How to Find the Proper Size of Wire & Cable In Metric & Imperial Systems
How to Determine the Suitable Size of Wire and Cable or Electrical Wiring Installation? Examples in Imperial and Metric Systems Based on NEC, IEC and IEEE.
The following stepbystep guide will show you how to calculate the correct size of cable and wire, or any other conductor, for electrical wiring installations with solved examples in both British or English and SI Systems, i.e., Imperial and Metric Systems, respectively.
Keep in mind that selecting the proper wire size is crucial when sizing a wire for electrical installations. An inappropriate size of wire for larger loads with high current may lead to chaos, resulting in the failure of electrical equipment, hazardous fires, and serious injuries.
This stepbystep guide explains how to calculate the correct wire and cable size for electrical wiring installations. We will use examples in both the British/English and SI systems (Imperial and Metric systems, respectively). The solved examples for wire sizing are based on wire ampacities and currentcarrying capacities according to NEC, IEC 60364, 60228, 608981, 609472 (International Standards), IET Wiring Regulations, BS 7671 (British Standards), and IEEE regulations.
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Why is Correct Wire Size Important?
Selecting the correct wire size is crucial when planning and sizing conductors for electrical installations. Using an improper wire size for highcurrent loads can lead to equipment failure, hazardous fires, and serious injuries.
using correct wire size is crucial in electrical wiring installations for several key reasons:
Safety
Undersized wires can overheat due to excessive current flow, which may lead to electrical fires or equipment damage. Choosing the correct wire size ensures that the wire can handle the expected load without overheating or failing.
CurrentCarrying Capacity
Each wire gauge has a specific currentcarrying capacity, known as ampacity. Using the proper wire size ensures that the wire can handle the required current without generating excessive heat or voltage drop, ensuring the efficient operation of electrical equipment.
Voltage Drop
Incorrect wire sizing can lead to excessive voltage drop, where electrical power is lost as heat along the wire. This can reduce the efficiency of appliances and lighting, cause motors to underperform, and increase energy consumption. Proper wire sizing minimizes voltage drop and ensures that the correct voltage reaches the end of the circuit.
Compliance with Standards
Electrical codes and standards, such as the NEC, IEC, and BS 7671, specify minimum wire sizes for different applications. Following these regulations ensures that installations are legal, safe, and meet industry standards.
Longevity of the Installation
Correctly sized wires have a longer lifespan because they are less likely to suffer from excessive heat, wear, or mechanical stress. This reduces maintenance costs and the need for early replacements.
Cost Efficiency
While undersized wires pose safety risks, oversizing wires can lead to unnecessary costs. Proper wire sizing strikes a balance between safety, performance, and cost.
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Factors Affecting the Wire Size
 Current Carrying Capacity: The amount of current the wire needs to carry (known as ampacity) influences its size, ensuring it can handle the load without overheating.
 Load Current and Breaker Size: The known value of load current in amperes and the rating of circuit breaker.
 Voltage Drop: For longer distances, consideration of voltage drop becomes crucial to maintain the efficiency and proper functioning of the electrical system.
 Ambient Temperature: The environmental temperature affects the wire’s ability to dissipate heat, impacting its currentcarrying capacity.
 Insulation Type: Different insulating materials have varying thermal and electrical properties, influencing the overall performance and size requirements.
 Installation Method: The manner in which the wire is installed, whether in conduit, cable trays, or exposed, can affect its ability to dissipate heat and, consequently, its size.
 Conductor Material: Copper and aluminum have different conductivity and temperature characteristics, influencing their suitability for specific applications and, consequently, their sizing.
 Code Guidelines: Adherence to electrical codes and standards is crucial, as they provide specifications and regulations for wire sizing to ensure safety and compliance.
 Type of Load: The nature of the electrical load, whether resistive, inductive, or capacitive, can impact the wire size needed for optimal performance.
 Cable Bundling: When multiple conductors are bundled together, there is a need to derate the ampacity of each conductor, affecting the overall wire size requirements.
 Frequency of Operation: For applications involving alternating current (AC), the frequency of operation can influence the skin effect, affecting the effective resistance and thus impacting wire sizing.
 Types of Cables: It also depends on types of wires and cables for copper and aluminum such as NMB (NonMetallic Sheathed Cable) also known as Romex, MC Metalclad Cables, THHN/THWN, UFB (Underground Feeder), USE (Underground Service Entrance), SE (Service Entrance):
Voltage Drop in a Cables
All conductors have resistance, which is directly proportional to their length and inversely proportional to their diameter:
R ∝ l/a … [Ohm’s laws of resistance R = ρ (L÷ a)]
As current flows through a conductor, a voltage drop occurs. While it can be neglected for short distances, longer or thinner wires require voltage drop considerations to ensure proper system function.
According to NEC 210.19(A), the maximum recommended voltage drop on a branch circuit is 3% and from the beginning of a feeder to the farthest outlet on a branch circuit should not exceed 5% (215.2(A). For longdistance runs over 50 feet(15.25 meters), consider upgrading to a larger gauge wire to compensate for voltage drop. According to NEC 31016, for every 100 feet (30.50 meters) of wire length, add 20% ampacity to account for voltage drop.
The IEEE B23 rule specifies that voltage drop should not exceed 2.5% of the supply voltage.
According to BS 7671 – TABLE 4Ab and IEC60364552, article 525, table G.52.1, the limit of maximum voltage drop for lighting circuits is 3% and 5% of other heating and power usage supplied by public LV distribution system. In case of private LV supply system, the voltage drop for lighting and other HVAC systems is 6% and 8% respectively. If the voltage drop exceeds the limits, larger conductor (cable and wires) must be used to compensate the condition.
According to AS3008, the standard allowable voltage drop from the supply to any point in the circuit should not exceed 5% (AS/NZS 3008). However, an exception is made when a lowvoltage substation is installed on the premises and used as a dedicated circuit. In this case, the allowable voltage drop can increase to 7%, as per AS3000.
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Example:
If the supply voltage is 230V AC, then the value of allowable voltage drop should be;
Allowable Voltage Drop = 230 $×$ (2.5/100) = 5.75V
Similarly, if the supply voltage is 120V AC, the allowable voltage drop should be no more than 3.6V for a separate branch circuit (120V × 3% = 3.6V) and 6V for feeder and branch circuits (120V × 5% = 6V). Refer to NEC Code 210.19(A) and 215.2(A), and Table 31016, which states that 20% additional ampacity for every 100 feet of distance should be added to counter the voltage drop in the circuit.
In electrical wiring circuits, voltage drops also occur from the distribution board to the various subcircuits and final subcircuits. For subcircuits and final subcircuits, the allowable voltage drop should be half of the general allowable voltage drop (i.e., 2.75V of 5.5V as calculated above).
Normally, voltage drop tables describe voltage drop in Amperes per meter (A/m), such as the voltage drop in a onemeter cable carrying one Ampere of current.
There are two methods to define voltage drop in a cable, which we will discuss below.
 For SI System (Metric): Voltage drop is defined in terms of per amp per meter.
 For FPS System (Imperial): Voltage drop is defined in volts per ampere per 100 feet (30 m) based on circuit run i.e. distance and length of wire.
Finding Voltage Drop in the Cable
To find the voltage drop in a cable for this tutorial, follow the simple steps below:
 First, find the maximum allowable voltage drop from the give tables.
 Next, determine the load current.
 According to the load current, select an appropriate cable (whose current rating is nearest to the calculated load current) from Table 1.
 From Table 1, find the voltage drop per meter or per 100 feet (depending on your preferred system) according to its rated current.
Now, calculate the voltage drop for the actual length of the wiring circuit based on its rated current using the following formulas:
 (Actual length of circuit × voltage drop per meter) / 100 = Voltage drop per meter.
 (Actual length of circuit × voltage drop per 100 feet) / 100 = Voltage drop per 100 feet.
Next, multiply the calculated voltage drop by the load factor, where:
Load factor = Load current / Rated current of the cable (from the table).
It will show the exact value of voltage drop in the cable when load current is flowing through it.
 If the calculated voltage drop is less than the maximum allowable voltage drop (step 1), then the selected cable size is correct.
 If the calculated voltage drop is greater than the maximum allowable voltage drop (step 1), calculate the voltage drop for the next larger cable size. Continue this process until the calculated voltage drop is less than the maximum allowable voltage drop from step 1.
You may use voltage drop calculator or manual methods using different voltage drop formulas to determine voltage drop and wire size. If the calculated voltage drop is less than the maximum allowable drop, the cable size is appropriate. If it is greater, select the next larger cable size.
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How to Determine the Correct Size of Cable & Wire for a Given Load?
Below are solved examples demonstrating how to find the proper cable size for a given load.
For a given load, cable size may be determined using various IEC and NEC tables (such Article 310 – Table – 310.15 (B) 16. However, it is crucial to keep in mind the role of ambient temperature and voltage drop for distance between main panel and subpanel.
When determining the size of cable for a given load, take into account the following general rules of thumbs.
 For a given load, apart from the known current value, there should be a 20% additional margin as safety factor for continuous load circuits and for future or emergency needs (required by both NEC and IEC/IEEE).
 In the NEC – 31016, add an additional 20% ampacity to the wire size if the length of the wire exceeds 100 ft (between the main panel and the subpanel).
 In IEC, the voltage drop should be limited to 1.25%, and for the final subcircuit, the voltage drop should not exceed 2.5% of the supply voltage from the energy meter to the distribution board.
 Consider the change in temperature, which affects the ampacity (current carrying capacity) of the wire. When needed, use the temperature factor (also known as Correction factors or Rating Factors) given in Table 3, Table 6 and related NEC table given at the end of this article).
 Additionally, take the load factor (diversity factor or demand factor (NEC – 220.42 and (220.45)) into consideration when determining the size of the cable.
 When calculating the cable size, consider the ambient temperature and type of wiring system; for example, in an open wiring system, temperature would be low, but in conduit wiring, temperature increases due to the absence of air.
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Solved Examples of Proper Cable & Wire Size
The following examples illustrate how to determine the proper size of cables for electrical wiring installations. These examples will make it easy to understand the method of determining the proper cable size for a given load in both single phase ad three phase wiring installations.
Example 1 – (Imperial System Followed by NEC)
What is the correct wire size for a 1,920W load circuit supplied by 120V AC at 60°C (140°F)?
Solution:
First of all, let’s determine the load current in amperes using basic Ohm’s Law that will flow from the 120V breaker to the 1.92kW load.
I (in Amps) = P (in Watts) ÷ V (in Volts)
I = 1,920 W ÷ 120 V
I = 16 Amp
Now, add a safety factor of 1.25 (based on 125% rule also known as continues load rule), as per NEC 210.20(A) for branch circuits, feeders, and service loads. This code specifies that only 80% of the branch circuit load should be connected to the circuit for the ampacity of the wire for any load.
In other words, the breaker should handle 125% of the rated load current amps. For instance, a 15ampere breaker should be used for a 12ampere load point. This way:
I = 16 amp $×$ 1.25
or 16 amp $×$ 125%
I = 20 Amp.
Now, If you see in the AWG wire size chart and NEC Table 31015B (16) Article 310.60 (given below), the right size for 20 amp circuit is #12 AWG copper or #10 AWG aluminum.
According to 240.4(D), you are permitted to use:
 14 AWG copper for 15 amperes breaker and outlets
 12 AWG copper 20 for amperes breaker and load circuits
 10 AWG copper 30 for amperes OCPD and load points
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Example 2 – (240V and Distance Involved – NEC)
Find the Proper wire size for 2,400 W load circuit supplied by 240V single phase at a distance of 100 feet?
Solution:
Find the current using the following formula (same as mentioned above)
Current = Power ÷ Voltage
I = 2,400 W ÷ 240 V
I = 10 Amperes
Now, multiply the safety factor of 1.25 (80% of load should be connected of the rated ampacity) to the calculated amperage. The same applies to the breaker size and outlet rating.
$×$
$I = 12.5 Amperes$
As the circuit is 100ft away, add additional 20% ampacity to the calculated value (according to the NEC Code – 31016) to counter the voltage drop in the circuit.
$20%of12.5A=0.20A×1.25A=2.5A.$
Total Amps = 12.5A + 2.5A = 15 Amperes.
According to NEC table 31015B and AWG wire size chart, the suitable wire size for 15 amp circuit is #14 AWG copper at 60°C (140°F) and #12 AWG aluminum.
Notes:
 NEC Table 310.15(B)(16) (formerly Table 310.16) with the help of 240.4(A) through (G) which shows the 14 AWG wire size is able to carry 15A at 60°C (140°F) and 20A at 75°C (167°F).
 The determined wire size, breaker rating, and ampacity in the above calculation for North America and Canada comply with the National Electrical Code (NEC) – Sections 210.19(A), 215.2, and 230.42(A) for continuous and noncontinuous loads and 110.14(C) for ambient temperature rating.
 This calculation are based on the NEC guidelines. For more details, refer to NEC 210.21, 210, 24, 220.110, 220.14, 220.42, 220.45, 220.53, 220.55, 31014 and 517.22.
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 #14 AWG can safely handle 15amp at 60°C (140°F) and 20amp at 75°C (167°F).
 #14 AWG can be used for maximum of 15Amp of noncontinuous load circuit and maximum of 12Amp of continuous load circuits (no more than 80% of the load should be connected to the rated of 15A circuit breaker).
 It is against the code to use lower size of wire than the recommended.
 It is always advisable to use copper wire instead of aluminum for better conductivity and reduced power consumption. If it is necessary to use aluminum, the wire size in AWG may differ compared to a copper conductor. Refer to the tables given below to select the wire size accordingly in the case of using aluminum.
 In addition, it is recommended to use solid wire instead of stranded wire for a better grip, reduced sparks, and a protected environment from shock and fire.
Example 3 – (BS 7671 (British Standards) in Imperial – IEC)
Example: For an electrical wiring installation in a building, where the total load is 5kW and the total length of cable from the energy meter to the subcircuit distribution board is 35 feet, with a supply voltage of 230V and a temperature of 40°C (104°F), find the most suitable size of cable if wiring is installed in conduits.
Solution:
 Total Load = 5kW = 5 $×$ 1,000W = 4,500W
 20% additional load = 5,000 $×$ (20/100) = 1,000W
 Total Load = 5000W + 1,000W = 5,400W
 Total Current = I = P ÷ V = 6,000W /230V = 26.08A
Now select the size of cable for load current of 26.08A (from Table 1 on the right side) which is 7/0.036″ (28 Amperes). It means we can use 7/0.036″ (4 mm^{2}) cable according to table 1.
Now, check the selected (7/0.036″) cable with the temperature factor in Table 3. The temperature factor is 0.94 (from Table 3) at 40°C (104°F), and the current carrying capacity of (7/0.036″) is 28A. Therefore, the current carrying capacity of this cable at 40°C (104°F) would be:
Current rating for 40°C (104°F) = 28 $×$ 0.94 = 26.32 Amp.
Since the calculated value (26.32 Amp) at 40°C (104°F) is less than the current carrying capacity of the (7/0.036″) cable, which is 28A, this size of cable 4 mm^{2} (7/0.036″) is also suitable with respect to temperature.
Now find the voltage drop for 100 feet for this (7/0.036″) cable from Table 4 which is 7V, But in our case, the length of cable is 35 feet. Therefore, the voltage drop for 35 feet cable would be;
Voltage Drop = Vd $×$ L $× I$
Actual Voltage drop for 35 feet = (7V $×$ 35/100) $×$ (26.08A/28A) = 2.28V
And Allowable voltage drop = 3% $×$ 230V = 6.9V
Here, the actual voltage drop (2.28V) is less than the maximum allowable voltage drop of 6.9V. Therefore, the most appropriate and suitable cable size for that given load in the electrical wiring installation is (7/0.036″) which is equal to 4 mm^{2}.
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Example 4 – (BS 7671 (British Standards ) in Metric – 18th Ed.)
If the singlephase supply voltage is 230V AC, how do you calculate the circuit current and cable size for each subcircuit and the main circuit for the following load to be connected in a residential building?
SubCircuit 1
 2 lamps each o 800W and
 3 fans each of 80W
 4 TV each of 120W
SubCircuit 2
 6 Lamps each of 80W and
 5 sockets each of 100W
 4 lamps each of 800W
Solution:
Total Load of SubCircuit 1
= (2 $×$ 800W) + (3 $×$ 80W) + (4 $×$120W)
= 1,600W + 240W + 480W = 2,320W
Current for SubCircuit 1 = I = P/V = 2,320W/230V = 10A
Total Load of SubCircuit 2
= (4 $×$ 80W) + (5 $×$ 100W) + (6 $×$ 500W)
= 320W + 500W + 3,000W= 3,820W
Current for SubCircuit 2 = I = P/V = 3,820W/230V = 16.6A
Therefore, Cable suggested for sub circuit 1 having 10A = 1 mm^{2} (13.5 Amp – Table 6 or Table 5) equivalent to 1/044″ or 3/.029″ (11A and 13 Amp respectively – Table – 4 ).
Cable suggested for SubCircuit 2 having 16.6A = 1.5 mm^{2} (17.5 Amp – Table 6) equivalent to 3/036″ or 7/.029″ (16A and 21 Amp respectively – Table – 4 ).
Total Current drawn by both SubCircuits = 10A + 16.6A = 26.6 A
Therefore, cable suggested for MainCircuit having 26.6A = 4 mm^{2} (32 Amp – Table 6) equivalent 7/.044″ ( 34 Amp – Table – 4 ).
 1mm^{2} twin and earth (T & E) cable size with 6A breaker is used for general lighting.
 2.5mm^{2} twin and earth (T & E) cable size with 16A breaker is used for power circuits.
 4mm^{2} twin and earth (T & E) cable size with 32A breaker is used for showers, water heaters and cookers.
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Example 5 – (IEC 60364 in Metric System – IEC)
Example: What XLPE or EPR size of cable suits for given situation?
 Load = 4.5kW
 Volts = 230V AC – Single Phase
 Length of Circuit = 20 meter
 Temperature = 35°C (95°F)
Solution:
Load = 4.5kW = 4,500 W
Voltage = 230V
Current = I = P/V = 4.500W / 230V = 19.56A
Total Load Current = 19.56A
Now select the size of cable for load current of 19.56A (from Table 6) which is 2.5mm^{2} (having current capacity of 24 Amperes). Hence, we are allowed to use 2.5mm^{2} XLPE or EPR cable according to the table 6.
Now check the selected (2.5mm^{2}) conductor with Correction factors or Rating Factors (C_{a}) for Ambient temperatures in Table 6. As the rating factor or correction factor is 0.96 (in table 6) at 35°C (95°F) for XLPE or EPR cable and current carrying capacity of (2.5mm^{2}) is 24A, therefore, current carrying capacity of this cable at 40°C (104°F) would be;
Current rating for 35°C (95°F) = 24 $×$ 0.96 = 23.04 Amp.
Since the calculated current (23.04A) at 35°C (95°F) is less than the currentcarrying capacity of the 2.5mm² cable, which is 24 A, this cable size is suitable for the temperature conditions
Now find the voltage drop for per ampere meter for (2.5mm²) cable from (Table 8) which is 19mV, But in our case, the length of cable is 35 meter. Therefore, the voltage drop for 20 meter cable would be:
Actual Voltage drop for 35 meters =
= Vd $×$ I $×$ L
= 19 $×$ 23.04A $× (20$/1000) = 8.75V
And 5% Allowable voltage drop = (5 $×$ 230V)/100 = 11.5V
Here the actual Voltage drop (8.75V) is less than that of maximum allowable voltage drop of 11.7V. Therefore, this is the correct size of cable conductor for that given load.
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Example 6 – Three Phase Motor – IEC
A 10H.P (7.46kW) three phase squirrel cage induction motor of continuous rating using StarDelta starting is connected through 400V supply by three single core PVC cables run in conduit from 250feet (76.2m) away from multiway distribution fuse board. Its full load current is 19A. Average summer temperature in Electrical installation wiring is 35°C (95°F). Calculate the size of the cable for the induction motor?
Solution:
 Motor load = 10H.P = 10 $×$ 746 = 7460W *(1H.P = 746W)
 Supply Voltage = 400V (3Phase)
 Length of cable = 250feet (76.2m)
 Motor full load Current = 19A
 Temperature factor for 35°C (95°F) = 0.97 (From Table 3)
Now select the size of cable for full load motor current of 19A (from Table 5) which is 2.5mm^{2} (21A). *(Remember that this is a 3phase system i.e. 3core cable) and the voltage drop is 13.2V per km (Table 8). It means, we can use 2.5mm^{2} cable according Table (6).
Now check the selected (2.5mm^{2}) cable with correction factor or rating factor in table (7), so the temperature factor is 0.94 and 0.96 for PVC an XPLE cables (in table 7) at 35°C (95°F) and current carrying capacity of (2.5mm^{2}) is 21 Amperes for 3phase, therefore, current carrying capacity of this cable at 40°C (104°F) would be:
Current rating for 35°C (95°F) = 21 $×$ 0.96 = 20.16 Amp.
Since the calculated value (20.16 Amp) at 35°C (95°F) is less than that of current carrying capacity of (2.5mm^{2}) cable which is rated for 21A, therefore this size of cable (2.5mm^{2}) is suitable with the ambient and operating temperature.
Load factor = 19A/23A = 0.826
Now find the voltage drop for 76.2 m (250 feet) for this (2.5mm^{2}) cable from table (8) which is 13.2V for three phase running motor at CosΦ 0.8. But in our case, the length of cable is 250 feet (76.2). Therefore, the voltage drop for this length would be;
Actual Voltage drop for 250 feet (76.2m) = (13.2 $×$ 19A $×$ 76.2/1000) $×$ 0.826 = 15.78V
(15.78V÷ 400V) $×$ 100 = 3.9 %
This value is less than that authorized (8%) and is satisfactory.
And maximum 8% Allowable voltage drop = (8/100) $×$ 400V= 32V
Here the actual Voltage drop (15.78V) is less than that of maximum allowable voltage drop of 32V. Therefore, this is the right wire size for the given 3phase load.
Wire CSA and Diameter Calculations
Wire Cross Section Area (CSA) Formulas
Wire Cross Sectional Area in kcmil (kilo circular mils)
A_{n} = 1000 × d_{n}^{2} = 0.025 × 92^{(36n)/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^{(}^{36n)/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^{(36n)/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}.
Wire Diameter Calculation

Wire Diameter in Inches Formula
d_{n} = 0.005 × 92^{(36n)/39} …. In inches
Where “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^{(36n)/39} …. In millimeters (mm).
Where “n” is number of the gauge size and “d” the wire diameter in mm.
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}).
Tables & Charts for Cable & Wire Sizes
Below are important tables and charts for currentcarrying capacity, voltage drop, correction/rating factors, temperature ratings, etc. Using these tables are useful to determine the proper cable size for electrical wiring installations in both singlephase and threephase supply systems.
BS 7671 – IET & 60364 – IEC Tables
 Table 1 – Current rating of copper cables in mm^{2} and Inche^{2} at 30°C (86°F).
Click image to enlarge
 Table 2 – Current rating of flexible cords – copper cables at 30°C (86°F).
Click image to enlarge
 Table 3 – Temperature factor
Click image to enlarge
 Table 4 – Cable Size in inches with current rating and voltage drop inside trucking and conduit.
Click image to enlarge
 Table 5 – Cable Size in mm^{2} with current rating and voltage drop inside trucking and conduit.
Click image to enlarge
 Table 6 – Current carrying capacity for 1Phase and 3Phase cables based on IEC 60364552 – Table B.52.4 and BS 7671 – Table 4D1A.
 Table 7 – Correction factors or rating factors (Ca) based on IEC 60364552 – Table B.52.14 and BS 7671 – Table 4B1
 Table 8 – Voltage drop values based on IEC 60364552 and BS 7671 – Table 4E1B
 Table 9 – Size of Circuit Breaker and Conductor Selections for Different Domestic Circuits
 Table 10 – IEE Recommended Current Demands and Diversity Factors for Various Loads
 Table 11: Conductor and breaker size selection for domestic and residential application in 230V, singlephase AC circuits for IEC following countries.
 Table 12: Typical current carrying capacity of cables with suggested cable size, current rating in amps and recommended circuit breaker rating in amperes for 230V, singlephase AC circuits for IEC following countries.
NEC Wire Size Table 310.15(B)(16) (formerly Table 310.16) & Chart
NEC (National Electrical Code) Table 310.15(B)(16) (formerly Table 310.16) – 310.60 – ARTICLE 310 – Conductors for General Wiring & Allowable Ampacities of Conductors & Wire Sizes based on AWG (American Wire Gauge).
310.60 ARTICLE 310 — CONDUCTORS FOR GENERAL WIRING  
Table 310.15(B)(16) (formerly Table 310.16) Allowable Ampacities of Insulated Conductors Rated Up to and Including 2000 Volts, 60°C Through 90°C (140°F Through 194°F), Not More Than Three CurrentCarrying Conductors in Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F)*  
Size AWG or kcmil  Temperature Rating of Conductor [See Table 310.104(A).]  Size AWG or kcmil  
60°C (140°F)  75°C (167°F)  90°C (194°F)  60°C (140°F)  75°C (167°F)  90°C (194°F)  
Types TW, UF  Types RHW, THHW, THW, THWN, XHHW, USE, ZW  Types TBS, SA, SIS, FEP, FEPB, MI, RHH, RHW2, THHN, THHW,
THW2, THWN2, USE2, XHH, XHHW, XHHW2, ZW2 
Types TW, UF  Types RHW, THHW, THW, THWN, XHHW, USE  Types TBS, SA, SIS, THHN, THHW,
THW2, THWN2, RHH, RHW2, USE2, XHH, XHHW, XHHW2, ZW2 

COPPER  ALUMINUM OR COPPERCLAD ALUMINUM  
18**  —  —  14  —  —  —  — 
16**  —  —  18  —  —  —  — 
14**  15  20  25  —  —  —  — 
12**  20  25  30  15  20  25  12** 
10**  30  35  40  25  30  35  10** 
8  40  50  55  35  40  45  8 
6  55  65  75  40  50  55  6 
4  70  85  95  55  65  75  4 
3  85  100  115  65  75  85  3 
2  95  115  130  75  90  100  2 
1  110  130  145  85  100  115  1 
1/0  125  150  170  100  120  135  1/0 
2/0  145  175  195  115  135  150  2/0 
3/0  165  200  225  130  155  175  3/0 
4/0  195  230  260  150  180  205  4/0 
250  215  255  290  170  205  230  250 
300  240  285  320  195  230  260  300 
350  260  310  350  210  250  280  350 
400  280  335  380  225  270  305  400 
500  320  380  430  260  310  350  500 
600  350  420  475  285  340  385  600 
700  385  460  520  315  375  425  700 
750  400  475  535  320  385  435  750 
800  410  490  555  330  395  445  800 
900  435  520  585  355  425  480  900 
1000  455  545  615  375  445  500  1000 
1250  495  590  665  405  485  545  1250 
1500  525  625  705  435  520  585  1500 
1750  545  650  735  455  545  615  1750 
2000  555  665  750  470  560  630  2000 

Here is the NEC table as a chart (image format to downloads as a reference)
Click image or open in a new tab to enlarge
Below is the general table based on NEC – 2020 with wire applications, rated ampacity and AWG wire size for given and specific uses.
Wire Applications  Rated Ampacity  Wire Gauge – AWG 
Lowvoltage lighting circuits  10 amps  #18 
Light duty Extension cords  13 amps  #16 
General lighting circuits, lamps and fixtures  15 amps  #14 
Kitchen, bathroom, and outdoor outlets and receptacles  20 amps  #12 
Electric water heaters, Electric ranges, stove, ovens, cooktops, dryers, air conditioners,  30 amps  #10 
Heavy duty Cooktops and ranges, EV Charging  4050 amps  #6 
Commercial Electric furnaces, large electric heaters  60 amps  #4 
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Thanks Sir,<br /><br />It Gives me all the answers for my question.<br /><br />Thanks Once Again.
Welcome Dear….and Thanks 4 appreciation…
For 10kv how much sqmm ug cable can put
4mm
Cable size cannot said with voltage alone. You have to tell the POWER in KVA OR KW of the load.ALSO THE CORCUIT IS 3 PHASE OR 3 PHASE IS TO BE CLEAR..
Thank you so much. Please don’t stop your work. Great.
Ya I need assistance
Sir
can you help me how to Know the cable size and suitable MCCB rating(example i have 120sqmm x4 core aluminium and 100MCCB is suitable for the above size cable if is not suitable what are possible problem may occureed)
Sir I could not get this "Number of wires & Thikness of each wires" (7 / 0.0229)……….. (70 / 0.0076)<br />Kindly give me answer.<br />
Dear @<br /> 7/0.029 means 7 numbers of wires ( Conductors in a cable) and diameter of each conductor is 0.029" inches.
I think it is: 7 wire strands, each with a 0.029 inch diameter, in a single conductor cable. Is this correct?
Dear sir.,<br />great valuable explanation . i am realy searching this stuff from past 3 years. not i got it just sake of ur study material.
Seems like I am doing exam's numerical, but thanks so much for this post. With high quality cable if you do not know how much the length should it be, then the resistance will be more. And these tables for calculating the Voltage drop really helped a lot
Thanks for Appreciation
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sir, can you tell me how to fish a wire through a conduit pipe..
I have a 130 watts panel charging a 120ah battery,and 500 watts inverter.I have connected 65watts TV,decoder,4 15watts lights and 100watts laptop.Looking at the load am I doing the right thing.
<a href="www.https://www.electricaltechnology.org/2013/05/acompletenoteonsolarpanel.html" rel="nofollow"> A Complete note on Solar panel Installation </a>
Great Work Boss.
Thanks for appreciation
Dear sir,
On above discussion, selection of cable size you are considered voltage drop should not be less than (+ or – 2.5%) of our supply voltage, But according to electricity distributor allowable voltage fluctuation is (+ or – 5%).
please clear my that confusion.
Thanks in advance
Sir,, Plz also explain in new assignments <br />1.how we calculate Breaker Size and its nature? as MCB, MCCB, ACB, VCB <br />2. Calculation of BusBar size w.r.t load ?<br />Thanx
Thanks for Positive feedback,,, Wait for the upcoming posts..
Thank you v much for this useful information…….. Stay blessed.
Welcome Dear, and bundle of thanks for best wishes..
Great Job,thanks a lot for the post.
Thanks for Appreciation
the cable the tech's glove is on has pulled down so much that it is in contact in two places; the upper right meter contact, and the cable guide to the right. <a href="http://500mcmcable.com/" rel="nofollow">750 mcm cable</a><br />
make a thread where show how to do home conduit wiring n all<br />btw grt work
This blog is very informative and I would like to see some more blogs on this topic.<br /><a href="http://www.kowelec.com.au/aboutus/" rel="nofollow">Experienced Industrial Electricians At Kowelec Bendigo</a>
gud job you gave the information about copper cables only..wat abt Alluminium armoured cables<br />
than you so much for this useful link
Great site. Wondering the source of tables, is it from any standards? ie. IEC, NEC, etc.<br />Thanks.
sir plz mention the standard according to which the tables have been made.<br /><br />
i want selection of flexible cable standard chart with electric load
If you look carefully at the calculations there are some basic mistake which would effect the final result. ie 2 TV's @ 150w ea. is the shown as 2 x 120w. Other mistake include the lenght calculated which starts at 250 feet then goes to 35 feet and then back to 250 feet.
These are just Examples…. You have to find the suitable size of cable for wiring according to your specification and load requirements… Any way thanks for correction.
sir plz mention the standard according to which the tables have been made.
Thanks for Positive and important Questions.<br />lets try to answer<br /><br />(1)<br />In clear words (for your first Question)<br /><br />Temperature factor means that when temperature increases, Cable/Wire or conductor resistance also increase, hence, Cable/Wire or conductor current carrying capacity decreases. For instance; <br />If the current carrying capacity of wire or cable is 28 Ampere
I am not an electrical engineer, but why does a higher temp in table 3 result in lower temp factor (which makes the calculated value less than the current carrying capacity)? For instance in example 1 – take 43 amps in the table times a temp factor of .94 to yield 40.42 amps. The results is always less than initial amps unless you use a lower temp. Therefore, the higher the temp, the lower the
Would you tell me about the IEC standard in Electrical Installation design and the importance of IEC standard. <br />Thanks for your nice post
if increase frequency value r decrease then what will happen?
Dear Sir.
Thank u very much much for deep explanation.
Do we need to consider power factor in the calculation of load current…..
But,when temprature of conductor increases then resistivty decreases.
When temperature increase resistivity increase too
Dear Sir,
I want to know that if we have no table and we do not remember the current caring capacity of conductor than how can we calculate the cable size? We know the total load, Temperature, Area from DB to load. Please guide.
Good day sir,
Great explanation, but the only aspect am confused of is the area of cable determination,i was expecting the sizes of cable in mm sq as against 7/0.036
thanks for the simple and straight forward language for communication. Please how did you arrive at 5800KW in example 2 above.
Thanks for Appreciation… But where is 5800kW in Example 2… It is 5.8kW = 5800W. :)
thankz a lot bro
nice it’s very helpful for me thanks
I have to install 2 hp single phase pump set load carrying 13 to 15 amps. voltage at site is 220 to 230 volts. Length of cable will be 500 mtrs. Can you suggest required cable size in Copper as well as alluminium ?
Reply me on my email
dear sir
i want to find cable size in mm for the load of 600 amp and the main breaker is 1000 amp area is 120 meters supply voltage is 380 v three phase .
Did You check this?
https://www.electricaltechnology.org/2014/12/advancevoltagedropcalculatorvoltagedropformula.html
i have a motor 3phases 227 kw i want to connect it to the soft starter which size of cable i have to use ?
also can i use 3cores instead of one core so the connection it will be inplace of 3cx…. 9 cores
each pahse will atke three cores Q!?
May Allah ( God ) Bless you for your this type of kind effort , it’s very helpful for me and for electrical related engineers
thanks
how to calculate 3 phase 400 v and 600 amp load’s cable size which must have a length of 25 meter
You can find it by using this calculator. You may also use this one.
how we can make selection of cable for our system
Dear sir, You have done a great job. It’s very helpful to me. You may live long.
Amarasiri from SL.
Thanks for your kind words… :)
dear sir,
please tell me relation between your cabe size(total coductor/ Diameter of each) and square mm cable size
eg. 7/085 cable= ……sq.mm
Dear any body feed back regarding cable selection depending load & length of cable 3 phase application
used cable size is 10 mm load 34.6 Amps 460 .3Ph 60 hz motor
Hi,
What size cable do I need to run 325metres from an 80amp transformer on the power pole so I still have 63amps at the end?
Thank you very much for your assistance.
Sorry, forgot to mention it needs to be 3phase
Are all made computations, in the samples above, AC or DC? Pleas I need answer.
I have responsibility to provide 150kVA genset to one of social concernt where they have 75kVA sound system with 20 nos of 250W Floodlightings. But my duty is to provide them above generator and power cable reaching to their Main DB panel.
Now, advice me which power cable should I go with the suitable size please???
Please, what size of cable can I use to wire an equipment of 3phase, 4wire 120KVA, 50Hz with power factor of 0.8?
This 7core configuration doesnt fit with NEC standards I think. We use 4c or 3c as per requirement. I am seeing for the first time like 7C..?
What formula is followed to make the table? I want to know in given load, length of cable which electrical calculation can be done to find out the size of cable?
Current rating for 40°C (104°F) = 28 x 0.94 = 26.32 Amp.
Since the calculated value (26.32 Amp) at 40°C (104°F) is less than that of current carrying capacity of (7/0.036) cable which is 28A, therefore this size of cable (7/0.036) is also suitable with respect to temperature.
I think this para is trying to say that if temperature increases the current carrying capacity or rating would decrease for the same size of cable. So it means that if my load current is less than the calculated rating of the wire then we can use the wire. So it should say “Since the calculated value (26.32 Amp) at 40°C (104°F) is less than that of load current which is 24.5A, therefore this size of cable (7/0.036) is also suitable with respect to temperature.”
This is with reference to example 1. Please correct if i am wrong.
state two factors to be considered in selection of cables sizes for a house wiring
sir, I have a farm for which I want to put 10 numbers of 25 watts CFL bulbs around the perimeter, the perimeter of the farm is around 1 KM. Can you please confirm what type of wire I need to use (gauge, insulation requirements, Copper vs Aluminum), Considering that I will be laying this underground through pipes (which might be exposed to water at times…I need wires that are resistant to moisture and soil interaction). I am planning on using drip pipes which are cheaper and pass the wires through them. Please advise. This will be run on solar power. I will size the solar power unit accordingly.
I want to know about it.
If you have a time or see this comment, Please reply me.
In example 1,
Why are you using that I = P/V?
Total Current = I = P/V = 5400W /220V =24.5A
Actually, It should be which is I = P/( V x PF)
Assume , Power Factor (PF) will be 0.8.
Total Current = I = P/( V x PF) = 5400W /( 220 x 0.8)V =30.68 A
bro I= P/V for DC and I= P/V x PF for AC Power
What is power factor?
You may read about
sir plz mention the standard according to which the tables have been made.
Total load 32 KW 3phase 4wire system (distribute)220 volt 1phase and length 140 Meter .what is cable size use in mm
sir, please provide below information with necessary calculations
for 200/150 HP/Kw 3phase induction motor
1. cable size from pannel to starter to motor and from HT transformer to LT pannel
2. type of starter to be used
3. Type of pannel to be used
thanks
Can you please tell me which Electrical Code did you use in above calculations?
I have 3phase 10 HP Moto ,16 amps,which size cable is sutiable,iam Diploma EEE,Fresher on maintenance Engineer on on eharama company,plz give me cable size flexible or armed cable.core and mm size plzz give,mee
you are calculating full load current of a 3 phase motor. that doesnot include sq.root of 3.
I Thank You Mr. Khan, for your great effort in helping your visitors with an easy to understand presentation. I must tell you you are the best online teacher and you understood what basics people must know before they would understand the subject matter. I guess you must be a generous human being who is helping others not to fail in life. keep up the effort!
Thanks again!
Thank you so much for appreciation,,,, :)
Honestly i enjoyed the tutorials, its quite interesting for all engineering student both graduates and under graduates… i appreciate the effort of electrical technology for her convenient idea of letting some techs into us. Thanks alot
Many thanks, for your outstanding efforts in educating and assisting all peoples to understand such valuable technical subjects.
Good explaination of calculating the size of the cable.
I got this (V.D of m = mV x I x L) a bit confusing. How is this formula derived? If you apply the values and their units, ypu dont get volts per ampere meter at the end.
Can you justify please.
Thanks for the information about electrical wiring
how to find out 240RM cable , which is actual system
great work
how did you find load factor for the last example ? we have replace the selected cable with new one, why do you choose the same load factor ?
Which questions are asking for industrial electrician in interview please reply
thank you very much, it is very useful, I hope you give us more examples of threephase circuits.
Best regards.
Eng. Abdolgabar Ahmed Mahmood.
Great article about wiring and installation.
Short and brief description thank you.
I want to ask about question number 4
In load factor, at first calculation of cable size(load factor)=19/23=0.82
But, after we change the cable size with higher size, why the load factor use in calculation still 0.82, isn’t that (load factor)=19/28?
Previously, i want to say thank you, this website is very incredible
First, i thank. In eg.1 the load factor is 24.5/ 28. In eg.2 it is change into I and is not 30.2/31 . Why is it. Please reply me if you have time. Pardon me if i wrote faulty because i am not fluency in english. Thank.
sir, my Q is how find / calculate the amps if the conductor size of cable is 32/0.2 mm and v 440
Dear sir
Please answer this
20 HP,415 v,0.8 of,3phase,0.85efficiency,1440 rpm and delta connected induction motor is to be connected to a motor control center by a cable of length 15 m,this cable is running with three other cables,ambient temperature is 45 degree Celsius and fault level is 20 KA,select the size of the cable from below
PVC cable 300mm.value of k/cu 103.value of k/al 68
XLPE cable <300mm.value of k/cu 114.value of k/al 92
It’s really great job to learn online from basic to till all experience fields i would like to have a request for lesson
1. Testing & commission of electrical equipment in substaion
2. CNC machines , types , work principles
3. Site Engineer work in Substaions
4. Single line diagram module and it’s implementation
5. QA or QC work and their Electrical Standards.
It will be very thankful if you can provide these important lessons to us to gain knowledge through your website atleast with video upload to have knowledge in theory as well as in practicle
Hello ….. could you please calculate the 3phase question using the metric table. The answer I am getting is unrealistic.
Hi Thomas ?
Can you please elaborate more in details ?
Sir,
I requesting to you please, make the blog and video of electrical wiring installation of any flate/home.
And your blogs are so useful for us.
Thanking you.
Having determine my total load requirements after considering all factors
for example say load current is 750A for a 3phase supply
If l want to use single core pvc/ pvc cable
Do l divide the 750A by 3 before chosen the corresponding cable size from the catalog or not
Regards
Bosun
Good way of describing, and nice piece of writing to obtain data about my presentation subject, which i am going
to convey in school.
Tanks you sir
How to connecting RCCB and RCB
Hi!
I use more simple way. I find R of cable R=ρ ×L/A, where
R is the resistance of the conductor in Ohms;
L is the length of the conductor in meters;
ρ is the electrical resistivity (also known as the specific electrical resistance) of a conductor, for cooper ρ=0,018 Ohm · mm²/m;
A is the crosssectional area, measured in square millimeters.
Then find voltage drop
for L+N voltage system, dU (V)= I×R×2,
for L+L voltage system, dU (V)= I×R,
Where I is a current in wire;
dU(V) is a voltage drop in volt.
And in the end I find dU in %
dU(%)=dU(V)×100/U
where U is a line voltage.
Sorry for my english.
Very good sir.
Thank you for this blog
Please add tutorials about MC, CB, O/L, FUSE, T/F, generators and motors suitable size calculations.
Hi, I am very impress of this blog . It is very much helpful for us as electrical engineer students in our respective areas of interest .
Please , for the information above , I really want a PDF copy of it .
Thanks once more …
Assalam alikum sir
Why power factor is not taken in calculation as it is AC circuit?
I am very impressed of ur blogs. pl post a table of cable & alu conductor current rating of different sizes to b use in ht/ lt lines in state utilities in day today practice.