Voltage Drop Calculator
Calculate voltage drop in an AC or DC circuit given wire gauge, voltage, current, and length. You can also determine the correct conductor size and length for a circuit given an allowable voltage drop.
Voltage Drop:
In Volts:  0 volts

As a Percentage:  0%

End of Circuit Voltage:  0 volts

Conductor Diameter
inches:  0 in

millimeters:  0 mm

Conductor Crosssectional Area
kcmil:  0 kcmil

square inches:  0 in^{2}

square millimeters:  0 mm^{2}

Expected Voltage Drop
feet  
meters 
On this page:
How to Calculate Voltage Drop
Voltage drop is the amount of voltage lost in a circuit due to the resistance of the conductor. Voltage drop is an important consideration when planning a circuit to allow equipment using the circuit to run as designed.
Voltage drop in a circuit means less voltage is delivered to connected devices. Electrical devices consume power, and when they get reduced voltages, the current flow increases to meet their power needs. This current increase can overheat the circuit, causing damage to the equipment and devices.
Using Ohm’s Law, the voltage can be calculated by multiplying the current by the resistance.
V_{(V)} = I_{(A)} × R_{(Ω)}
Ohm’s Law states that the voltage is equal to the product of the current and the resistance. You can find the resistance for various conductor sizes in the chart below, or use one of the following voltage drop formulas.
Voltage Drop Formula
Thus, you can calculate voltage drop for singlephase AC and DC circuits using the following formula derived from Ohm’s Law:^{[1]}
V_{drop (V)} = 2 × I_{(A)} × ρ_{(Ω·m)} × L_{(m)} / A_{m²}
Where:
V = allowable voltage drop in volts
I = current in amps
ρ = conductor resistivity in ohmmeters (Ω·m)
L = conductor length in meters
A = conductor crosssectional area in square meters
So, the voltage drop V in volts is equal to twice the product of the current I in amps, the conductor resistivity ρ in ohmmeters, and the conductor length L in meters, divided by the crosssectional area of the conductor A in square meters.
The resistivity of copper conductors is 1.7241 × 10^{8} Ω·m and the resistivity of aluminum conductors is 2.6548 × 10^{8} Ω·m. Since copper has lower resistivity than aluminum, the voltage drop across a copper wire would be lower than an aluminum wire, for a given length and crosssectional area.
Alternate Voltage Drop Formula
You may also see voltage drop calculated using an alternate formula, which calculates the conductor resistance in a slightly different way:^{[2]}
V_{drop (V)} = I_{(A)} × 2 × K × L_{(m)} / A_{mm²}
Where:
V = allowable voltage drop in volts
K = direct current constant – use 12.9 for copper conductors and 21.2 for aluminum conductors. This is equal to the resistance of a conductor that is one thousand circular mils in area and one thousand feet in length.
I = current in amps
L = conductor length in meters
A = conductor crosssectional area in square millimeters
For example: let’s calculate the voltage drop of a 120V circuit drawing 15A using a 25′ long 14AWG (American Wire Gauge) copper conductor.
A 14 AWG wire has a crosssectional area of 2.081 mm² per the table below, and you can convert this to square meters by multiplying by 1,000,000. Recall that the resistivity of a copper conductor is 1.7241×10^{8} Ω·m. You can convert the length in feet to meters by multiplying by 0.3048.
V_{drop (V)} = 2 × 15A × 1.7241×10^{8} × (25 ft × 0.3048 m⁄ft) / 2.081 m² × 1,000,000
V_{drop (V)} = 1.894 V
So, we can expect a voltage drop of 1.894 volts in this circuit.
Voltage Drop Formula for ThreePhase Systems
The formula to calculate voltage drop in threephase systems is
V_{drop (V)} = √3 × I_{(A)} × ρ_{(Ω·m)} × L_{(m)} / A_{m²}
In a threephase system, the voltage drop V in volts is equal to the square root of 3 times the product of the current I in amps, the conductor resistivity ρ in ohmmeters, and the conductor length L in meters, divided by the crosssectional area of the conductor A in square meters.
How to Calculate the Minimum Conductor Size
Depending on the application, it is important to use conductors of the proper crosssectional area for economical and reliable power delivery. A large conductor would be a waste of money, and an undersized conductor might not be able to handle the required current flows.
Using the equation for voltage drop and rearranging the variables, the minimum conductor size in square meters for a circuit can be found using the following:
A_{(m²)} = 2 × ρ_{(Ω·m)} × L_{(m)} × I_{(A)} / V_{drop (V)}
Where:
A = crosssectional area in square meters
ρ = conductor resistivity in ohmmeters (Ω·m)
L = length in meters
I = current in amps
V_{drop} = allowable voltage drop in volts
The minimum crosssectional area A for a conductor in square meters is equal to twice the product of the conductor resistivity ρ in ohmmeters, the conductor length L in meters, and the current I in amps, divided by the maximum allowable voltage drop V in volts.
You can multiply the resulting crosssectional area by 1,000,000 to calculate the area in square millimeters. Then, use our wire size calculator to find the wire gauge with the correct crosssectional area.
You’ll also need to consider the current rating when determining the minimum conductor size. The National Electric Code has several tables to determine the maximum allowable current for a conductor. You can also use our wire ampacity calculator to calculate the maximum current rating for a conductor.
You should always consult with an electrician or electrical professional when choosing your wire to ensure it meets the code requirements and will work safely for your application.
For example: let’s find the minimum wire gauge needed for a 120V circuit drawing 20A using a 40′ long copper conductor with a max voltage drop of 3%.
A 3% voltage drop would be 3.6 volts. Recall that the resistivity of a copper conductor is 1.7241×10^{8} Ω·m. You can convert the length in feet to meters by multiplying by 0.3048.
A_{(m²)} = 2 × 1.7241×10^{8} × (40 ft × 0.3048 ft⁄m) × 20A / 3.6V
A_{(m²)} = 2 × 1.7241×10^{8} × 12.192 m × 20A / 3.6V
A_{(m²)} = 8.408×10^{6} / 3.6V
A_{(m²)} = 2.3356×10^{6}m²
Then, multiply by 1,000,000 to convert the crosssectional area in square meters to square millimeters.
A_{(mm²)} = 2.3356×10^{6} m² × 1,000,000
A_{(mm²)} = 2.3356 mm²
Finally, refer to the table below to find the minimum wire gauge for this crosssectional area. From the table, the wire gauge required for a crosssectional area of 2.081 mm² is 14 AWG and the wire gauge required for a crosssectional area of 2.624 mm² is 13 AWG. So, the minimum wire gauge required for our example (crosssectional area of 2.3356 mm²) is 13 AWG.
How to Calculate the Maximum Conductor Length
It is essential to use conductors of proper lengths to ensure minimal voltage drop in the conductive element. A longer conductor means the current must travel a longer distance through the conductive material and against more resistance.
This increase in resistance produces more voltage drops across the length of the circuit and causes it to overheat.
The maximum length of a conductor in a circuit can be determined by rewriting the formula for voltage drop like this:
L_{(m)} = V_{drop (V)} × A_{m²} / 2 × ρ_{(Ω·m)} × I_{(A)}
Where:
V_{drop} = allowable voltage drop in volts
I = current in amps
ρ = conductor resistivity in ohmmeters (Ω·m)
L = conductor length in meters
A = conductor crosssectional area in square meters
The maximum conductor length is equal to the product of the maximum allowable voltage drop V_{drop} in volts and the crosssectional area of the conductor in square meters, divided by twice the product of the conductor resistivity ρ in ohmmeters and the current I in amps.
Conductor Size and Resistance Table
AWG  CrossSectional Area  Resistance  

(kcmil)  (mm²)  Ohms per 1000ft  Ohms per 1000m  
0000 (4/0)  211.6  107.22  0.049  0.1608 
000 (3/0)  167.81  85.029  0.0618  0.2028 
00 (2/0)  133.08  67.431  0.0779  0.2557 
0 (1/0)  105.53  53.475  0.0983  0.3224 
1  83.693  42.408  0.1239  0.4066 
2  66.371  33.631  0.1563  0.5127 
3  52.635  26.67  0.197  0.6464 
4  41.741  21.151  0.2485  0.8152 
5  33.102  16.773  0.3133  1.028 
6  26.251  13.302  0.3951  1.296 
7  20.818  10.549  0.4982  1.634 
8  16.51  8.366  0.6282  2.061 
9  13.093  6.634  0.7921  2.599 
10  10.383  5.261  0.9988  3.277 
11  8.234  4.172  1.26  4.132 
12  6.53  3.309  1.588  5.211 
13  5.178  2.624  2.003  6.571 
14  4.107  2.081  2.525  8.285 
15  3.257  1.65  3.184  10.448 
16  2.583  1.309  4.015  13.174 
17  2.048  1.038  5.063  16.612 
18  1.624  0.823  6.385  20.948 
19  1.288  0.6527  8.051  26.415 
20  1.022  0.5176  10.152  33.308 
21  0.8101  0.4105  12.802  42.001 
22  0.6424  0.3255  16.143  52.962 
23  0.5095  0.2582  20.356  66.784 
24  0.404  0.2047  25.668  84.213 
25  0.3204  0.1624  32.367  106.19 
26  0.2541  0.1288  40.814  133.9 
27  0.2015  0.1021  51.466  168.85 
28  0.1598  0.081  64.897  212.92 
29  0.1267  0.0642  81.833  268.48 
30  0.1005  0.0509  103.19  338.55 
31  0.0797  0.0404  130.12  426.9 
32  0.0632  0.032  164.08  538.32 
33  0.0501  0.0254  206.9  678.8 
34  0.0398  0.0201  260.9  855.96 
35  0.0315  0.016  328.98  1,079.3 
36  0.025  0.0127  414.84  1,361 
37  0.0198  0.01  523.1  1,716.2 
38  0.0157  0.007967  659.62  2,164.1 
39  0.0125  0.006318  831.77  2,728.9 
40  0.009888  0.00501  1,048.8  3,441.1 
Also check out our electricity cost calculator to see how much it will cost to power a device.
References
 LibreTexts  Physics, University Physics II  Thermodynamics, Electricity, and Magnetism  9.4: Resistivity and Resistance, https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_II__Thermodynamics_Electricity_and_Magnetism_(OpenStax)/09%3A_Current_and_Resistance/9.04%3A_Resistivity_and_Resistance
 IAEI Magazine, Voltage Drop Formulas, https://iaeimagazine.org/electricalfundamentals/voltagedropformulas/