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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. Owais
Apr 20, 2021

Sir, what is the figure 1.732?

2. B BAIKUNTHO RAO SUBUDHI
Apr 19, 2021

Thanks for sharing.

3. MOHAMMED
Mar 24, 2021

Sir, what will be the cable derated ampacity incase if i have 3 runs of 1C cable with base ampacity 170amps & cable derating 0.8. do i have to times with 1.732, say 170×0.8×1.732=235.5A. please guide.

4. Devdutt Vin
Mar 09, 2021

Sir, please let us know from which standard (IEC / BS / IS) you have taken the cable ratings.

5. TJ Khawaja
Feb 18, 2021

(1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 2.2%
the above calculation does not give 2.2% it gives 4.3% please check again. If you use two runs then it will be 2.2%

6. Parikshit Ningavale
Jan 31, 2021

Dear sir, how did you take system short circuit capacity as 6kA?

7. MAYSSARA
Jan 26, 2021

GREAT INFORMATIONS

8. kome makoge elvise
Jan 21, 2021

Good day,

9. Yogendra Singh Singhvi
Oct 09, 2020

How we got 139 amperes after getting 170*0.93=158.10, calculation for derated current?
Some step is missing.

10. val
Aug 18, 2020

Sir. can you please give an idea on how did you get the value of .8 and .6. can you show to me the computation

11. Waqas
Aug 09, 2020

i need clarification how sin Value is 0.6? any calculation behind it?

• MAHENDRA KUMAR
Aug 11, 2020

yes there is calculation ,with formula that is sinǾ = SQRT(1-( cosǾ * cosǾ))

• irfan
Mar 10, 2021

cos(fi)= o.8
cos inv. of 0.8 is 36.86
then sin of 36.86 is 0.599986 i.e approx 0.6

12. Mizan Al Aqaid
Jul 21, 2020

What the distance between load and supply ? What is the voltage value ? Pf ? And more. Are you serious about your question?

• Dipankar Rai
Jul 28, 2020

He has considered everything 200m length, 415 V voltage, Pf for converting Kw to KVA.

13. Mundiya Mulozi
Jul 20, 2020

how do i calculate the size of the cable, given the power and voltage, taking installation methods and correction factors into account. e.g. an electrical load is 80KW located at a distance of 200m from supply, voltage is 415V three phase, power is 0.8, permissible voltage drop is 5% and demand factor is 1.

14. Harvinder Paul Singh
Jun 19, 2020

Sir pl supply a cable size calculator EXEL Sheet.

15. solomon
May 02, 2020

generate 1250 kva and transformer 1200 kva the power haws 1600 A bracer what cable size