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Home » Basic/Important Electrical Formulas » How to Find The Suitable Size of Cable & Wire for Electrical Wiring Installation (Solved Examples in British and SI System)

How to Find The Suitable Size of Cable & Wire for Electrical Wiring Installation (Solved Examples in British and SI System)

How to determine the suitable size of cable for Electrical Wiring Installation

Voltage drop in Cables

We know that all conductors and cables (except Super conductor) have some amount of resistance.
This resistance is directly proportional to the length and inversely proportional to the diameter of conductor R ∝ L/a [Laws of resistance R = ρ (L/a)]
Whenever current flows through a conductor, a voltage drop occurs in that conductor. Generally, voltage drop may neglect for low length conductors but in a lower diameter and long length conductors, we cannot neglect that voltage drops.
According to IEEE rule B-23, at any point between power supply terminal and installation, Voltage drop should not increase above 2.5% of provided (supply) voltage.
Example: if the Supply voltage is 220V, then the value of allowable voltage drop should be;
Allowable Voltage Drop = 220 x (2.5/100) = 5.5V
In electrical wiring circuits, voltage drops also occur from the distribution board to the different sub circuit and final sub circuits, but for sub circuits and final sub circuits, the value of voltage drop should be half of that allowable voltage drops (i.e. 2.75V of 5.5V in the above case)
Normally, Voltage drop in tables is described in Ampere per meter (A/m) e.g. what would be the voltage drop in a one meter cable which carrying one Ampere current?
There are two methods to define the voltage drop in a cable which we will follow.
In SI (System international and metric system) voltage drop is described by ampere per meter.

In FPS (foot pound system) voltage drop is described in 100feet. 

  • Update: Now you can also use the following Calculators to find Voltage drop & the wire size in American wire gauge.
  1. Electrical Wire & Cable Size Calculator (Copper& Aluminum)
  2. Wire & Cable Size Calculator in AWG
  3. Voltage Drop in Wire & Cable Calculator
Tables & Charts for Suitable Cable & Wire Sizes
Below are the important Tables which you should follow for determining the proper size of cable for Electrical Wiring Installation.
 Click image to enlargeTable-1-current-rating-of-Copper-cables-at-86F-or-30C
Click image to enlargeTable-Chart-current-rating-of-flexible-cords-Copper-cables-at-86F-30CClick image to enlargeHow to Find The Suitable Size of Cable & Wire for Electrical Wiring Installation (Solved Examples in British and SI System)
Click image to enlargeHow to Find The Suitable Size of Cable & Wire |Solved Examples steb by step Click image to enlargeTable-Cable-Size-Current-Rating-with-voltage-drop-Metric-Decimal-SI-System

 To find voltage drop in a cable, follow these simple steps

  • First of all, find the maximum allowable voltage drop
  • Now, Find load current
  • Now, according to load current, select a proper cable (which current rating should be nearest to the calculated load current) from table 1
  •  From Table 1, find the voltage drop in meter or 100feet (what system you prefer) according its rated current

(Stay cool 🙂 we will follow both methods and system for finding voltage drops (in meter and 100feet) in our solved example for whole electrical installation wiring).

  • Now, calculate the voltage drop for the actual length of wiring circuit according to its rated current with the help of following formulas.

(Actual length of circuit x volt drop for 1m) /100 —->       to find Volt drop in per meter.
(Actual length of circuit x volt drop for 100ft) /100—> to find volt drop in 100feet.

  • Now multiply this calculated value of volt drop by load factor where;

Load factor = Load Current to be taken by Cable/ Rated Current of Cable given in the table.

  • This is the value of Volt drop in the cables when load current flow through it.
  • If the calculated value of voltage drop is less than the value calculated in step (1) (Maximum allowable voltage drop), than the size of selected cable is proper
  • If the calculated value of voltage drop is greater than the value calculated in step (1) (Maximum allowable voltage drop), than calculate voltage drop for the next (greater in size) cable and so on until the calculated value of voltage drop became less than the maximum allowable voltage drop calculated in step (1)

How to determine the proper Cable Size for Given Load (with Examples)

For a given load, cable size may be found with the help of different tables but we should keep in mind and follow the rules about voltage drop.
Determining the size of cable for a given load, take into account the following rules.
For a given load except the known value of current, there should be 20% extra scope of current for additional, future or emergency needs.
From Energy meter to Distribution board, Voltage drop should be 1.25% and for final sub circuit, voltage drop should not exceed 2.5% of Supply voltage.
Consider the change in temperature, when needed, use temperature factor (Table 3)
Also, consider the load factor when finding the size of cable
When determining the cable size, consider the wiring system i.e. in open wiring system, temperature would be low but in conduit wiring, temperature increases due to the absence of air.

Solved Examples of Proper Wire & Cable Size

Following are the examples of determining the proper Size of cables for electrical wiring installation which will make it easy to understand the method of “how to determine the proper size of cable for a given load”.
Example 1 ……….. (British / English System)
For Electrical wiring installation in a building, Total load is 4.5kW and total length of cable from energy meter to sub circuit distribution board is 35 feet. Supply voltages are 220V and temperature is 40°C (104°F). Find the most suitable size of cable from energy meter to sub circuit if wiring is installed in conduits.
Solution:-
  • Total Load = 4.5kW = 4.5 x1000W = 4500W
  • 20% additional load = 4500 x (20/100) = 900W
  • Total Load = 4500W + 900W = 5400W
  • Total Current = I = P/V = 5400W /220V =24.5A
Now select the size of cable for load current of 24.5A (from Table 1) which is 7/0.036 (28 Amperes) it means we can use 7/0.036 cable according table 1.
Now check the selected (7/0.036) cable with temperature factor in Table 3, so the temperature factor is 0.94 (in table 3) at 40°C (104°F) and current carrying capacity of (7/0.036) is 28A, therefore, current carrying capacity of this cable at 40°C (104°F) would be;
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.
Now find the voltage drop for 100feet 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 35feet cable would be;
Actual Voltage drop for 35feet = (7 x 35/100) x (24.5/28) = 2.1V
And Allowable voltage drop = (2.5 x 220)/100 = 5.5V
Here The Actual Voltage Drop (2.1V) is less than that of maximum allowable voltage drop of 5.5V. Therefore, the appropriate and most suitable cable size is (7/0.036) for that given load for Electrical Wiring Installation.
Example 2 ……. ( SI / Metric / Decimal System )
What type and size of cable suits for given situation
Load = 5.8kW
Volts = 230V
Length of Circuit = 35meter
Temperature = 35°C (95°F)
Solution:-
Load = 5.8kW = 5800W
Voltage = 230V
Current = I = P/V = 5800 / 230 = 25.2A
20% additional load current = (20/100) x 5.2A = 5A
Total Load Current = 25.2A + 5A = 30.2A
Now select the size of cable for load current of 30.2A (from Table 1) which is 7/1.04 (31 Amperes) it means we can use 7/0.036 cable according table 1.
Now check the selected (7/1.04) cable with temperature factor in Table 3, so the temperature factor is 0.97 (in table 3) at 35°C (95°F) and current carrying capacity of (7/1.04) is 31A, therefore, current carrying capacity of this cable at 40°C (104°F) would be;
Current rating for 35°C (95°F) = 31 x 0.97 = 30 Amp.
Since the calculated value (30 Amp) at 35°C (95°F) is less than that of current carrying capacity of (7/1.04) cable which is 31A, therefore this size of cable (7/1.04) is also suitable with respect to temperature.
Now find the voltage drop for per ampere meter for this (7/1.04) cable from (Table 5) which is 7mV, But in our case, the length of cable is 35 meter.  Therefore, the voltage drop for 35 meter cable would be:
Actual Voltage drop for 35meter =
= mV x I x L
(7/1000) x 30×35 = 7.6V
And Allowable voltage drop = (2.5 x 230)/100 = 5.75V
Here the actual Voltage drop (7.35V) is greater than that of maximum allowable voltage drop of 5.75V. Therefore, this is not suitable size of cable for that given load. So we will select the next size of selected cable (7/1.04) which is 7/1.35 and find the voltage drop again. According to Table (5) the current rating of 7/1.35 is 40Amperes and the voltage drop in per ampere meter is 4.1 mV (See table (5)). Therefore, the actual voltage drop for 35 meter cable would be;
Actual Voltage drop for 35meter =
= mV x I x L
(4.1/1000) x 40×35 = 7.35V = 5.74V
This drop is less than that of maximum allowable voltage drop. So this is the most appropriate and suitable cable or wire size.
Example 3
Following Loads are connected in a building:-
Sub-Circuit 1
  • 2 lamps each o 1000W and
  • 4 fans each of 80W
  • 2 TV each of 120W
Sub-Circuit 2
  • 6 Lamps each of 80W and
  • 5 sockets each of 100W
  • 4 lamps each of 800W
If supply voltages are 230V then calculate circuit current and Cable size for each Sub-Circuit?
Solution:-
Total load of Sub-Circuit 1
= (2 x 1000) + (4 x 80) + (2×120)
= 2000W + 320W + 240W = 2560W
Current for Sub-Circuit 1 = I = P/V = 2560/230 = 11.1A
Total load of Sub-Circuit 2
= (6 x 80) + (5 x 100) + (4 x 800)
= 480W + 500W + 3200W= 4180W
Current for Sub-Circuit 2 = I = P/V = 4180/230 = 18.1A
Therefore, Cable suggested for sub circuit 1 = 3/.029” (13Amp) or 1/1.38mm (13Amp)
Cable suggested for Sub-Circuit 2 = 7/.029” (21Amp) or 7/0.85mm (24Amp)
Total Current drawn by both Sub-Circuits = 11.1A + 18.1A = 29.27
So cable suggested for Main-Circuit = 7/.044” (34Amp) 0r 7/1.04mm (31Amp)
Example 4
A 10H.P (7.46kW) three phase squirrel cage induction motor of continuous rating using Star-Delta starting is connected through 400V supply by three single core PVC cables run in conduit from 250feet (76.2m) away from multi-way 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 motor?
Solution:-
  • Motor load = 10H.P = 10 x 746 = 7460W     *(1H.P = 746W)
  • Supply Voltage = 400V (3-Phase)
  • 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 4) which is 7/0.36” (23 Amperes) *(Remember that this is a 3-phase system i.e. 3-core cable) and the voltage drop is 5.3V for 100Feet. It means we can use 7/0.036 cable according Table (4).
Now check the selected (7/0.036) cable with temperature factor in table (3), so the temperature factor is 0.97 (in table 3) at 35°C (95°F) and current carrying capacity of (7/0.036”) is 23 Amperes, therefore, current carrying capacity of this cable at 40°C (104°F) would be:
Current rating for 40°C (104°F) = 23 x 0.97 = 22.31 Amp.
Since the calculated value (22.31 Amp) at 35°C (95°F) is less than that of current carrying capacity of (7/0.036) cable which is 23A, therefore this size of cable (7/0.036) is also suitable with respect to temperature.
Load factor = 19/23 = 0.826
Now find the voltage drop for 100feet for this (7/0.036) cable from table (4) which is 5.3V, But in our case, the length of cable is 250 feet.  Therefore, the voltage drop for 250 feet cable would be;
Actual Voltage drop for 250feet = (5.3 x 250/100) x 0.826 = 10.94V
And maximum Allowable voltage drop = (2.5/100) x 400V= 10V
Here the actual Voltage drop (10.94V) is greater than that of maximum allowable voltage drop of 10V. Therefore, this is not suitable size of cable for that given load. So we will select the next size of selected cable (7/0.036) which is 7/0.044 and find the voltage drop again. According to Table (4) the current rating of 7/0.044 is 28Amperes and the volt drop in per 100feet is 4.1V (see Table 4). Therefore, the actual voltage drop for 250feet cable would be;
Actual Voltage drop for 250feet =
= Volt drop per 100feet x length of cable x load factor
(4.1/100) x 250 x 0.826 = 8.46V
And Maximum Allowable voltage drop = (2.5/100) x 400V= 10V
The actual voltage drop is less than that of maximum allowable voltage drop. So this is the most appropriate and suitable cable size for Electrical wiring Installation of given situation.
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72 comments

  1. Thanks Sir,<br /><br />It Gives me all the answers for my question.<br /><br />Thanks Once Again.

  2. Sir I could not get this &quot;Number of wires &amp; Thikness of each wires&quot; (7 / 0.0229)……….. (70 / 0.0076)<br />Kindly give me answer.<br />

  3. Seems like I am doing exam&#39;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

  4. Located in Tianchang city, Anhui province, Tianchang Zhengjie Wire Drawing Die Factory is a professional manufacturer of diamond wire drawing dies in China. <a href="http://www.philatron.com/wire-cable-manufacturer/&quot; rel="nofollow">SBR cable</a><br />

  5. sir, can you tell me how to fish a wire through a conduit pipe..

  6. 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.

    • Mateeullah Khan

      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

  7. 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

  8. Thank you v much for this useful information…….. Stay blessed.

  9. Great Job,thanks a lot for the post.

  10. the cable the tech&#39;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/&quot; rel="nofollow">750 mcm cable</a><br />

  11. make a thread where show how to do home conduit wiring n all<br />btw grt work

  12. 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/about-us/&quot; rel="nofollow">Experienced Industrial Electricians At Kowelec Bendigo</a>

  13. gud job you gave the information about copper cables only..wat abt Alluminium armoured cables<br />

  14. than you so much for this useful link

  15. Great site. Wondering the source of tables, is it from any standards? ie. IEC, NEC, etc.<br />Thanks.

  16. sir plz mention the standard according to which the tables have been made.<br /><br />

  17. i want selection of flexible cable standard chart with electric load

  18. If you look carefully at the calculations there are some basic mistake which would effect the final result. ie 2 TV&#39;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.

  19. sir plz mention the standard according to which the tables have been made.

  20. 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

  21. 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

  22. Would you tell me about the IEC standard in Electrical Installation design and the importance of IEC standard. <br />Thanks for your nice post

  23. if increase frequency value r decrease then what will happen?

  24. trimurthy reddy

    Dear Sir.

    Thank u very much much for deep explanation.
    Do we need to consider power factor in the calculation of load current…..

  25. But,when temprature of conductor increases then resistivty decreases.

  26. Muhammad Atif Shehzad

    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.

  27. Adekeye Lukman

    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

  28. thanks for the simple and straight forward language for communication. Please how did you arrive at 5800KW in example 2 above.

  29. thankz a lot bro

  30. nice it’s very helpful for me thanks

  31. Chandrakant borhade

    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

  32. 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 .

  33. 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!?

  34. May Allah ( God ) Bless you for your this type of kind effort , it’s very helpful for me and for electrical related engineers

    thanks

  35. how to calculate 3 phase 400 v and 600 amp load’s cable size which must have a length of 25 meter

  36. how we can make selection of cable for our system

  37. Dear sir, You have done a great job. It’s very helpful to me. You may live long.
    Amarasiri from SL.

  38. 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

  39. Dear any body feed back regarding cable selection depending load & length of cable 3 phase application

  40. used cable size is 10 mm load 34.6 Amps 460 .3Ph 60 hz motor

  41. 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.

  42. Sorry, forgot to mention it needs to be 3-phase

  43. Are all made computations, in the samples above, AC or DC? Pleas I need answer.

  44. Tshering Dukpa

    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???

  45. Please, what size of cable can I use to wire an equipment of 3-phase, 4-wire 120KVA, 50Hz with power factor of 0.8?

  46. 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..?

  47. DiptiPrakash Palai

    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?

  48. 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.

  49. state two factors to be considered in selection of cables sizes for a house wiring

  50. 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.

  51. 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

  52. sir plz mention the standard according to which the tables have been made.

  53. Total load 32 KW 3phase 4wire system (distribute)220 volt 1phase and length 140 Meter .what is cable size use in mm

  54. sir, please provide below information with necessary calculations
    for 200/150 HP/Kw 3-phase 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

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