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# An example how to calculate voltage drop and size of electrical cable

Home / Technical Articles / An example how to calculate voltage drop and size of electrical cable

## Input information

### Electrical details:

Electrical load of 80KW, distance between source and load is 200 meters, system voltage 415V three phase, power factor is 0.8, permissible voltage drop is  5%, demand factor is 1.

### Cable laying detail:

Cable is directed buried in ground in trench at the depth of 1 meter. Ground temperature is approximate 35 Deg. Number of cable per trench is 1. Number of run of cable is 1 run.

### Soil details:

Thermal resistivity of soil is not known. Nature of soil is damp soil.

## Ok, let’s dive into calculations…

Consumed Load in KW = 80 · 1 = 80 KW
• Consumed Load in KVA = KW/P.F.:
Consumed Load in KVA = 80/0.8 = 100 KVA
• Full Load Current = (KVA · 1000) / (1.732 · Voltage):
Full Load Current = (100 · 1000) / (1.732 · 415) = 139 Amp.

Calculating Correction Factor of Cable from following data:

#### Temperature Correction Factor (K1) When Cable is in the Air

 Temperature Correction Factor in Air: K1 Ambient Temperature Insulation PVC XLPE/EPR 10 1.22 1.15 15 1.17 1.12 20 1.12 1.08 25 1.06 1.04 35 0.94 0.96 40 0.87 0.91 45 0.79 0.87 50 0.71 0.82 55 0.61 0.76 60 0.5 0.71 65 0 0.65 70 0 0.58 75 0 0.5 80 0 0.41

#### Ground Temperature Correction Factor (K2)

 Ground Temperature Correction Factor: K2 Ground Temperature Insulation PVC XLPE/EPR 10 1.1 1.07 15 1.05 1.04 20 0.95 0.96 25 0.89 0.93 35 0.77 0.89 40 0.71 0.85 45 0.63 0.8 50 0.55 0.76 55 0.45 0.71 60 0 0.65 65 0 0.6 70 0 0.53 75 0 0.46 80 0 0.38

#### Thermal Resistance Correction Factor (K4) for Soil (When Thermal Resistance of Soil is known)

 Soil Thermal Resistivity: 2.5 KM/W Resistivity K3 1 1.18 1.5 1.1 2 1.05 2.5 1 3 0.96

#### Soil Correction Factor (K4) of Soil (When Thermal Resistance of Soil is not known)

 Nature of Soil K3 Very Wet Soil 1.21 Wet Soil 1.13 Damp Soil 1.05 Dry Soil 1 Very Dry Soil 0.86

#### Cable Depth Correction Factor (K5)

 Laying Depth (Meter) Rating Factor 0.5 1.1 0.7 1.05 0.9 1.01 1 1 1.2 0.98 1.5 0.96

#### Cable Distance correction Factor (K6)

 No of Circuit Nil Cable diameter 0.125m 0.25m 0.5m 1 1 1 1 1 1 2 0.75 0.8 0.85 0.9 0.9 3 0.65 0.7 0.75 0.8 0.85 4 0.6 0.6 0.7 0.75 0.8 5 0.55 0.55 0.65 0.7 0.8 6 0.5 0.55 0.6 0.7 0.8

#### Cable Grouping Factor (No of Tray Factor) (K7)

 No of Cable/Tray 1 2 3 4 6 8 1 1 1 1 1 1 1 2 0.84 0.8 0.78 0.77 0.76 0.75 3 0.8 0.76 0.74 0.73 0.72 0.71 4 0.78 0.74 0.72 0.71 0.7 0.69 5 0.77 0.73 0.7 0.69 0.68 0.67 6 0.75 0.71 0.7 0.68 0.68 0.66 7 0.74 0.69 0.675 0.66 0.66 0.64 8 0.73 0.69 0.68 0.67 0.66 0.64

According to above detail correction factors:

– Ground temperature correction factor (K2) = 0.89
Soil correction factor (K4) = 1.05
– Cable depth correction factor (K5) = 1.0
– Cable distance correction factor (K6) = 1.0

Total derating factor = k1 · k2 · k3 · K4 · K5 · K6 · K7

– Total derating factor = 0.93

### Selection of Cable

For selection of proper cable following conditions should be satisfied:

1. Cable derating amp should be higher than full load current of load.
2. Cable voltage drop should be less than defined voltage drop.
3. No. of cable runs  (Full load current / Cable derating current).
4. Cable short circuit capacity should be higher than system  short circuit capacity at that point.

### Selection of cable – Case #1

Let’s select 3.5 core 70 Sq.mm cable for single run.

• Current capacity of 70 Sq.mm cable is: 170 Amp,
Resistance = 0.57 Ω/Km and
Reactance = 0.077 mho/Km
• Total derating current of 70 Sq.mm cable = 170 · 0.93 = 159 Amp.
• Voltage Drop of Cable =
(1.732 · Current · (RcosǾ + jsinǾ) · Cable length · 100) / (Line voltage · No of run · 1000) =
(1.732 · 139 · (0.57 · 0.8 + 0.077 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 5.8%

Voltage drop of cable = 5.8%

Here voltage drop for 70 Sq.mm Cable (5.8 %) is higher than define voltage drop (5%) so either select higher size of cable or increase no of cable runs.

If we select 2 runs, than voltage drop is 2.8% which is within limit (5%) but to use 2 runs of cable of 70 Sq.mm cable is not economical, so it’s necessary to use next higher size of cable.

### Selection of cable – Case #2

Let’s select 3.5 core 95 Sq.mm cable for single run, short circuit capacity = 8.2 KA.

• Current capacity of 95 Sq.mm cable is 200 Amp,
Resistance = 0.41 Ω/Km and
Reactance = 0.074 mho/Km
• Total derating current of 70 Sq.mm Cable = 200 · 0.93 = 187 Amp.
• Voltage drop of cable =
(1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 2.2%

To decide 95 Sq.mm cable, cable selection condition should be checked.

1. Cable derating Amp (187 Amp) is higher than full load current of load (139 Amp) = O.K
2. Cable voltage Drop (2.2%) is less than defined voltage drop (5%) = O.K
3. Number of cable runs (1) ≥ (139A / 187A = 0.78) = O.K
4. Cable short circuit capacity (8.2KA) is higher than system short circuit capacity at that point (6.0KA) = O.K

95 Sq.mm cable satisfied all three condition, so it is advisable to use 3.5 Core 95 Sq.mm cable.

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### Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

1. Eric Perry
Jan 18, 2022

I’m developing a multifamily subdivision. (12 buildings ) consists of 18 units each buildind.
I’m planning on running unground eletric, 8 feet deep.

Each build will have a inside transformation. 1,100kva. Each building unit will have the strength of 60,000 volts.

I need to know what size wire I use to jump style of the transformers to each build from one power source of the main supply. So basically a line that can support 13,000 kva. Can someone help?

2. Alejandro
Jan 16, 2022

In which standard can I find these tables for the correction factors?

3. Zahirul islam
Dec 07, 2021

Different topics on improvement of electric distribution system and related calculations etc.

4. Benjamin Rich
Oct 26, 2021

A 2 core cable supplies current to a 240v single phase load of 18kw at 0.78 power factor. The cable is 40m long and each conductor has a cross-sectional area of 35mm.calculate the voltage drops in the cable at the load,ignoring the reactance of the cable?

5. hema
Oct 22, 2021

Voltage drop of cable =
(1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 2.2%

• Swarup
Oct 30, 2021

Yeah, voltage drop Calculation for 3.5C x 95 mm² is incorrect. it should be 4.3% instead of 2.2 %

6. Avishek Banerjee
Sep 25, 2021

This information is fantastic.

Thanks

7. S K MANDAL
Sep 03, 2021

Allowable voltage drop : 5%, Cable Length: 1300M, Motor power: 110KW. cable to be laid on tray in air
Cable size?

8. Hennah S. Kamara
Sep 03, 2021

From the above calculations of Full Load current, I still have a doubt.
Why did we ignored the PF in getting our full Load current?

KVA/ 1.732*415 =amp

Amp =KVA/1.732*415*0.8

• Marifat Majid
Nov 02, 2021

KVA/1.732*VL =amp , if we include PF then it wont be kVA which is Apparent power, but Real Power kW.
so we will have, kW/1.732*VL*PF = amp
hope this clears the confusion.

• Akshay
Nov 12, 2021

Amp = KVA/(1.732*415) or
Amp = KW/(1.732*415*0.8)

Hope you understand the difference.

9. Afnitha
Sep 01, 2021

How we can calculate number of cables that can accommodate in a cable tray ?

10. ABHIMANYU BHARDWAJ
Jul 27, 2021

how to calculate cable sizes when the total load is 18 kw but it is divided into 6 towers each at a distance of 1 km and each tower has a load of 3 kw ,with 3 towers on one side and 3 towers on the other side , and placing one feeder pillar between them (i.e . 3 towers on one side then feeder pillar and then 3 towers on other side), which will feed them supply ,this feeder will given main supply from LT PANEL placed at a distance of 700 mtrs . please do tell . this is the problem i am facing .

11. Dirshe
Jul 16, 2021

95sq.mm 3.5 core cable which standard is this?

Jul 07, 2021

I love this platform. Looking forward to learning more from you genius.

13. NASIR SHAH
Jul 02, 2021

I WANT TO BA AN ENGINEER AND WANT TO MAKE HUGE MACHINERY AND MAKE FREE ENERGY GENERATOR SYSTEM I LIVE IN GULSHAN RAVI

14. Durga
May 04, 2021

How to identify the value for R and j
For 95 sq mm R=0.41 · 0.8 + j=0.074 · 0.6
Where as
For 70 sq mm R=0.57 · 0.8 + j=0.077 · 0.6)

• Nuwan
Jun 03, 2021

It’s tabulated value. BS 7671:2018

• SURAJ
Jul 06, 2021

U CAN FIND IT IN CURRENT RTING CHART

Jul 19, 2021

Why I couldn’t find current ratings chart

15. siva
Apr 26, 2021

(1.732 · 139 · (0.57 · 0.8 + 0.077 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 5.8%

sir from above how we got 139

• Shivaram
Apr 29, 2021

Consumed Load in KW = 80 · 1 = 80 KW
Consumed Load in KVA = KW/P.F.:
Consumed Load in KVA = 80/0.8 = 100 KVA
Full Load Current = (KVA · 1000) / (1.732 · Voltage):
Full Load Current = (100 · 1000) / (1.732 · 415) = 139 Amp.