**Advance Voltage Drop Calculator and Voltage Drop Formula.**

According to the NEC (National Electric Code) [210.19 A (1)] FPN number 4 and [215.2 A (3)] FPN number 2, the allowable Voltage drop for feeders is 3% and the acceptable voltage drop for final sub circuit and branch circuit is 5% for proper and efficient operation.

For example, if the Supply voltage is 110V, then the value of allowable voltage drop should be;

Allowable Voltage Drop = 110 x (3/100) = 3.3V.

We have already discussed the selection of proper size of cable for electrical wiring installation in SI and British system with examples. In the above article, we have also explained the voltage drop calculation and voltage drop formula and online cable size calculator.

Today, we are going to share Online Advance Voltage Drop Calculator and Voltage Drop formulas in detail.

*Good to know*: Read the full description below the voltage drop calculator for better explanation as there are many voltage drop calculation formulas with example. in addition, there is also a

**very simple method to calculate the voltage drop**.

→Also check

Enter value and click on calculate. Result will be displayed |

Conductor Temp: | ||

Select Wire Type: | ||

Enter Wire Dia size: | ||

Enter Wire Length: | ||

Select Current Type: | ||

Enter Voltage in Volts: | V | |

Enter Current in Amps: | A | |

Voltage Drop in Volts: | V | |

Percentage of Voltage Drop: | % | |

Wire Resistance: | Ω | |

Voltage at the End: | Ve |

**Voltage Drop Formulas and Voltage Drop Calculation.**

**1.**

The basic electrical Voltage drop formula is;

V

_{D}= IR ……. (Ohm’s Law).Where;

V

_{D}= Voltage Drop in Volts.I = Current in Amperes.

R = Resistance in Ohms (Ω).

But this is not always the case, and we can’t run the wheel of the system with this basic formulaJ.

**2.**

This is the approximate voltage drop formula at unity power factor, cabal temperature 75˚C, and cable conductors in steel conduit.

**V**for Single Phase._{D}= (2xkxQxIxD) / cm**V**for Three Phase._{D}= (1.732xkxQxIxD) / cm

Where;

Cm = the cross sectional area of the conductor in circular mils.

D = the one way distance in feet.

I = the circuit current in amperes.

Q = the ratio of AC resistance and DC resistance (R

_{AC}/R/_{DC}) for conductor larger than 2/0 for skin effect.k = specific resistivity = 21.2 for Aluminum and 12.9 for Copper.

**3.**

**Voltage Drop formula for Single Phase and DC Circuits.**

** 1) When wire length is in feet.**

V

_{D}= I × RV

_{D}= I × (2 × L × R / 1000)Where ;

V

_{D}= Voltage Drop in Volts.I = Wire Current in Amperes.

R = Wire Resistance in Ohms (Ω) [Ω/kft].

L = wire distance in feet.

And;

** 2) When Wire length is in meters.**

V

_{D}= I × (2 × L × R / 1000)Where;

V

_{D}= Voltage Drop in Volts.I = Wire Current in Amperes.

R = Wire Resistance in Ohms (Ω) [Ω/km].

L = wire distance in meters.

**4.**

**Voltage Drop Calculation and formula for Three Phase system.**

**For 3 Phase 3 Wire system.**

*V*= 0.866 ×

_{D }*I*×

*R*

*V*= 0.866 ×

_{D }*I*× 2 ×

*L*×

*R*/ 1000

**For 3 Phase 4 Wire systems.**

*V*

_{D }*=*0.5

*×*

*I*×

*R*

*V*

_{D }*=*0.5

*×*

*I*× 2 ×

*L*×

*R*/ 1000

Where;

V

_{D}= Voltage Drop in Volts.I = Wire Current in Amperes.

R = Wire Resistance in Ohms (Ω) [Ω/km or] or (Ω/kft).

L = wire distance in meters or feet.

**5.**

**Wire cross sectional area calculations**

**Wire cross sectional area in kcmil (kilo circular mils)**

**A**_{n}**= 1000×**

*d*_{n}^{2}= 0.025 × 92^{(36-n)/19.5}Where;

An = cross sectional area of “n” gauge wire size in kcmil.

kcmil = kilo circular mils.

n = the number of gauge size.

d = wire square diameter in in

^{2}.**Wire cross sectional area in****square inches (in**^{2}).

** A**_{n} *= ***(π/4)***×*d_{n}^{2} = 0.000019635* × *92^{(}^{36–n)/19.5}

**A**

_{n}*=*

*×*d

_{n}^{2}= 0.000019635

*×*92

^{(}^{36–n)/19.5}

Where;

An = cross sectional area of “n” gauge wire size in square inches (in

^{2}).n = the number of gauge size.

d = wire square diameter in in

^{2}.**Wire cross sectional area in kcmil (kilo circular mils)**

* A*_{n} = (π/4) ×d_{n}^{2} = 0.012668 × 92^{(36-n)/19.5}

*A*= (π/4) ×d

_{n}

_{n}^{2}= 0.012668 × 92

^{(36-n)/19.5}

Where;

An = cross sectional area of “n” gauge wire size in square millimeters (mm

^{2})n = the number of gauge size.

d = wire square diameter in mm

^{2}.**6**

**Wire diameter Calculation**

**Wire diameter in Inches formula**

**d**_{n}**= 0.005 × 92**

^{(36-n)/39}…. In inchesWhere “n” is number of the gauge size and “d” the wire diameter in inches.

**Wire diameter in mm (millimeters) formula**

**d**_{n}**= 0.127 × 92**

^{(36-n)/39}…. In millimeters (mm).Where “n” is number of the gauge size and “d” the wire diameter in mm.

**7**

**Wire resistance calculations formula**

*(1).*

**R**_{n }**= 0.3048 × 10**

^{9}×*ρ*/ (25.4^{2}×*A*)_{n}Where;

R = Resistance of the wire conductors (in Ω/kft).

n = # of Gauge size.

ρ = rho = resistivity in (Ω·m).

An = the cross sectional area of n #gauge in square inches (in

^{2}).*Or;*

*(2).*

**R**_{n }**= 10**

^{9}×*ρ*/*A*_{n}Where;

R = Resistance of the wire conductors (in Ω/km).

n = # of Gauge size.

ρ = rho = resistivity in (Ω·m).

An = the cross sectional area of n #gauge in square millimeters (mm

^{2}).**8.**

**Voltage at the End of the cable formula and calculation.**

**V**

_{End}= V – V_{D}Where;

V

_{End}= Supply voltage at the end of the cable.V = Supply voltage.

V

_{D}= Voltage Drop in the cable conductors.**9.**

**Voltage Drop Calculation formula for Circular milsCircular mils and Voltage drop**

**V**

_{D}= ρ P L I / AWhere;

V

_{D}= Voltage drop in voltsρ = rho = specific resistivity in (

*Ω – circular mils/foot*).P = Phase Constant = 2 (for single phase and dc system) = √3 = 1.732 (for three phase system)

L = wire length in feet.

A = wire area in circular mils

**10**

**Easy and simple method to calculate voltage drop in copper conductor (1 & 3 phase)**

The voltage drop in copper conductors can be calculated via the below simple and easy formula with the help of the following table .

**V**

_{D}= f x I … L = 100 feetWhere;

f = factor from the table below.

I = Current in amperes.

L = conductor length in feet (100 feet) .

(See the solved example below the table for better understanding)

**Example**: Suppose, Voltage is 220V single phase, current is 5A, Conductor length is 100 feet, and Wire gauge (AWG) is #8. Calculate the voltage drop.

**Solution:**

The voltage drop can be found by the following formula

VD = f x I … L = 100 feet

Since the factor for #8 AWG conductor is 0.125 (from the above table). Now, putting the values in the above formula.

VD = 0.125 x 5A x ( for 100 feet)

VD = Voltage drop = 0.625V.

**P.S:**The above voltage drop calculator provides approximate values and we do not guarantee of 100% accurate results as results may change with different and real cables, conductors, wires and different resistivity of material, number of strands in wire, temperature and weather condition, conduit and PVC etc .

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Really great detailed post, thanks for that!

thank you very muh

Update any new formula please informed me on my e – mail Id

thank you very much

The voltage drop formula must include the reactance impedance as well. Too many design software programs omit voltage drop and other safety code requirements, resulting in design deficiencies and fiascos like the Apollo 13 disaster. Contractors make multiple profits over eir original bid with change orders.