Hi everyone today i am here to guide you about the

Before we start talking about the selection of cable according to load or selection of circuit breaker according to load, first we need to discuss some important points. As you know that i already publish a post and video tutorial about the laws of resistance. And we can not ignore the resistance.

In law's of Resistance we known that the resistance is directly proportional to Length , so if we in cress the cable length then our cable resistance will also increase and due to this we face some voltage drop in electric cables.

So in low length of electric cable we can neglects the voltage drop but in long length of electric cable we can not neglects the electric cable, so before we calculate the load or to chose electric cable , we need to talk about the voltage drop.

According IEEE role B-23 voltage drop should not be in cress form 2.5 % of incoming supply, If the provided electric supply is 220 volts.

Then 220 x (2.5/100) = 5.5 volts.

So the available voltage drop is 5.5 volts for 220 volts supply.

Here is the electric cable sizes and it's rated load table in which i shown the wire size , it's current carrying capacity and voltage drop. This table help you to find out the suitable cable size for your house wiring.

Temperature is also affect on electric wire , so we can't neglects the temperature , a electric cable have can get different current load capacity on different temperature. Here the temperature factor table which help to chose the correct size of electric wire.

For load calculation we use formula power/pressure = load

for example 100W/220V = 0.45A (note that this formula can be used for resistive load or DC Load)

So lets take a house electrical wiring load calculation

For example our house have 2 rooms and one kitchen, and we want to find out it's complete load. First we need to find out the one room total load.

In one room we have one ceiling fan, 4 light bulb or energy sever light, one Air conditioner and outlets. So we divide the load in two parts, Load 1 and Load 2,

(Ceiling fan, lights and outlets will be numbering in load 1)

(Air conditioner and room heater will be numbering in load 2)

In outlets we can use iron, stand fan, laptop or mobile charger etc. First we need to plus the complete load.

Load calculation of Room (load 1)

Ceiling fan 80 x 1 = 80 watt

Lights 4 x 25 = 100 watt

Iron = 1x 1000 = 1000 watt

Laptop and Mobile charger etc = 200 watt

These are load 1 of a room, so first we need to plus all loads

80+100+500+200 + 880 watts

We need to put in 20 % additional load so 1380 x(20/100) = 276 watt

The complete load is 1380 + 276 = 1656 watt

Our incoming supply is 220 volts so 1656/220 = 7.52 amps

So our complete load 1 appliances is 7.52 ampere.

(Here is the ceiling fan not a resistive load , it's a inductive load, for inductive load we also do caluclation of power factor, but ceiling fan watt is not to high that we do inductive load calculation , however if we have high number of fan in building then we must fallow the inductive load formula.)

Load 2 of a room are air conditioner or electric heater, Nowadays air conditioner are mostly 1650 watts or above , or it's can be lover or high then this , However if your air conditioner is 1650 watts, and electric heater 2000 watts .

So As we did not use both on same time then we need to calculate the only one load , so we select the high load , so 2000 watts is the high load.

So 2000 watt / 220 volts = 9.09 ampere, but the other load is Air conditioner and air conditioner have internally install a induction motor in compressor , so it's take double current in starting,

Therefor

According to the inductive load formula

power / (pressure x Power factor)

So Watt/ (volts x PF)

1650 / (220 x 0.95) = 7.89 Ampere

To plus the starting current of Air conditioner then our total load is 7.8 x 2 = 15.6 amps

So our load 2 of a room is 15.6 amps

(Room 2 have also same load like Room one there for we did not calculate the room 2 load)

Now come to calculate the kitchen load,

Ceiling fan 80 x 1 = 80W

lights 25 x 4 = 100W

Outlets Load 1200W

So the total load is 1380 same like a room, now plus 20 % additional load then complete load is

1380 x (20/100) = 276

1380 + 276 = 1656 W

So the load is 1656 watt / 220 volts = 7.5

Outlet load can be a fridge / refrigerator , or other electrical appliances , however we did not fallow the inductive load formula , we use the resistive load formula.

As you know that we also use the water heater in our house, in Pakistan we use mostly use 1200 watt element in water heater geyser, so if we have to water heater in our house or one water heater in which the element is install 2400.

So this is resistive load therefore we use resistive load calculation formula.

2400 / 220 = 10.9

Water Pump Motor Load Calculation.

Mostly we use 1 HP water pump motor in our house , we know that 1 HP motor is 746 watt. So this is also a induction motor there fore we use the inductive load calculation formula.

746 / (220 x 0.95) = 3.5 So to plus the staring current the total load is = 7 A

If your environment temperature is 40 Centigrade then according temperature factor current carrying capacity of 2.5 mm cable on 104 Fahrenheit.

So 40 °C is 140 °F therefor according cable temperature factor table is 0.94

So the 2.5mm (7/0.29) current carrying capacity is 18 Ampere there for

18 x 0.94 = 16.9 amps

And our total (load 1) is 7.5 amps , and the 2.5 mm current carrying capacity on 40 °C is 16.1 amps therefor the 2.5 mm or 7/0.29 wire is suitable wire for this load.

Now come to Load 2 in which the air conditioner get 15.6 ampere of current in starting time, so we chose the cable regarding starting current.

So the total current is 15.6 amps , so according the suitable cable table the suitable wire size 2.5 mm but when it's come to temperature factor then

(if temperature is 40 °C) 18 x 0.94 = 16.9 amps

Our starting current of air conditioner is 15.6 amps , and 2.5 mm wire current carrying capacity on 40 C is 16.9 there for 2.5 mm wire is suitable for this load.

(The 2.5 mm cable is best for this load according the load calculation and formula but when you change your Air conditioner or if air conditioner get more current on starting time, then your cable will not working , so if you use 4 mm wire for your air conditioner then this will best)

So we find out that 2.5 mm of cable is best for room load 1 and 4 mm wire for room load 2. The 2nd room have also same load like room one therefore we use the same size of cable for 2nd room.

Now come to kitchen load, the kitchen load 1 is same like a room load 1 therefore we use same wire size as we chose for room load 1.

Water heater total load is 10.9, according to the temperature factor and cable table the 2.5 mm wire is best for this load.

Water Pump Motor Supply Cable Selection

Our staring current of water pump motor is 7 A and according to the table 1.5 mm wire is best for this load, but when it's come to the temperature factor table , so the 1.5 wire current carrying capacity on 40 °C is

7 x 0.94 = 6.5 A

After temperature factor calculation , we know that the 1.5 mm wire is not suitable for water pump motor , so we use the 2nd wire on table. The next wire is 2.5 mm wire, so suitable cable size for single phase water pump induction motor is 2.5 mm.

How to chose suitable size of circuit breaker for room (MCB)

Room one (load 1) total load is 7.5 ampere, there for we chose 10 ampere circuit breaker

Room one (load 2) total load is 15.6 ampere there for we chose 20 ampere circuit breaker

Room two is same load like room one therefore we chose same MCB breakers for this room

Kitchen (load 1) total load is 7.5 ampere there for we select 10 ampere CB.

Water heater total load is 10.90 ampere therefor we select 16 ampere Circuit breaker.

If we have water pump motor of 1 HP in our house then

1 HP = 746 Watt

746 / (220 x 0.95) = 3.56

So it's will take is starting time double current 3.56 x 2 = 7.12

So the suitable size of electric cable is 2.5 mm according table 1.

So Our Total load is 52 Amps, so the suitable size of RCD circuit breaker is 63 Amps and double pole main circuit breaker is also 63 amps.

As you know that all electrical appliances are not running on same time and mostly 70% load running on all time

So our total load is

52 amps

52 x 70 / 100 = 36 amps So the suitable size of electric cable for main supply according table 1 is 16 mm (7.064). However if you running all the load on same time then you must chose the electric cable according you total load.

(Note that the above electric current carrying capacity load table / chart is for number one quality wires)

I try my level best to make you understood about the selection of cable size and circuit breaker , but still if you did not understood it completely then watch the below video tutorial which in your own mother language Urdu & Hindi. One thing more that in video tutorial is discuss some more important point which i did not write in this post text. So for better understanding kindly watch the below video tutorial

I hope after watching the above video tutorial about the

**selection of cable size circuit breaker for single phase house wiring**, form this post you will completely understand to find out the suitable size of circuit breaker and electric cable size with complete examples and cable size chart. In this post i will completely guide you that how to calculate the load of electrical house appliances. I will discuss each step and also a video tutorial which well be best for you to make you complete understood.## Selection of cable size - wire size and Selection of suitable size of Circuit breaker for house wiring

Before we start talking about the selection of cable according to load or selection of circuit breaker according to load, first we need to discuss some important points. As you know that i already publish a post and video tutorial about the laws of resistance. And we can not ignore the resistance.

In law's of Resistance we known that the resistance is directly proportional to Length , so if we in cress the cable length then our cable resistance will also increase and due to this we face some voltage drop in electric cables.

So in low length of electric cable we can neglects the voltage drop but in long length of electric cable we can not neglects the electric cable, so before we calculate the load or to chose electric cable , we need to talk about the voltage drop.

According IEEE role B-23 voltage drop should not be in cress form 2.5 % of incoming supply, If the provided electric supply is 220 volts.

Then 220 x (2.5/100) = 5.5 volts.

So the available voltage drop is 5.5 volts for 220 volts supply.

Here is the electric cable sizes and it's rated load table in which i shown the wire size , it's current carrying capacity and voltage drop. This table help you to find out the suitable cable size for your house wiring.

Temperature is also affect on electric wire , so we can't neglects the temperature , a electric cable have can get different current load capacity on different temperature. Here the temperature factor table which help to chose the correct size of electric wire.

### House Electrical Load Calculation (resistive and inductive)

Now come to load calculation that how can we calculate the electrical load. Here i try my best to make you understood about the load calculation using a simple load calculation formula.For load calculation we use formula power/pressure = load

for example 100W/220V = 0.45A (note that this formula can be used for resistive load or DC Load)

### How to Calculate the Complete House Wiring Load

So lets take a house electrical wiring load calculation

For example our house have 2 rooms and one kitchen, and we want to find out it's complete load. First we need to find out the one room total load.

In one room we have one ceiling fan, 4 light bulb or energy sever light, one Air conditioner and outlets. So we divide the load in two parts, Load 1 and Load 2,

(Ceiling fan, lights and outlets will be numbering in load 1)

(Air conditioner and room heater will be numbering in load 2)

In outlets we can use iron, stand fan, laptop or mobile charger etc. First we need to plus the complete load.

Load calculation of Room (load 1)

Ceiling fan 80 x 1 = 80 watt

Lights 4 x 25 = 100 watt

Iron = 1x 1000 = 1000 watt

Laptop and Mobile charger etc = 200 watt

These are load 1 of a room, so first we need to plus all loads

80+100+500+200 + 880 watts

We need to put in 20 % additional load so 1380 x(20/100) = 276 watt

The complete load is 1380 + 276 = 1656 watt

Our incoming supply is 220 volts so 1656/220 = 7.52 amps

So our complete load 1 appliances is 7.52 ampere.

(Here is the ceiling fan not a resistive load , it's a inductive load, for inductive load we also do caluclation of power factor, but ceiling fan watt is not to high that we do inductive load calculation , however if we have high number of fan in building then we must fallow the inductive load formula.)

Load 2 of a room are air conditioner or electric heater, Nowadays air conditioner are mostly 1650 watts or above , or it's can be lover or high then this , However if your air conditioner is 1650 watts, and electric heater 2000 watts .

So As we did not use both on same time then we need to calculate the only one load , so we select the high load , so 2000 watts is the high load.

So 2000 watt / 220 volts = 9.09 ampere, but the other load is Air conditioner and air conditioner have internally install a induction motor in compressor , so it's take double current in starting,

Therefor

According to the inductive load formula

power / (pressure x Power factor)

So Watt/ (volts x PF)

1650 / (220 x 0.95) = 7.89 Ampere

To plus the starting current of Air conditioner then our total load is 7.8 x 2 = 15.6 amps

So our load 2 of a room is 15.6 amps

(Room 2 have also same load like Room one there for we did not calculate the room 2 load)

Now come to calculate the kitchen load,

Ceiling fan 80 x 1 = 80W

lights 25 x 4 = 100W

Outlets Load 1200W

So the total load is 1380 same like a room, now plus 20 % additional load then complete load is

1380 x (20/100) = 276

1380 + 276 = 1656 W

So the load is 1656 watt / 220 volts = 7.5

Outlet load can be a fridge / refrigerator , or other electrical appliances , however we did not fallow the inductive load formula , we use the resistive load formula.

As you know that we also use the water heater in our house, in Pakistan we use mostly use 1200 watt element in water heater geyser, so if we have to water heater in our house or one water heater in which the element is install 2400.

So this is resistive load therefore we use resistive load calculation formula.

2400 / 220 = 10.9

Water Pump Motor Load Calculation.

Mostly we use 1 HP water pump motor in our house , we know that 1 HP motor is 746 watt. So this is also a induction motor there fore we use the inductive load calculation formula.

746 / (220 x 0.95) = 3.5 So to plus the staring current the total load is = 7 A

### Selection of cable size / wire size selection for electrical load

The complete load 1 of room is 7.5 ampere , so according the wire size table the suitable size of cable is 2.5mm wire (7/.029), because it's can take 18 amps.If your environment temperature is 40 Centigrade then according temperature factor current carrying capacity of 2.5 mm cable on 104 Fahrenheit.

So 40 °C is 140 °F therefor according cable temperature factor table is 0.94

So the 2.5mm (7/0.29) current carrying capacity is 18 Ampere there for

18 x 0.94 = 16.9 amps

And our total (load 1) is 7.5 amps , and the 2.5 mm current carrying capacity on 40 °C is 16.1 amps therefor the 2.5 mm or 7/0.29 wire is suitable wire for this load.

Now come to Load 2 in which the air conditioner get 15.6 ampere of current in starting time, so we chose the cable regarding starting current.

So the total current is 15.6 amps , so according the suitable cable table the suitable wire size 2.5 mm but when it's come to temperature factor then

(if temperature is 40 °C) 18 x 0.94 = 16.9 amps

Our starting current of air conditioner is 15.6 amps , and 2.5 mm wire current carrying capacity on 40 C is 16.9 there for 2.5 mm wire is suitable for this load.

(The 2.5 mm cable is best for this load according the load calculation and formula but when you change your Air conditioner or if air conditioner get more current on starting time, then your cable will not working , so if you use 4 mm wire for your air conditioner then this will best)

So we find out that 2.5 mm of cable is best for room load 1 and 4 mm wire for room load 2. The 2nd room have also same load like room one therefore we use the same size of cable for 2nd room.

Now come to kitchen load, the kitchen load 1 is same like a room load 1 therefore we use same wire size as we chose for room load 1.

Water heater total load is 10.9, according to the temperature factor and cable table the 2.5 mm wire is best for this load.

Water Pump Motor Supply Cable Selection

Our staring current of water pump motor is 7 A and according to the table 1.5 mm wire is best for this load, but when it's come to the temperature factor table , so the 1.5 wire current carrying capacity on 40 °C is

7 x 0.94 = 6.5 A

After temperature factor calculation , we know that the 1.5 mm wire is not suitable for water pump motor , so we use the 2nd wire on table. The next wire is 2.5 mm wire, so suitable cable size for single phase water pump induction motor is 2.5 mm.

How to chose suitable size of circuit breaker for room (MCB)

### Selection of suitable circuit breaker for house wiring (MCB & RCD)

Room one (load 1) total load is 7.5 ampere, there for we chose 10 ampere circuit breaker

Room one (load 2) total load is 15.6 ampere there for we chose 20 ampere circuit breaker

Room two is same load like room one therefore we chose same MCB breakers for this room

Kitchen (load 1) total load is 7.5 ampere there for we select 10 ampere CB.

Water heater total load is 10.90 ampere therefor we select 16 ampere Circuit breaker.

If we have water pump motor of 1 HP in our house then

1 HP = 746 Watt

746 / (220 x 0.95) = 3.56

So it's will take is starting time double current 3.56 x 2 = 7.12

So the suitable size of electric cable is 2.5 mm according table 1.

So Our Total load is 52 Amps, so the suitable size of RCD circuit breaker is 63 Amps and double pole main circuit breaker is also 63 amps.

As you know that all electrical appliances are not running on same time and mostly 70% load running on all time

So our total load is

52 amps

52 x 70 / 100 = 36 amps So the suitable size of electric cable for main supply according table 1 is 16 mm (7.064). However if you running all the load on same time then you must chose the electric cable according you total load.

(Note that the above electric current carrying capacity load table / chart is for number one quality wires)

### How to find out suitable size of electric cable / wires and Circuit breakers (Urdu & Hindi Video Tutorial)

I try my level best to make you understood about the selection of cable size and circuit breaker , but still if you did not understood it completely then watch the below video tutorial which in your own mother language Urdu & Hindi. One thing more that in video tutorial is discuss some more important point which i did not write in this post text. So for better understanding kindly watch the below video tutorial

I hope after watching the above video tutorial about the

*finding of suitable size of electric cable and circuit breaker according to the load*, now you will be complete understood. However if still you have question or any suggestion regarding the post or video tutorial then you can use the below comment system box.