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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.
*** For wire length of 2x10ft, wire length should be 10ft.
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:
Vdrop (V) = Iwire (A) × Rwire(Ω)
= Iwire (A) × (2 × L(ft) × Rwire(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:
Vdrop (V) = Iwire (A) × Rwire(Ω)
= Iwire (A) × (2 × L(m) × Rwire (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:
Vdrop (V) = √3 × Iwire (A) × Rwire (Ω)
= 1.732 × Iwire (A) × (L(ft) × Rwire (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:
Vdrop (V) = √3 × Iwire (A) × Rwire (Ω)
= 1.732 × Iwire (A) × (L(m) × Rwire (Ω/km) / 1000(m/km))
The n gauge wire diameter dn in inches (in) is equal to 0.005in times 92 raised to the power of 36 minus gauge number n, divided by 39:
dn (in) = 0.005 in × 92(36-n)/39
The n gauge wire diameter dn in millimeters (mm) is equal to 0.127mm times 92 raised to the power of 36 minus gauge number n, divided by 39:
dn (mm) = 0.127 mm × 92(36-n)/39
The n gauge wire's cross sercional area An in kilo-circular mils (kcmil) is equal to 1000 times the square wire diameter d in inches (in):
An (kcmil) = 1000×dn2 = 0.025 in2 × 92(36-n)/19.5
The n gauge wire's cross sercional area An in square inches (in2) is equal to pi divided by 4 times the square wire diameter d in inches (in):
An (in2) = (π/4)×dn2 = 0.000019635 in2 × 92(36-n)/19.5
The n gauge wire's cross sercional area An in square millimeters (mm2) is equal to pi divided by 4 times the square wire diameter d in millimeters (mm):
An (mm2) = (π/4)×dn2 = 0.012668 mm2 × 92(36-n)/19.5
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 An in square inches (in2):
Rn (Ω/kft) = 0.3048 × 109 × ρ(Ω·m) / (25.42 × An (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 An in square millimeters (mm2):
Rn (Ω/km) = 109 × ρ(Ω·m) / An (mm2)
| AWG # | Diameter (inch) |
Diameter (mm) |
Area (kcmil) |
Area (mm2) |
|---|---|---|---|---|
| 0000 (4/0) | 0.4600 | 11.6840 | 211.6000 | 107.2193 |
| 000 (3/0) | 0.4096 | 10.4049 | 167.8064 | 85.0288 |
| 00 (2/0) | 0.3648 | 9.2658 | 133.0765 | 67.4309 |
| 0 (1/0) | 0.3249 | 8.2515 | 105.5345 | 53.4751 |
| 1 | 0.2893 | 7.3481 | 83.6927 | 42.4077 |
| 2 | 0.2576 | 6.5437 | 66.3713 | 33.6308 |
| 3 | 0.2294 | 5.8273 | 52.6348 | 26.6705 |
| 4 | 0.2043 | 5.1894 | 41.7413 | 21.1506 |
| 5 | 0.1819 | 4.6213 | 33.1024 | 16.7732 |
| 6 | 0.1620 | 4.1154 | 26.2514 | 13.3018 |
| 7 | 0.1443 | 3.6649 | 20.8183 | 10.5488 |
| 8 | 0.1285 | 3.2636 | 16.5097 | 8.3656 |
| 9 | 0.1144 | 2.9064 | 13.0927 | 6.6342 |
| 10 | 0.1019 | 2.5882 | 10.3830 | 5.2612 |
| 11 | 0.0907 | 2.3048 | 8.2341 | 4.1723 |
| 12 | 0.0808 | 2.0525 | 6.5299 | 3.3088 |
| 13 | 0.0720 | 1.8278 | 5.1785 | 2.6240 |
| 14 | 0.0641 | 1.6277 | 4.1067 | 2.0809 |
| 15 | 0.0571 | 1.4495 | 3.2568 | 1.6502 |
| 16 | 0.0508 | 1.2908 | 2.5827 | 1.3087 |
| 17 | 0.0453 | 1.1495 | 2.0482 | 1.0378 |
| 18 | 0.0403 | 1.0237 | 1.6243 | 0.8230 |
| 19 | 0.0359 | 0.9116 | 1.2881 | 0.6527 |
| 20 | 0.0320 | 0.8118 | 1.0215 | 0.5176 |
| 21 | 0.0285 | 0.7229 | 0.8101 | 0.4105 |
| 22 | 0.0253 | 0.6438 | 0.6424 | 0.3255 |
| 23 | 0.0226 | 0.5733 | 0.5095 | 0.2582 |
| 24 | 0.0201 | 0.5106 | 0.4040 | 0.2047 |
| 25 | 0.0179 | 0.4547 | 0.3204 | 0.1624 |
| 26 | 0.0159 | 0.4049 | 0.2541 | 0.1288 |
| 27 | 0.0142 | 0.3606 | 0.2015 | 0.1021 |
| 28 | 0.0126 | 0.3211 | 0.1598 | 0.0810 |
| 29 | 0.0113 | 0.2859 | 0.1267 | 0.0642 |
| 30 | 0.0100 | 0.2546 | 0.1005 | 0.0509 |
| 31 | 0.0089 | 0.2268 | 0.0797 | 0.0404 |
| 32 | 0.0080 | 0.2019 | 0.0632 | 0.0320 |
| 33 | 0.0071 | 0.1798 | 0.0501 | 0.0254 |
| 34 | 0.0063 | 0.1601 | 0.0398 | 0.0201 |
| 35 | 0.0056 | 0.1426 | 0.0315 | 0.0160 |
| 36 | 0.0050 | 0.1270 | 0.0250 | 0.0127 |
| 37 | 0.0045 | 0.1131 | 0.0198 | 0.0100 |
| 38 | 0.0040 | 0.1007 | 0.0157 | 0.0080 |
| 39 | 0.0035 | 0.0897 | 0.0125 | 0.0063 |
| 40 | 0.0031 | 0.0799 | 0.0099 | 0.0050 |
Voltage drop is an essential concept in both residential and industrial electrical installations. When current flows through a conductor, a voltage drop occurs due to the resistance of the wire. In long circuits or circuits carrying heavy loads, this drop can become significant, leading to poor performance or even equipment damage. Our online Voltage Drop Calculator is designed to help electricians, engineers, and hobbyists determine voltage loss in wiring, ensuring safe and efficient electrical design.
This in-depth guide explains the theory, formulas, causes, and practical consequences of voltage drop, along with examples and usage scenarios of our Voltage Drop Calculator. By the end, you’ll have a thorough understanding of how to manage voltage drops in your circuits effectively.
Voltage drop refers to the reduction in voltage as electrical current moves through the resistive elements of a circuit—typically the conductors themselves. This is due to the fact that all conductors have some inherent resistance, however small.
For DC and single-phase AC circuits:
Vdrop = I × R
Where:
For three-phase circuits:
Vdrop = √3 × I × R
Resistance is calculated as:
R = ρ × (2 × L) / A
For copper: ρ ≈ 1.724 × 10−8 Ω·m
For aluminum: ρ ≈ 2.65 × 10−8 Ω·m
Our calculator will also estimate power loss and final load voltage based on the input voltage.
National and international electrical codes specify acceptable limits:
Staying within these limits ensures energy efficiency and safety.
Here is a sample table of common AWG sizes with corresponding resistance per 1000 feet:
These values are essential when calculating R in the voltage drop formula.
Let’s say you’re running a 120V light circuit over a distance of 150 feet using 12 AWG copper wire.
Total resistance = (1.588 / 1000) × 300 = 0.4764 ohms
Voltage drop = 10 × 0.4764 = 4.764 V
Percentage drop = (4.764 / 120) × 100 = 3.97%
An electric vehicle charger rated for 240V at 30A is placed 100 feet from the panel using 8 AWG aluminum wire.
R = (0.778 / 1000) × 200 = 0.1556 ohms
Voltage drop = 30 × 0.1556 = 4.668V
Percent drop = (4.668 / 240) × 100 ≈ 1.95%
Well within acceptable range for such a heavy load.
DC circuits are especially sensitive to voltage drop, especially in automotive, solar, and battery-powered systems. Every volt counts when working at low voltages (e.g., 12V or 24V).
Rule of thumb: For low-voltage systems, keep the voltage drop under 2%.
Three-phase systems are common in industrial and commercial setups. Voltage drop calculations must include the √3 multiplier:
Vdrop = √3 × I × R
Underground circuits often require longer runs, increasing the risk of voltage drop. You may also need to derate due to temperature or conduit fill. Our calculator allows input adjustments for these conditions.
Excess voltage drop means wasted power. Power loss in the cable is:
Ploss = I² × R
Even small drops in high-current systems can result in significant power losses over time, reducing overall efficiency.
LED strips and drivers are very sensitive to voltage drops, especially in 12V and 24V systems. Use minimal lengths or heavier gauge wires to prevent brightness inconsistencies.
For each circuit during planning, particularly when:
Yes. It can lead to overheating, equipment malfunction, or even fire in extreme cases. It is both a performance and safety concern.
Yes, but you must use a larger gauge than copper and account for higher resistance. Connections should be made with approved anti-oxidation compounds.
240V results in less current for the same power, thereby reducing voltage drop. Higher voltages are more efficient for long cable runs.
Absolutely. Dim lights and sluggish appliances are common symptoms of voltage drop, especially in overloaded or poorly designed circuits.
Voltage drop is a critical design consideration in any electrical system. It affects performance, safety, and energy efficiency. Whether you're working on a small LED lighting project or a large commercial installation, accounting for voltage drop ensures your equipment operates within safe and optimal parameters.
Use our Voltage Drop Calculator to quickly and accurately estimate voltage losses across your wiring. Simply input a few parameters and let the tool do the math for you—saving time, preventing errors, and ensuring code compliance. With the right cable sizing and design strategies, you can easily keep voltage drop within acceptable limits.
Make voltage drop calculation a regular part of your electrical planning and prevent surprises down the road. Try our calculator now to streamline your next installation.