

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 (mm) = 0.127 mm × 92^(36-n)/39

Wire cross-sectional area calculations

The n gauge wire’s cross-sectional area A_n in kilo-circular mils (kcmil) is equal to 1000 times the square wire diameter d in inches (in):

A_n (kcmil) = 1000×d_n² = 0.025 in² × 92^(36-n)/19.5

The n gauge wire’s cross-sectional area A_n in square inches (in²) is equal to pi divided by 4 times the square wire diameter d in inches (in):

A_n (in2₎ = (π/4)×d_n² = 0.000019635 in² × 92^(36-n)/19.5

The n gauge wire’s cross-sectional area A_n in square millimeters (mm²) is equal to pi divided by 4 times the square wire diameter d in millimeters (mm):

A_n (mm2₎ = (π/4)×d_n² = 0.012668 mm² × 92^(36-n)/19.5

Wire resistance calculations   (voltage drop resistor calculated)

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 25.4² times the cross-sectional area A_n in square inches (in²):

R_n (Ω/kft) = 0.3048 × 10⁹ × ρ_(Ω·m) / (25.4² × A_n (in2₎)

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 (mm²):

R_n (Ω/km) = 10⁹ × ρ_(Ω·m) / A_n (mm2₎

Other wire & cable technical resources you may find useful

• MM2 to AWG
• MM2 to MCM
• How to Size a Cable according to the NEC

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