# How to Find The Suitable Size of Cable & Wire for Electrical Wiring Installation? – Examples in British and SI System

**How to Determine the Proper Size of Wire & 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 i.e.

**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 small length of conductors but in case of a lower diameter and long length conductors, we have to take into account the considerable voltage drops for proper wiring installation and future load managment.

According to **IEEE rule B-23**, at any point between power supply terminal and installation, **Voltage drop should not increase above 2.5% of the provided (supply) voltage**.

- Related Post
*:***How to calculate the Cable size for LT & HT Motors?**

**Example:**

if the Supply voltage is 220V AC, 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 as calculated above)

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 discuses below.

In **SI** (**System international and metric system**) voltage drop is described by **ampere per meter (A/m)**.

In **FPS (foot pound system)** voltage drop is described in length based which is **100feet.**

**Update**: Now you may also use the following electrical Calculators to find**Voltage drop & the wire size in American wire gauge**system.

**Electrical Wire & Cable Size Calculator (Copper& Aluminum)****Wire & Cable Size Calculator in AWG****Voltage Drop in Wire & Cable Calculator**

**Tables & Charts for Proper Cable & Wire Sizes**

Below are the important Tables which you should follow for determining the proper size of cable for Electrical Wiring Installation.

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**How to Find the Voltage Drop in a Cable?**

**To find voltage drop in a cable, follow the simple steps given below.**

- 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 100 feet (what system you prefer) according its rated current

(Stay cool 🙂 we will follow both methods and system for finding voltage drops (in meter and 100 feet) 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 100 feet.

- 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).

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**How to Determine the Proper Cable & Wire Size for Given Load?**

**Below are solved examples showing how to find the proper Cable Size for Given Load.**

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.

- Note: Keep in mind Diversity Factor in Electrical Wring Installation while selecting the proper size of cable for electrical wiring installation

Related Posts:

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- How To Locate Faults In Cables? Cable Faults, Types & Causes
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**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)*

*(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 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;

Actual Voltage drop for 35 feet = (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 AV

Length of Circuit = 35 meter

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 35 meter =

**= 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 230 V AC, 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**” (

**13 Amp**) or

**1/1.38 mm**(

**13 Amp**)

** Cable suggested for Sub-Circuit 2** =

**7/.029**” (

**21 Amp**) or

**7/0.85 mm (24 Amp)**

Total Current drawn by both Sub-Circuits = 11.1A + 18.1A = **29.27 A**

So ** cable suggested for Main-Circuit** =

**7/.044” (34 Amp)**or

**7/1.04 mm (31 Amp**)

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

Welcome Dear….and Thanks 4 appreciation…

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/a-complete-note-on-solar-panel.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/about-us/" 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/advance-voltage-drop-calculator-voltage-drop-formula.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 3-phase

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 3-phase, 4-wire 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 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

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 three-phase 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 3-phase 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 cross-sectional 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 …