This is a calculator for the estimation of the voltage drop of an electrical circuit based on the wire size, distance, and anticipated load current. The actual voltage drop can vary depend on the condition of the wire, the conduit being used, the temperature, the connector, the frequency etc. It is recommended that the voltage drop should be less than 5% under the fully loaded condition.

## Voltage drop calculator

* @ 68°F or 20°C

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

### Voltage drop calculations

#### DC / single phase calculation

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 calculation

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)})

#### 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 sercional 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*

_{n}<= 0.025 in

^{2}× 92

^{(36-n)/19.5}

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}*

_{ (in}2

_{)}= (π/4)×

*d*

_{n}

^{2}= 0.000019635 in

^{2}× 92

^{(36-n)/19.5}

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}*

_{ (mm}2

_{)}= (π/4)×

*d*

_{n}

^{2}= 0.012668 mm

^{2}× 92

^{(36-n)/19.5}

#### Wire resistance calculations

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

^{2}):

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

_{ (in}2

_{)})

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

^{2}):

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

_{ (mm}2

_{)}

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