Enter the wiring system, the wire size (or its resistance per length), the source voltage, the one-way length and the current. The voltage drop, the percent drop and the voltage at the load end are calculated.
"From wire size" uses a simplified formula for copper wire only (resistance of copper about 0.0178 Ω per meter for 1 mm²). For aluminum and other wire, choose "from wire resistance" and use the conductor resistance from the specs. All values must be numbers greater than 0.
Result and graph
Enter the wiring system, wire size, one-way length, current and other values on the left and press "Calculate". The result and a graph will appear here.
Table of Contents
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What you can do on this page
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What is this calculation used for?
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How to Use
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Formulas and figures
-
Symbols and terms
-
Good to know before you start
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How to calculate it in Excel
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How to calculate it in Google Sheets
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How to calculate it in Python
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How to write it in LaTeX and other math languages (copy and paste)
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How to have ChatGPT do the calculation
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DataChef Features
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Related Features
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NumberChef Calculators List
What you can do on this page
- From the wire size (AWG, or the area in circular mils), the one-way length and the current, find the voltage drop (V), the percent voltage drop (%) and the voltage at the load end, all at once
- Uses the standard US method for copper wire, \(e = \dfrac{2 K L I}{A}\) for a single-phase circuit, with \(K = 12.9\) Ω·cmil/ft. Single-phase 2-wire, single-phase 3-wire, three-phase 3-wire, three-phase 4-wire and DC are supported
- Just pick the wire size from the "common wire sizes" list (14 AWG to 500 kcmil) and the area in circular mils is filled in for you
- For aluminum or special wire, calculate directly from the conductor resistance in the specs (Ω/1000 ft, Ω/ft or mΩ/ft) with the "from wire resistance" mode. Two to four parallel conductors are also supported
- Along with the result, a graph shows how the voltage drop grows with length (with a guide line at 5% of the source voltage)
- To work in mm² and meters, switch "Units" above the calculator to Metric (the metric side uses the simplified copper formula \(e = \dfrac{35.6 L I}{1000 A}\))
This calculator is a guide for learning and for planning. The US National Electrical Code (NEC) recommends, in informational notes rather than as a requirement, a voltage drop of no more than 3% on a branch circuit and no more than 5% in total for the feeder and branch circuit together. Electrical work on a building's wiring should be done by a licensed electrician, following local codes and permits.
What is this calculation used for?
Preparing for electrician licensing exams (study)
"What is the voltage drop on a 120 V circuit with 12 AWG copper, a one-way length of 100 ft and a load of 16 A?" Voltage drop problems like this come up again and again on journeyman and master electrician exams. The answer is \(2 \times 12.9 \times 16 \times 100 \div 6530 \approx 6.32\) V, about 5.27%, which is over the recommended 3%. Going up to 10 AWG gives about 3.98 V (3.31%), and 8 AWG gives about 2.50 V (2.08%).
Working through a few versions by hand builds a feel for the structure of the formula: the drop is proportional to the length and the current, and inversely proportional to the wire area.
A first estimate of the wire size for a long run (electrical design)
When you feed a detached garage or workshop far from the panel, the voltage drop can get too large if the wire stays small. For example, a 240 V, 30 A circuit run 150 ft one way with 10 AWG copper drops about 11.2 V (4.66%). With 6 AWG copper, the drop is about 4.42 V (1.84%).
Real designs also check ampacity, the installation conditions and local codes, but this formula is what people use to get a first idea of whether a wire size is enough for a given distance and current.
Low-voltage DC wiring for solar and batteries
In 12 V and 24 V systems for RVs, boats and solar setups, voltage drop matters much more than in 120 V house wiring. For example, 10 AWG wire run 10 ft one way at 20 A drops about 0.50 V, which is about 4.1% of 12 V (on a 120 V circuit the same drop would be only about 0.41%).
You can see in numbers the golden rule of DC wiring: the lower the voltage, the thicker the wire you need.
Why long extension cords lower the voltage
Run 10 A (a device of about 1200 W) through a 50 ft extension cord with 16 AWG wire (2,580 cmil), and the voltage drop is \(2 \times 12.9 \times 10 \times 50 \div 2580 = 5.0\) V. Even with 120 V at the outlet, the device gets only about 115 V. This is why power tools and heaters lose power on long, thin cords.
A thin cord carrying a large current can also overheat, so use a heavier cord (a lower AWG number) for high-current tools and keep cords as short as you can.
Planning a 240 V circuit for EV charging
Installing a 240 V circuit for a Level 2 EV charger often involves a long run from the panel to the garage or driveway. For example, a 40 A charger fed with 6 AWG copper over a one-way length of 75 ft has a drop of about 2.95 V (1.23%), well within 3%.
The work itself is a job for a licensed electrician, but this calculation helps you understand a quote or compare where to put the charger.
Formulas and figures
Voltage drop formula (single-phase 2-wire and DC)
Figure
Standard notation
(the usual math form)
\(e\)
\(=\)
\(2\)
\(\times\)
\(K\)
\(\times\)
\(L\)
\(\times\)
\(I\)
\(\div\)
\(A\)
In words
(symbols replaced with words)
⑥
\(e\): voltage drop
\(=\)
①
multiplier 2
\(\times\)
②
\(K\): 12.9 for copper
\(\times\)
③
\(L\): one-way length
\(\times\)
④
\(I\): current
\(\div\)
⑤
\(A\): wire area
The formula in words
①
For single-phase 2-wire and DC, take the
multiplier 2 (out and back)
②
multiply it by the
\(K\): 12.9 Ω·cmil/ft for copper
③
the
\(L\): one-way length (ft)
④
and the
\(I\): current (A)
⑤
divide by the
\(A\): wire area (cmil)
⑥
and you get the
\(e\): voltage drop (V)
Quick example
For a 120 V single-phase circuit wired with 12 AWG copper (6,530 cmil), a one-way length of 50 ft and a current of 15 A, the voltage drop is
\(e\): voltage drop
\(=\)
multiplier 2
\(\times\)
\(K\) (12.9)
\(\times\)
length (50 ft)
\(\times\)
current (15 A)
\(\div\)
area (6,530 cmil)
\(2 \times 12.9 \times 50 \times 15 = 19350\)
\(19350 \div 6530 \approx 2.96\)
Key idea
\(K = 12.9\) is the resistance, in ohms, of a copper wire 1 ft long with an area of 1 circular mil (cmil), at a conductor temperature of 75°C. It is the value used for copper in the NEC Handbook (for aluminum, \(K = 21.2\)). The circular mil is the US unit for wire area: a round wire 1 mil (0.001 in) across has an area of 1 cmil, and the AWG sizes are listed in cmil in NEC Chapter 9, Table 8 (12 AWG = 6,530 cmil). This formula only looks at the resistance of the wire (power factor 1, no reactance), so for motor loads with a low power factor or very large wires, the real voltage drop can differ a little.
Enter \(L\) as the one-way length of the run. The "2" in front already covers the wire going out and the wire coming back.
If you switch "Units" to Metric, the same idea is written as \(e = \dfrac{35.6 L I}{1000 A}\), with \(L\) in meters and \(A\) in mm² (35.6 = 2 × 17.8, where 17.8 comes from the resistance of copper, about 0.0178 Ω per meter for 1 mm², near room temperature).
Voltage drop for three-phase 3-wire and single-phase 3-wire (three-phase 4-wire)
Figure
Standard notation
(the usual math form)
\(e\)
\(=\)
\(\sqrt{3}\)
\(\times\)
\(K\)
\(\times\)
\(L\)
\(\times\)
\(I\)
\(\div\)
\(A\)
\(e\)
\(=\)
\(1\)
\(\times\)
\(K\)
\(\times\)
\(L\)
\(\times\)
\(I\)
\(\div\)
\(A\)
In words
(symbols replaced with words)
②
\(e\): voltage drop
\(=\)
①
multiplier \(\sqrt{3}\)
\(\times\)
\(K\): 12.9 for copper
\(\times\)
\(L\): one-way length
\(\times\)
\(I\): current
\(\div\)
\(A\): wire area
④
\(e\): voltage drop
\(=\)
③
multiplier 1
\(\times\)
\(K\): 12.9 for copper
\(\times\)
\(L\): one-way length
\(\times\)
\(I\): current
\(\div\)
\(A\): wire area
The formula in words
①
For three-phase 3-wire, change the multiplier to
multiplier \(\sqrt{3}\)
(\(K\), the length \(L\), the current \(I\) and the area \(A\) are used the same way as in the single-phase formula). This gives the
②
\(e\): voltage drop (line to line)
③
For single-phase 3-wire and three-phase 4-wire (balanced load), change the multiplier to
multiplier 1
instead. This gives the
④
\(e\): voltage drop (line to neutral)
Quick example
For a 208 V three-phase 3-wire circuit wired with 8 AWG copper (16,510 cmil), a one-way length of 100 ft and a current of 30 A, the voltage drop is
\(e\): voltage drop
\(=\)
multiplier \(\sqrt{3}\)
\(\times\)
\(K\) (12.9)
\(\times\)
length (100 ft)
\(\times\)
current (30 A)
\(\div\)
area (16,510 cmil)
\(\sqrt{3} \times 12.9 \times 100 \times 30 \approx 67030\)
\(67030 \div 16510 \approx 4.06\)
Key idea
The three formulas differ only in the multiplier. The drop in one wire is \(K L I \div A\). Single-phase 2-wire and DC count both the outgoing and the returning wire, so the multiplier is 2. Three-phase 3-wire gives the drop between two lines, which is \(\sqrt{3} \approx 1.732\) times the drop in one wire. For single-phase 3-wire and three-phase 4-wire, measured between a line wire and the neutral, no current flows in the neutral when the load is balanced, so only one wire counts and the multiplier is 1. In the example, the drop of about 4.06 V is about 1.95% of 208 V.
Formula from the wire resistance
Figure
Standard notation
(the usual math form)
\(e\)
\(=\)
\(M\)
\(\times\)
\(I\)
\(\times\)
\(R\)
\(\times\)
\(L\)
In words
(symbols replaced with words)
⑤
\(e\): voltage drop
\(=\)
①
\(M\): system multiplier
\(\times\)
②
\(I\): current
\(\times\)
③
\(R\): resistance per ft
\(\times\)
④
\(L\): one-way length
The formula in words
①
Take the
\(M\): multiplier for the wiring system
②
multiply it by the
\(I\): current (A)
③
the
\(R\): wire resistance per foot (Ω/ft)
④
and the
\(L\): one-way length (ft)
⑤
and you get the
\(e\): voltage drop (V)
Quick example
For a single-phase 2-wire circuit (\(M = 2\)) wired with 6 AWG aluminum, whose resistance is 0.808 Ω/1000 ft (= 0.000808 Ω/ft), a one-way length of 150 ft and a current of 40 A, the voltage drop is
\(e\): voltage drop
\(=\)
multiplier (2)
\(\times\)
current (40 A)
\(\times\)
resistance (0.000808 Ω/ft)
\(\times\)
length (150 ft)
\(2 \times 40 \times 0.000808 \times 150 = 9.696\)
Key idea
The multiplier \(M\) depends on the wiring system: \(M = 2\) for DC 2-wire and single-phase 2-wire (the wire out and the wire back), \(M = \sqrt{3} \approx 1.73\) for three-phase 3-wire, and \(M = 1\) for single-phase 3-wire and three-phase 4-wire (between a line wire and the neutral, balanced load).
This formula is Ohm's law \(V = I \times R\) itself, so it works for any wire, copper or aluminum, as long as you know the conductor resistance from the specs (NEC Chapter 9, Table 8 lists it per 1000 ft). When \(N\) identical wires are run in parallel, divide the resistance \(R\) by \(N\).
Percent voltage drop
Figure
Standard notation
(the usual math form)
\(\varepsilon\)
\(=\)
\(e\)
\(\div\)
\(V\)
\(\times\)
\(100\)
In words
(symbols replaced with words)
④
\(\varepsilon\): percent drop
\(=\)
①
\(e\): voltage drop
\(\div\)
②
\(V\): source voltage
\(\times\)
③
constant 100
The formula in words
①
Take the
\(e\): voltage drop (V)
②
divide it by the
\(V\): source voltage (V)
③
multiply by the
constant 100
to make it a percentage,
④
and you get the
\(\varepsilon\): percent voltage drop (%)
Quick example
For a 120 V circuit with a voltage drop of 2.96 V, the percent voltage drop is
\(\varepsilon\): percent drop
\(=\)
voltage drop (2.96 V)
\(\div\)
source voltage (120 V)
\(\times\)
constant 100
\(2.96 \div 120 \times 100 \approx 2.47\ \ (2.47\%)\)
Key idea
Voltage drop is judged by "what percent of the source voltage was lost" rather than by "how many volts were lost". A 1 V drop is less than 1% on a 120 V circuit but more than 8% on a 12 V battery circuit, so the effect is completely different.
The NEC recommends, in informational notes, no more than 3% on a branch circuit and no more than 5% in total for the feeder and branch circuit together. The example above, at about 2.47%, is within 3%. Follow local codes and the equipment maker's instructions in real designs.
Voltage drop is "multiplier × K × one-way length × current ÷ wire area", with \(K = 12.9\) Ω·cmil/ft for copper and the area in circular mils. The multiplier depends on the wiring system (2 for single-phase 2-wire and DC, \(\sqrt{3}\) for three-phase 3-wire, 1 for single-phase 3-wire and three-phase 4-wire). The keys are to enter the one-way length and to judge the result as a percentage of the source voltage (the NEC recommends 3% or less for a branch circuit).
Symbols and terms
Symbols
| \(e\) | e | Voltage drop (V). The voltage lost between the source and the load because of the resistance of the wire itself. It is also often written VD. |
| \(L\) | L | One-way length (ft). The length of the run from the source to the load, one way only. The multiplier in the formula already covers the two wires, out and back. |
| \(I\) | I | Current (A). The load current flowing in the circuit. |
| \(A\) | A | The cross-sectional area of the wire (conductor) in circular mils (cmil); in mm² with Metric units. The larger the area, the lower the resistance and the smaller the voltage drop. |
| \(K\) | K | The resistance of copper in the US formula: 12.9 Ω·cmil/ft, the resistance of a copper wire 1 ft long with an area of 1 cmil at 75°C (21.2 for aluminum). |
| \(R\) | R | Wire resistance per foot (Ω/ft). Wire specs and NEC Chapter 9, Table 8 usually list it as "conductor resistance" per 1000 ft. |
| \(M\) | M | The multiplier set by the wiring system: 2 for DC 2-wire and single-phase 2-wire, \(\sqrt{3}\) for three-phase 3-wire, and 1 for single-phase 3-wire and three-phase 4-wire (line to neutral). |
| \(N\) | N | The number of parallel conductors, that is, how many identical wires are run in parallel for each phase. Paralleling makes the wire resistance \(\dfrac{1}{N}\), so the voltage drop also becomes \(\dfrac{1}{N}\). |
| \(V\) | V | Source voltage (V). The voltage at the sending end (such as the panel). |
| \(\varepsilon\) | epsilon | Percent voltage drop (%). What percent of the source voltage the voltage drop is. |
| \(\sqrt{3}\) | square root of 3 | About 1.732. It appears in line-to-line calculations for three-phase 3-wire systems, because the currents in the three wires are 120 degrees apart from each other. |
Terms
| voltage drop | Wires have a small resistance of their own, so when current flows, some voltage is lost along the wire and the load (an outlet or a device) gets a lower voltage than the source. It grows when the wire is thinner or longer, or when the current is larger. |
| one-way length | The length of the run from the source to the load, measured one way (also called the length of run or circuit length). In voltage drop formulas it is always the one-way length; the multiplier covers the round trip. |
| circular mil (cmil) | The US unit for wire area. A round wire 1 mil (0.001 in) in diameter has an area of 1 cmil, and a wire's area in cmil is its diameter in mils squared. Large sizes are given in kcmil (1000 cmil), such as 250 kcmil. |
| AWG (American Wire Gauge) | The standard system of wire sizes in the US. The smaller the number, the thicker the wire (14 AWG is thinner than 12 AWG). Sizes larger than 1 AWG are 1/0, 2/0, 3/0 and 4/0, and then kcmil sizes. |
| single-phase | A way of supplying electricity over two (or three) wires, used for outlets and appliances in homes. |
| three-phase | A way of supplying electricity for large motors in factories and similar places, sending three AC waves that are 120 degrees apart over three wires. |
| single-phase 3-wire | A system with two line (hot) wires and one neutral wire. The 120/240 V split-phase service to US homes is this system; 120 V is available between a hot wire and the neutral, and 240 V between the two hot wires. |
| balanced load | A single-phase 3-wire or three-phase system where the loads on each line (each phase) are the same size. When the load is balanced, no current flows in the neutral, so for single-phase 3-wire and three-phase 4-wire systems the multiplier is 1 (one wire only). |
| stranded wire | Wire whose conductor is many thin copper strands twisted together. It is more flexible than solid wire and is used for larger sizes and for cords. |
| solid wire | Wire whose conductor is one single copper wire. Small branch-circuit sizes such as 14 AWG and 12 AWG in home wiring are usually solid. |
| NM cable | Nonmetallic-sheathed cable, the most common cable for wiring inside US homes. For example, "12/2" is two 12 AWG insulated wires plus a bare ground wire, typically used for 20 A circuits. |
| National Electrical Code (NEC) | The US standard for safe electrical installation (NFPA 70), adopted by most states and cities, often with local changes. Its Chapter 9 tables list wire areas and resistances, and its informational notes give the recommended voltage drop limits (3% and 5%). |
| licensed electrician | In the US, electrical wiring work in buildings generally has to be done by a licensed electrician (rules vary by state and city, and permits and inspections are usually required). Use this calculator only as a guide for learning and planning. |
Good to know before you start
Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
| Proportional and inversely proportional relationships (Grades 7–8) |
|
| Ohm's law (middle school physical science) |
|
| Percentages (Grade 6) |
|
| Square roots (Grade 8) |
|
How to calculate it in Excel
Copy the whole table below and paste it into cell A1 in Excel. It works as is.
Table for the voltage drop in single-phase 2-wire and DC circuits
| One-way length L (ft) | 50 |
| Current I (A) | 15 |
| Wire area A (cmil) | 6530 |
| Voltage drop e (V) | =2*12.9*B1*B2/B3 |
Table for the voltage drop in three-phase 3-wire circuits
| One-way length L (ft) | 100 |
| Current I (A) | 30 |
| Wire area A (cmil) | 16510 |
| Voltage drop e (V) | =SQRT(3)*12.9*B1*B2/B3 |
Table for the voltage drop in single-phase 3-wire and three-phase 4-wire circuits
| One-way length L (ft) | 100 |
| Current I (A) | 10 |
| Wire area A (cmil) | 6530 |
| Voltage drop e (V) | =12.9*B1*B2/B3 |
Table for the voltage drop from the wire resistance
| System multiplier M | 2 |
| Current I (A) | 40 |
| Resistance per ft R (Ω/ft) | 0.000808 |
| One-way length L (ft) | 150 |
| Voltage drop e (V) | =B1*B2*B3*B4 |
Table for the percent voltage drop
| Voltage drop e (V) | 2.96 |
| Source voltage V (V) | 120 |
| Voltage drop (%) | =B1/B2*100 |
After pasting, the upper rows in column B are your inputs and the last row is calculated automatically. "*" is multiplication, "/" is division and SQRT(3) is the square root of 3.
The first table shows about 2.96 (V) in B4. The second shows about 4.06 (V), the third about 1.98 (V), the fourth about 9.70 (V) and the fifth about 2.47 (%).
Just replace the numbers in column B with the values for your circuit. The single-phase, three-phase and line-to-neutral tables differ only in the multiplier (2, SQRT(3) or 1) in front of 12.9. For aluminum wire, change 12.9 to 21.2.
The first table shows about 2.96 (V) in B4. The second shows about 4.06 (V), the third about 1.98 (V), the fourth about 9.70 (V) and the fifth about 2.47 (%).
Just replace the numbers in column B with the values for your circuit. The single-phase, three-phase and line-to-neutral tables differ only in the multiplier (2, SQRT(3) or 1) in front of 12.9. For aluminum wire, change 12.9 to 21.2.
How to calculate it in Google Sheets
Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.
Table for the voltage drop in single-phase 2-wire and DC circuits
| One-way length L (ft) | 50 |
| Current I (A) | 15 |
| Wire area A (cmil) | 6530 |
| Voltage drop e (V) | =2*12.9*B1*B2/B3 |
Table for the voltage drop in three-phase 3-wire circuits
| One-way length L (ft) | 100 |
| Current I (A) | 30 |
| Wire area A (cmil) | 16510 |
| Voltage drop e (V) | =SQRT(3)*12.9*B1*B2/B3 |
Table for the voltage drop in single-phase 3-wire and three-phase 4-wire circuits
| One-way length L (ft) | 100 |
| Current I (A) | 10 |
| Wire area A (cmil) | 6530 |
| Voltage drop e (V) | =12.9*B1*B2/B3 |
Table for the voltage drop from the wire resistance
| System multiplier M | 2 |
| Current I (A) | 40 |
| Resistance per ft R (Ω/ft) | 0.000808 |
| One-way length L (ft) | 150 |
| Voltage drop e (V) | =B1*B2*B3*B4 |
Table for the percent voltage drop
| Voltage drop e (V) | 2.96 |
| Source voltage V (V) | 120 |
| Voltage drop (%) | =B1/B2*100 |
The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the numbers in column B with the values for your circuit.
How to calculate it in Python
multiplier = 2 # wiring system (single-phase 2-wire and DC = 2, three-phase 3-wire = 3 ** 0.5, line to neutral = 1)
k_copper = 12.9 # K for copper (ohm-cmil/ft at 75 C; 21.2 for aluminum)
length_ft = 50 # one-way length of the run (ft)
current_a = 15 # current (A)
area_cmil = 6530 # wire area (cmil); 12 AWG = 6530
source_voltage = 120 # source voltage (V)
voltage_drop = multiplier * k_copper * length_ft * current_a / area_cmil # voltage drop (V)
drop_rate = voltage_drop / source_voltage * 100 # percent voltage drop (%)
end_voltage = source_voltage - voltage_drop # voltage at the load end (V)
print(f"Voltage drop: {voltage_drop} V")
print(f"Percent voltage drop: {drop_rate} %")
print(f"Voltage at the load end: {end_voltage} V")
Runs with the standard library only. Replace the multiplier and the circuit values at the top and run it. This example prints a voltage drop of about 2.96 V (about 2.47%). To calculate from the resistance, for aluminum or other wire, change the voltage_drop line to "multiplier * current * resistance per ft * one-way length".
How to write it in LaTeX and other math languages (copy and paste)
Voltage drop formula (single-phase 2-wire and DC)
e = 2 × K × L × I ÷ A
e = \dfrac{2 K L I}{A}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>e</mi>
<mo>=</mo>
<mfrac>
<mrow><mn>2</mn><mi>K</mi><mi>L</mi><mi>I</mi></mrow>
<mi>A</mi>
</mfrac>
</mrow>
</math>
e = (2 K L I)/A
2*k*l*i/a
e := 2*k*l*i/a;
e = 2*K*L*I/A;
e = (2 K L I)/A
Voltage drop for three-phase 3-wire and single-phase 3-wire (three-phase 4-wire)
e = √3 × K × L × I ÷ A (three-phase 3-wire), e = K × L × I ÷ A (single-phase 3-wire, three-phase 4-wire)
e = \dfrac{\sqrt{3} K L I}{A}, \quad e = \dfrac{K L I}{A}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>e</mi>
<mo>=</mo>
<mfrac>
<mrow><msqrt><mn>3</mn></msqrt><mi>K</mi><mi>L</mi><mi>I</mi></mrow>
<mi>A</mi>
</mfrac>
</mrow>
</math>
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>e</mi>
<mo>=</mo>
<mfrac>
<mrow><mi>K</mi><mi>L</mi><mi>I</mi></mrow>
<mi>A</mi>
</mfrac>
</mrow>
</math>
e = (sqrt(3) K L I)/A, e = (K L I)/A
Sqrt[3]*k*l*i/a (* three-phase 3-wire. Single-phase 3-wire and three-phase 4-wire: k*l*i/a *)
e := sqrt(3)*k*l*i/a; # three-phase 3-wire. Single-phase 3-wire and three-phase 4-wire: k*l*i/a
e = sqrt(3)*K*L*I/A; % three-phase 3-wire. Single-phase 3-wire and three-phase 4-wire: e = K*L*I/A;
e = (√3 K L I)/A, e = (K L I)/A
Formula from the wire resistance
e = M × I × R × L
e = M I R L
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>e</mi>
<mo>=</mo>
<mi>M</mi>
<mi>I</mi>
<mi>R</mi>
<mi>L</mi>
</mrow>
</math>
e = M I R L
m*i*r*l
e := m*i*r*l;
e = M*I*R*L;
e = M I R L
Percent voltage drop
ε = e ÷ V × 100
\varepsilon = \dfrac{e}{V} \times 100
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>ε</mi>
<mo>=</mo>
<mfrac>
<mi>e</mi>
<mi>V</mi>
</mfrac>
<mo>×</mo>
<mn>100</mn>
</mrow>
</math>
epsilon = e/V * 100
e/v*100
epsilon := e/v*100;
rate = e/V*100;
ε = (e/V) × 100
How to have ChatGPT do the calculation
You are a calculation assistant for electrical installations. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing). A 120 V single-phase 2-wire circuit uses 12 AWG copper wire (6,530 cmil), with a one-way length of 50 ft and a load current of 15 A. Using the formula e = 2 × K × L × I ÷ A with K = 12.9 ohm-cmil/ft for copper, find each of the following: 1. The voltage drop (V) 2. The percent voltage drop (%) 3. The voltage at the load end (V) Show the formulas you used and the numbers from the execution result.
How to Use
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1Enter your numbersType the numbers you want to calculate with into the input fields
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2CalculatePress the "Calculate" button
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3Check the resultThe result appears on the spot. The same page also explains the idea behind the calculation and the formula
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