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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. Sheik
Apr 24, 2020

Thank u so much sir, such a wonderful messages.And please sir I need to know about how prepared load calculation and schedule?

• Mizan Al Aqaid
Jul 21, 2020

2. teja
Mar 04, 2020

Hello sir, is there any formula to calculate the spare load in a electrical panel from the amperage values of branch circuit breakers.

3. Kalpesh
Feb 07, 2020

Hello sir, are this factors applicable for UNINYVIN cables also.

4. Uvaosudheen m
Feb 02, 2020

Hlo sir,. I understood all these calculations and I studied well.
But now I need the table which have the datas of cable like current carrying capacity, etc. Plz share iny email

5. salahodeen
Dec 31, 2019

Control cable for high voltage s/s laid in a cable tray in a trench, is it a must to be armored or not, and what does IEC said in this issue.

thanks a lot

6. Rahul Kumar Dubey
Dec 24, 2019

415V, 10kw is Light Power load and if we calculate amp so it will be 18amp. Then why we select 10 or 16asqmm cu or Alar cable, but this cable current capacity is 3 times greater to load. Please Explain

7. SATYANJIBI
Nov 26, 2019

sir kindly reply me how I simply calculate, how many amp.current taken by any ht or lt cable

Oct 27, 2019

Dear Sir
Please tell how to calculate the size of 11 kV distribution overhead aluminum conductor (ACSR) for particular load and distance.
Thanks

9. satu
Oct 01, 2019

Hello Sir, Kindly help…I’m going to install lightning fixture for tunnel with a total lenght of 7500meters. the lights consist of 2x36watts a distance of every 10meters.
Circuit breaker – ?
Cable size – ?
Transformer type to be install about 4kms. – ?

10. Mohamed Gaber
Sep 24, 2019

Dear Mr. Jignesh Parmar;

You used The demand factor for the calculation of consumed load but as I checked all other sources and calculation programs no use for the demand factor of loads , Kindly explain and provide your source for this formula

11. MOHAMED
Sep 12, 2019

how to take the de-rating value for K1,K2,K3 K4,K5

12. Enkali jafet Nyanyukweni
Sep 10, 2019

Hi, help me please: A single phase 10kw motor is connected 50m from the distribution board by an underground armoured copper cable. USE different cable sizes to determine the correct cable suitable for this motor starting with smallest cable for the load and calculate the correct cable that can be used? Calculate the % volt drop for the correct cable?

13. Patience
May 30, 2019

I want to supply a pump 8m away from the point of supply. The cable will be through a cable tray. The pump specs are:
Rated Voltage: 230 Vac single phase
Rated Power: 0.25kW
Locked Rotor Current:4.8A

How do you calculate the size of the circuit breaker (in Amps) and cable (in mm2).

14. siddharth jain
May 17, 2019

Hi All,
How to select the value of cosǾ & sinǾ in the formula.
Voltage Drop of Cable= (1.732x Current x (RcosǾ+jsinǾ) x Cable Lengthx100) / (Line Voltage x No of Runx1000)
Can anyone help me in this please?

• shilpa
Jul 30, 2019

While calculating the voltage drop, cosǾ = starting p.f.(0.2), which is constant.
sinǾ = SQRT(1-( cosǾ * cosǾ)).
Example ; cosǾ=0.2 , then sinǾ = SQRT(1-( 0.2 * 0.2)) =0.979

• shilpa
Jul 30, 2019

Voltage Drop of cable at Full load = (1.732*current *((resistance*cosǾ)+(reactance *sinǾ)) * cable length * 100) / (line voltage * No.of runs * 1000)

Voltage Drop of cable at transient = (1.732*current *((resistance*cosΦst)+(reactance *sinǾst)) * cable length * 100) / (line voltage * No.of runs * 1000)

where, cosΦst = 0.2
sinǾst = SQRT (1-(0.2*0.2))

Aug 06, 2019

cospi is power factor from there u can find sinpi using inverse cos formula

• Premdas
Dec 08, 2019