# How to Calculate Voltage Drop in Cables with Examples

Table of Contents

**Calculating Voltage Drop in Cables**

We know that every cable has some type of resistance. This resistance is inversely proportionate to a conductor’s diameter and directly proportionate to its length. When current flows through a cable, a voltage drop occurs on it due to the resistance of that cable. The voltage drop is generally ignored in a cable having a short length; however, a voltage drop is never ignored in a thin cable having a low diameter and larger length. According to the IEE regulation B-23, voltage drops at any point between the power supply terminal and installation, should not exceed 2.5 percent of the supplied voltage. For example, if the supply voltage is 250V, the value of the maximum permissible voltage drop tends to be as below;

Maximum allowable voltage drop = 250 x 2.5 / 100 = 6 Volts

The voltage drop also occurs in a circuit wiring at different points between a distribution board and different sub-circuits. The value of this drop taking place on the sub-circuit should always be half the value of the aforementioned drop (i.e., 3V). Voltage drop is generally expressed as an ampere/meter in tables. That’s if the length of any cable is one meter, and if one-ampere current transmits through it, then in such a situation how many voltage drops occur? In the old system (i.e., Foot, Pound system, or FPS) voltage drop is expressed based on per 100 feet.

The following steps should be adopted to determine the voltage drops in any cable;

(1). First, maximum allowable voltage drops at the rate of 2.5 percent are determined with the help of the following formula;

Maximum allowable voltage drop = 2.5 / 100 x supply voltage

(2). Then load current is determined with the help of the following formula;

Load current IL = Total watt / Supply voltage x power factor = W / V Cos

Remember that if the power factor has not been given, then its value is also considered 0.85. Moreover, if output power and efficiency have been given, then load current can be drawn by means of determining the input power (in watts) with the help of the following formula;

Input = Output (watts) / efficiency

Output power has always been mentioned on the nameplate of a motor (in horsepower or watts). If the output has mentioned in horsepower, it can be converted to watts by multiplying it by 746. However, in the case efficiency is not known, efficiency is always treated as 90 %.

(3). After this keeping in view the load current determined with the help of a suitable table, the size of a suitable cable (i.e., a cable having a current rating nearest to the calculated current rating) should be found.

(4). Then note down the voltage drop taking place on the cable per 100 feet according to the rated current of the cable through the given table.

(5). Then calculate the voltage drops on a cable’s rated current for the actual length (in feet or meters) of that cable, or circuit with the help of the following formulae;

(i). If the length of a cable has been given in feet, the voltage drops occurring in that cable can be determined with the help of the following formula;

Voltage drops for actual length of the circuit = Length of the circuit run x Volt drop for 100 feet/ 100

(ii). If the length of a cable has been given in meters, voltage drops taking place in this cable can be determined as follows;

Permissible voltage drops V_{d} = mV x I x L / 1000

Where I = Current in amperes

L = Route length of cable in meters

mV = Approximate milli volt drop / ampere/ meter of cable given in table.

(6). Then multiply this value of the voltage drop by the following ratio (i.e., load factor)

Load factor = Load current to be taken by cable / Rated current of cable given in the table

Remember that this value reflects the actual voltage drop of the cable on the passing of this load current.

(7). If this calculated voltage drop value exceeds the extreme permissible voltage drop value, then calculate the voltage drop again according to the aforementioned procedure by means of taking the next large size of the cable and the rated current of this cable. However, if this calculated voltage drop value is less than the allowable voltage drops, then the selected cable size is perfectly correct.

## Example 1: **Calculating **Voltage Drop in Cable

Declared single-phase supply voltage is 250V, 40 yards run of one twin core cable installed in conduit, required to supply a 3.5 KW load straight from the main distribution board. Current taken by load is 17 amperes. Average summer temperature in building where wiring is installed 40°C. Find the size of single core PVC cable run in conduit that has a voltage drop within the recommended limit.

Solution:

Declared supply voltage = 250 V

Length of twin core cable = 40 yards = 40 x 3 = 120 feet

Load current = 17 Amp.

Ambient temp. = 40°C

Size of single core PVC cable run in conduit =?

**At Ambient Temperature 30°C **

As cable is twin core run in conduit, we refer Pakistan Cable & Table 14 under column 3 & 4.

The nearest current ratio = 19 A

Size of PVC cable = 7/ .029” (0.0045in^{2})

**At Ambient Temperature 40°C**

The same cable will be derated by multiplying the approximately / suitable factor which is 0.94 at 40°C (as table is for coarse excess protections).

Now rated current of the cable at 30°C x Rating factor at 40°C

= 19 x .094 = 17.86 Amp.

Since load current is less than the rated current, so 7/.029” cable is suitable if voltage drop in cable is ignored. This however should not be done and the suitability of the cable from the voltage drop consideration must be checked because the distance between D.F.B & load is more than 100 ft (i.e., 40 yards = 120 feet)

Maximum permissible voltage drop = 2.5 % of declared voltage drop

= 2.5 x 350 / 100 = 6.25 volts

For 19 A this voltage drops for 100 ft, run in 7/ .029” cable – 7.7 V

For 19 A voltage drop for 120 ft

= Actual load current of cable / Rated current of cable given in table = Actual length of cable / 100 x Voltage drop for 100 ft = 17 / 19 x 120 /100 x 7.7 = 8.3 V

This voltage drop is greater than maximum permissible voltage drops, so the calculation must be repeated for next large size of cable.

Next large size of cable = 7 / .036”

Rated current = 26 A

For 26 A voltage drop per 100 ft = 7 V

For 17 A voltage drop per 120 ft = 17 / 26 x 120 / 100 x 7 = 5.5 V

This is now within the acceptable limits

The suitable cable size =7/.036″ (.007 in** ^{2})** PVC single core.

## Example 2: **Calculating **Voltage Drop in Cable

Declared single-phase supply voltage = 230 V, 40 yard run one twin core cable installed in conduit, required to supply a 1.5 KW load straight from main distribution board. Current taken by load is 7 Amp. Average summer temperature in building where wiring is installed is 104° Fahrenheit. Find the size of single-core PVC cable that has a voltage drop within the recommended limits.

Solution;

Declared supply voltage =230 V

Maximum permissible voltage drop = 2.5% of declared voltage

i.e., 2.5 / 100 x230 V = 5.75 V

Current taken by the load = 7 Amp.

**At Ambient Temperature 30°C**

A PVC 1/ 044″ is rated at 10 Amp. (See table 14) for conduit wiring with an ambient temperature of 30°C (86°F)

**At Ambient Temperature 40°C**

The same cable in an ambient temperature of 40°C (104°F) will have to be declared by multiplying by the temperature factor 0.94 (as table is for coarse excess protection).

Now rated current of the cable = Current at 30°C x Rating factor at 40°C

= 10x.94 =9.4 A

Since load current is less than the rated current, so 1/ 0.44″ PVC cable is suitable if voltage drop in cable is ignored. This however, should not be done and the suitability of the cable from the voltage drop consideration must checked, because the distance between DFB & load, is more than 100 ft. (i.e., 40 yards =120 ft)

For 10 amperes voltage drop for 100 ft run in 1/ .044″ (120ft) cable is 13 V.

7/10 x 120/ 100x 13=10.92 V

This is above the maximum permissible voltage drop. So, the calculation must be repeated for the next large size cable i.e., 3/.029″. The volt drop per 100 ft of this cable is 12 V at 12 Amp.

Therefore, at 7Ampere for 120 ft, it will be

7/12 x 120/100 x 12= 8.4 V

This is also above the maximum permissible limit of voltage. So the next large size of cable is 3/.036″. The volt drop per 100 ft length of this cable is 9.8 V at 16 amperes.

Therefore, at 7 amperes, for 120 ft, its volt drop is

7/16 x120/100 x 9.8 = 5.145 V

Answer; This voltage drop is below to the permissible limit, so the selected size of cable is 3/.036″ single core having rating of 16 Amp.

## Example 3: **Calculating **Voltage Drop in Cable

A pair of mains 50 yards long are connected to a supply of 230 volts. The total load is 230 4KW. The diversity factor is 60% and mains are installed in such conduit which contains single core PVC cable. Find the size of the main cable if temperature is 40°C.

Solution;

Length of cable= 50 yards = 50 x 3 = 150 ft

Total connected load = 4 KW = 4000 W

Diversity factor = 60 % of 4000

Working load = 60 / 100 x 4000 = 2400 V

Supply voltages = 230 V

Load current I= W/ V = 2400 / 230 = 10. 43 amps.

Assume 20% additional load for demand in future.

Then total load current = 20 / 100 x 10.43 +10.43 = 12.51 amp.

**At Ambient Temperature 30°C**

According to table 14 for 12.51 amps. Suitable size of cable is 3/.036″ (16 amp).

**At Ambient Temperature 40°C**

The same cable in an ambient temperature of 40°C will have to be derated by multiplying by the temperature factor 0.94 (as table is for coarse excess protection).

Current rating of cable for 40°C= 0.94 x 16= 15.04 amps.

Since load current is less than the rated current, so 3/.036″ PVC cable is suitable if voltage drop in cable is ignored. This however, should not be done and the suitability of the cable from the voltage drop consideration must be checked because the distance between DFB & load is more than 100 ft (i.e., 150 ft).

Permissible voltage drop = 2.5/ 100 x 230 = 5.75 V

Actual voltage drops for 150 ft = 10.43 / 16 x 150 / 100 x 9.8 = 9.58 V

This drop is above the maximum permissible voltage drop, so the calculation must be repeated for the next large size cable i.e., 7 / .029″. The volt drop per 100 ft length of this cable is 7.7 V at 19 amp. Therefore, it will be

Actual voltage drops for 150 ft = 10.43 / 19 x 150 / 100 x 7.7 = 6.34 V

This is also above the maximum limit of voltage drop. So, the next large size of cable is 7 / .036″. The volt drop per 100 ft length of this cable is 7 V at 26 amp.

Actual voltage drops for 150 ft = 10.43 / 26 x 150 / 100 x 7 = 4.21 V

Answer; This voltage drop is below to the permissible limit, so the selected size of cable is 7 / .036″ PVC single core having rating of 26 Amp.

## Example 4: **Calculating **Voltage Drop in Cable

A pair of mains 25m long are connected to a supply of 230 volts. Total light load at the far end of the main is 3600 W. if diversity factor is 60% and mains are installed in conduit, find the size of twin core PVC insulated main cable when ambient temperature is 40°C.

Solution;

Total light load = 3600 W

Load diversity factor = 60% = 0.6

Working light load = 3600 x 0.6 = 2160 W

10% additional load = 10 / 100 x 2160 = 216 W

Total estimated load = 2160 + 216 = 2376 W

Load current, I = W/V = 2376 / 230 = 10 .33 A

Length of cable = 25 m = 25 x 3.28 = 82 ft

**At Ambient Temp. 30°C**

According to table 14 for 10.33 amps, suitable size of cable is 3 / .029” (12 amp)

**At Ambient Temp. 40°C**

The same cable in an ambient temperature of 40°C will have to be derated by multiplying by the temperature factor 0.94 (as table is for coarse excess protection)

Current rating of cable for 40°C = 0.94 x 12 = 11.28 amps

Since load current is less than the rated current, so 3/.029” PVC cable is suitable.

The length of the cable is less than 100 ft (i.e., 82 ft), therefore voltage drop calculation should not be done.

Answer; The suitable cable size is 3/.029” PVC main core.

## Example 5: **Calculating **Voltage Drop in Cable

Determine the size of conductor for two-core cable required to carry the maximum current of 50A. it is given that the length of the cable is 60 meter and declared supply voltage is 200 volt**.**

Solution;

Declared supply voltage (V) = 200 V

Load current (I) = 50 A

Length of cable (L) = 60 m

Suitable size of conductor =?

Permissible voltage drop @ 2.5% = 2.5/100 x 200 = 5 Volt

According to table X under column 2 core (given in the data tables).

Nearest current rating of conductor = 53 A

Size of the conductor = 7/ 1.626 mm (14.50 mm** ^{2}**)

Amp – meter / Volt drop given in the table = 820

Amp – meter of the required cable = 50 x 60 = 3000 Amp. Meter

Voltage drops per 1 Amp – meter = 1 / 820 V

Voltage drops per 3000 Amp – meter = 1 / 820 x 3000 = 3.66 Volt

This voltage drop is within permissible limits.

So, the suitable size of cable is 7/1.626 mm (14.50 mm^{2})

## Example 6: **Calculating **Voltage Drop in Cable

Determine the size of sub-main cable run in conduit supplying a load of 3 KW at a distance of 30 m from D.F.B. Declared supply is 230 V, 50 Hz. The maximum ambient temperature is 40°C.

Solution;

Load power (W) = 3 KW = 3000 W

Supply voltage (V) = 230 V

Power factor (assumed) = 0.85

Length of cable = 30 m

Maximum ambient temp. 40°C

Permissible voltage drop = 2.5 / 100 x 230 = 5.75 V

Load current (**IL)** = W/V. Cos = 3000 / 230 x 0.85 = 15.34 A

Voltage drops in cable = Permissible voltage drop x 1000 / I**L** x L

= 5.75 x 1000 / 15.34 x 30 = 12.5 mV

We shall select that cable which has equal or less rating of mV /Amp / meter from table 9.

**At Ambient Temp. 30°C**

Refer table 9 under column 4 & 3

Voltage drops = 10 mV / Amp / meter

Current rating = 24 A

Cable size = 7 / 0.85 mm

**At Ambient Temp. 40°C**

The same cable will be derated by multiplying the appropriate rating factor, which is 0.94 (as table is for coarse excess protection). ** **

so, rated current of this table = Current rating at 30°C x Rating factor

= 24 x 0.94 = 22. 56 A

Since load current is less than this rated current.

So, cable 7/ 0.85 mm is suitable.

## Example 7: **Calculating **Voltage Drop in Cable

A load of 3.4 KW at 250 Volt A.C is to be supplied on Batten at a distance of 25 meter from D.F.B. Average summer temperature 45°C. Find the size of twin core PVC insulated cable keeping in view the recommended voltage drop.

Solution;

Supply voltage = 250 volts, Ambient temp. = 45°C

Power factor (assumed) = 0.8

Load power = 3.4 KW = 3.4 x 1000 = 3400 Watts

Load current (I)= W / V. Cos = 3400 / 250 x 0.8 = 17 A

Length of twin core cable = 25 m = 25 x 3.28 = 82 ft

Size of twin core cable run on batten =?

**At Ambient Temp. 30°C**

As cable is twin core run on batten, we refer Pak. Cable & Table 14 under column 7 & 8.

The nearest current rating = 19 A

Size of PVC cable = 3 / .036”

**At Ambient Temp. 45°C**

The same cable will be derated by multiplying the appropriate rating factor, which is 0.91 at 45°C (as cable is for coarse excess protection)

Now rated current of the cable = Current rating at 30°C x Rating factor 45°C

= 19 x 0.91 = 17.29 A

Since the load current rating is less than the rated current, so 3 / .036” PVC cable is suitable.

The distance between D.F.B and load is less than 100 ft. Therefore, voltage drop calculation should not be done.

Answer; The suitable cable size = 3 / .036” PVC twin core

## Example 8: **Calculating **Voltage Drop in Cable

50 yard run of single core cable installed in a concealed conduit to supply 5KW load from main supply of 220 V. Current taken by the load is 18 Ampere. Building temperature is 20°C. Find the size of single-core PVC cable that has voltage drop within the recommended limit. Find the size of circuit breaker too.

Solution;

Length of cable = 50 yard = 50 x 3 = 150 ft

Supply voltage = 220 V

Load current = 18 A

Ambient temp. = 20°C

Method of installation = Concealed wiring

Size of single core PVC cable =?

Size of circuit breaker =?

Maximum permissible voltage drop = 2.5 % of declared supply voltage

= 2.5 x 220 / 100 = 5.5 Volt.

**At Ambient Temp. 30°C**

As cable is single core run in conduit, we refer Pak. Cable & table 12 under column 2 & 4.

The nearest current rating = 21 A

Size of PVC cable = 7 / .029” (0.0045 inch^{2})

**At Ambient Temp. 20°C**

As ambient temperature 20°C is not given in the table, hence we shall select the nearest temperature 25°C. The same cable will be up rated by multiplying the appropriate rating factor, which is 1.02 9as this table is for coarse excess protection)

So current rating of this cable = 21 x 1.02 = 21.42 A

Since load current is less than the rated current, therefore 7 / .029” PVC cable is suitable if voltage drop is ignored. But it should not be done, because the distance is over 100 ft.

Voltage drops of cable 7 / .029” for 100 ft = 8.4 volt

Voltage drops of cable 7 / .029” for 150 ft

= Load current / Rated current x Actual length of cable / 100 x Volt dropper 100 ft

= 18 / 21.42 x 150 / 100 x 8.4 = 10.588 volt.

The voltage drop is greater than the maximum permissible voltage drops. Hence it is not suitable. So, the calculation must be repeated for the next large size cable i.e., 7/.036” having current rating 28 V at 30°C.

Current rating at 25°C = 28 x 1.02 = 28.56 A

Voltage drops of 7/.036” per 100 ft = 7 Volt

Voltage drops of 7/.036” per 150 ft = 18 / 28.56 x 150 / 100 x 7 = 6.617 Volt

This voltage drop is also greater than the maximum permissible voltage drops. Hence, this cable is not recommended. So, the calculation must be repeated for the next large size cable i.e., 7/.044” having current rating of 34 A at 30°C.

Current rating at 25°C = 34 x 1.02 = 34.68 A

Voltage drops of 7/.044” per 100 ft. = 5.7 Volt

Voltage drops of 7/.044” per 150 ft. = 18 / 34.68 x 150 / 100 x 5.7 = 4.437 Volt

This is now within the acceptable limit.

Size of circuit breaker = 1.2 x Load current = 1.2 X 18 = 21.6 A

Circuit breaker of such rating is not available, so the nearest rating of circuit breaker is 25 Amp.

## Example 9: **Calculating **Voltage Drop in Cable

4 single-phase power circuit 3.5 KW at a power factor of 0.95 lagging, 60 m from the main distribution fuse board which has the declared voltage of 220 volts. Use the table and determine the size of cable conductor.

Data;

Supply voltage = 220 volts

Power factor = 0.95 lagging

Load power = 3.5 KW

Power = 3.5 x 1000 = 3500 W

Distance = 60 m

Total distance of cable = 60 x 2 = 120 m

Size of cable =?

Solution;

Maximum permissible volt drops = 220 x 2.5 / 100 = 5.5 V

Total load current = I_{L}

**= **P/ V Cos

= 3.5 x 1000 / 220 x 0.95 = 17 Amp.

From table 9, we choose 1/1.78 mm cable for which volt drop is 16 mV/Amp/m and its current carrying capacity is 18 Amperes.

Actual volt drop = mV x Load current x Length / 1000

= 16 x 17 x 120 / 1000 = 32.64 Volts

These voltage drops are above the permissible volt. drop, so we will choose the next cable.

Next cable 7/0.85, its current carrying capacity is 24 Amperes and voltage drop/Amp/meter is 10 volt.

Voltage drops at 24 Amp. Load current = 10 x 17 x 120 /1000 = 20.4 Volts

This cable is also rejected. Now we consider cable 7/1.04 mm. its current carrying capacity is 31 Amperes and voltage drop 6.8 mV/Amp/meter.

Voltage drops at 31 Amp. load current = 6.8 x 17 x 120 / 1000 = 13.87 Volts

This cable is also rejected.

Answer:

Hence for next larger size of cable 7/1.35 mm. its current carrying capacity is 42 Amperes and voltage drop 4 mV/Amp/meter.

Voltage drops at 42 Amp. = 4 x 17 x 120 / 1000 = 8.16 Volt

This cable is also rejected.

Hence for next larger size of cable 7/1.70 mm. its current carrying capacity is 56 Amperes and voltage drop 2.6 mV/Amp/meter.

Voltage drops at this load = 2.6 x 17 x 120 / 1000 = 5.3 Volt

This voltage drop is within the acceptable range. So, 7/1.70 mm single core PVC cable is recommended.

Answer; (7/ 1.70 mm)

## Example 10: **Calculating **Voltage Drop in Cable

What type and size of cable suits for given situation.

(i). Load = 5 KW (ii). Volts = 230 V (iii). Length of circuit = 30 m (iv). Temperature = 40°C, dry atmosphere

Solution:

Temperature factor for 40°C = 0.94

Load = 5 KW = 5000 W

Voltage = 230 V

Current I = W/V = 5000 / 230 = 21.74 A

20% additional load current = 20 / 100 x 21. 74 = 4.35 A

Total load current = 21.74 + 4.35 = 26 A

**At Ambient Temp. 30°C**

For selecting the size of cable, we refer Pak. Cable & table 12 under column 3 & 4.

The nearest current rating = 28 A

Size of PVC cable = 7/.036” (.007 inch^{2})

**At Ambient Temp. 40°C**

The same cable will be derated by multiplying the appropriate rating factor, which is 0.94 at 40°C (as table is for coarse excess protection).

Current rating of cable for 40°C = 0.94 x 28 = 26.32 amps.

Since load current is less than the rated current, so 7/.036” PVC cable is suitable if voltage drop is ignored. This however, should not be done and the suitability of the cable from the voltage drop consideration must be checked because the length of the circuit is 100 ft. (i.e., 30 m = 100 ft.)

Voltage drops for 100 ft. = 7V

Actual voltage drops = 26 / 28 x 100 / 100 x 7 = 6.5 V

Allowable voltage drops = 2.5 x 230 / 100 = 5.75 V

This voltage drop is greater than the maximum allowable voltage drops, hence this cable is not suitable. So, the calculation must be repeated for the next large size cable i.e., 7/.044” having current rating 34 A at 30°C, and voltage drop for 100 ft. run is 5.7 V.

Actual voltage drop = 26 / 34 x 100 /100 x 5.7 = 4.35 V

This voltage drop is within the acceptable range, so 7/.044” cable is recommended.

## Example 11: **Calculating **Voltage Drop in Cable

A room is to be wired in conduit for single phase A.C supply directly taken from mains which has declared voltage of 200 volts. The length of wire from main switch to length & plug point is 30 meter. If the wire is to carry 5 Amp, determine the size of conductor. The ambient temperature is 30°C.

Solution:

Declared supply voltage (V) = 200

Load current (I) = 5 A

Length of circuit (L) = 30 m

Ambient temp. = 30°C

Suitable size of conductor =?

Permissible voltage drops @ 2.5 % = 2.5/ 100 x 200 = 5V

From table 9 Pakistan Cables & Tables (metric) under column 3 & 4

Minimum current of conductor = 11 Amp.

Size of conductor = 1.0 sq. mm (I /1.13 mm)

Voltage drops at 1 amp. current for 1 m length = 40 mV

Voltage drops at 5 amp. current for 30 m length = mV x I xL / 1000

= 40 x 5 x 30 / 1000 = 6V

The voltage drop is above the maximum permissible voltage drop. The calculation must be repeated for the next large size of cable.

Next large size of cable = 1.5 sq. mm (I / 1.38 mm)

Current rating of this cable = 13 A

Voltage drops at 1 amp current for 1 m length = 27 mV

Voltage drops at 5 amp current for 30 m length = 27 x5 x30 / 1000 = 4.05 V

Now this is permissible limit

Answer: Thus, the suitable size of cable is 1.5 sq. mm (I/ 1.38 mm)

**Alternative Method**

The size of cable can also be calculated within permissible voltage drops as given below:

Solution;

Permissible V drops = 200 x 2.5 / 100 = 5 V

Formula mV = Permissible V drops x 1000 / I_{L} x L

= 5 x 1000 / 5 x 30 = 33.33 mV / Amp / meter

We shall select the cable which has equal or less rating of mV from table. Refer table 9 under column 3 & 4.

Voltage drop = 27 mV / Amp / meter

Current rating = 13 A

Cable size = I / 1.38 mm

Answer; PVC cable single core = I / 1.38 mm (1.5 sq. mm)

## Example 12: **Calculating **Voltage Drop in Cable

Find the suitable size of cable, run in conduit to deliver 40 A at 230 V and a distance of (i). 10 m (ii). 40 meters

Solution;

Load current = 40 V, Supply voltage = 230 V

(i). Length of cable = 10 m, (ii). Length of cable = 40 m

Permissible voltage drop = 2.5% of supply voltage = 2.5 / 100 x 230 = 5.75 V

According to Pak. Cables & Tables 9 (metric) under column 3 & 4

The nearest current rating = 42 A

Size of conductor = 10 mm^{2 }(7 / 1.35 mm)

**(i). Length of cable = 10 m**

Voltage drops at 1A current for 1m length = 4mV

Voltage drops at 40 A current for 10 m length = mV x I x L / 1000 = 4 x 40 x 10 / 1000 = 1.6 V

This voltage drop is within permissible limits

So, the suitable size of cable is 10 mm^{2} (7/1.35 mm)

**(ii). Length of cable = 40 m**

Voltage drops at 40 A current for 40 m length

= mV x I x L / 1000 = 4 x 40 x 40 / 1000 = 6.40 V

The voltage drop is much greater than the maximum permissible voltage drops, we try next large size of cable.

The next large size of cable = 16 mm^{2} (7 / 1.7 mm)

Current rating = 56 A

Voltage drops at 1A current for 1 m length = 2.6 mV

Voltage drops at 40 V current for 40 m length = 2.6 x 40 x 40 / 1000 = 4.16 V

Now this is within permissible limit

Thus, the suitable size is 16 mm^{2} (7 / 1.7 mm)

**Alternative Method**

The size of cable can also be calculated within permissible voltage drops as below;

Formula; mV = Permissible Voltage Drops x 1000 / I_{L} x L

Where, mV = milli volt drop / Amp / meter of cable given in the table

I_{L} = Load current

L = Length of cable (go & return for a single – phase)

Solution;

Permissible voltage drops = 2.5 / 100 x 230 = 5.75 V

**(i). Length of Cable 10 m**

Voltage drops in cable (mV) = Permissible voltage drops x 1000 / I_{L} x L

= 5.75 x 1000 / 40 x 10 = 14. 375 mV / Amp / meter

We shall select that cable from table, which has an equal or less rating of mV / Amp / meter than voltage drop in cable (14. 375)

Refer table 9 under column 3 & 4

Voltage drops in cable = 10 mV / Amp / meter

Current rating = 24 Amp.

Cable size = 4 mm^{2} (7 / 0.85 mm)

The current rating of this cable is less than load current, so it is not suitable

The next large size = 6 mm^{2 }

Current rating 31 A

The current rating of this cable is less than load current, so it is not suitable.

The next large size = 10 mm^{2 }

Current rating = 42 A

Voltage drop = 4 mV /Amp / meter

Cable size = 10 mm^{2 }(7 / 1.35 mm)

So, this cable size is suitable

Note; solve part (ii) Length of cable = 40 m in the same way by yourself

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