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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.

An example how to calculate voltage drop and size of electrical cable
An example how to calculate voltage drop and size of electrical cable (photo credit: 12voltplanet.co.uk)

Soil details:

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


Ok, let’s dive into calculations…

  • Consumed Load = Total Load · Demand Factor:
    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 TemperatureInsulation
PVCXLPE/EPR
101.221.15
151.171.12
201.121.08
251.061.04
350.940.96
400.870.91
450.790.87
500.710.82
550.610.76
600.50.71
6500.65
7000.58
7500.5
8000.41

Ground Temperature Correction Factor (K2)

Ground Temperature Correction Factor: K2
Ground TemperatureInsulation
PVCXLPE/EPR
101.11.07
151.051.04
200.950.96
250.890.93
350.770.89
400.710.85
450.630.8
500.550.76
550.450.71
6000.65
6500.6
7000.53
7500.46
8000.38

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

Soil Thermal Resistivity: 2.5 KM/W
ResistivityK3
11.18
1.51.1
21.05
2.51
30.96

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

Nature of SoilK3
Very Wet Soil1.21
Wet Soil1.13
Damp Soil1.05
Dry Soil1
Very Dry Soil0.86

Cable Depth Correction Factor (K5)

Laying Depth (Meter)Rating Factor
0.51.1
0.71.05
0.91.01
11
1.20.98
1.50.96

Cable Distance correction Factor (K6)

No of CircuitNilCable diameter0.125m0.25m0.5m
111111
20.750.80.850.90.9
30.650.70.750.80.85
40.60.60.70.750.8
50.550.550.650.70.8
60.50.550.60.70.8

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

No of Cable/Tray123468
1111111
20.840.80.780.770.760.75
30.80.760.740.730.720.71
40.780.740.720.710.70.69
50.770.730.70.690.680.67
60.750.710.70.680.680.66
70.740.690.6750.660.660.64
80.730.690.680.670.660.64

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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author-pic

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 13 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

172 Comments


  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,
    Please what is your reference for selecting resistance and reactance values


  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.
      So what are you asking????


  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

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