# Voltage drop formula [Step by Step Calculations and tables]

Voltage drop formula is used to calculate the voltage dropped across the conductor. Three equations are often used for doing calculations.

where

I = Current

R = Resistance

K = Resistance in ohms in one CM of conductors

L = Length of the circuit in feet

CM = Circular mills

K = 12.9 for copper conductors

K = 21.2 for Aluminum Conductors

————————————-

For other temperatures use the equation

R

_{2}= R

_{1}[1 + α (T

_{2}− 75)]

Where

α

_{cu}= 0.00323

α

_{AL}= 0.00330

## Circular Mills table

The CM (circular mills) are calculated from the solid copper wire American wire gauge table:

Size
(AWG) |
Diameter
(inches) |
Area (CM) |
---|---|---|

4/0 | 0.46 | 211600 |

3/0 | 0.4096 | 167800 |

2/0 | 0.3648 | 133100 |

1/0 | 0.3249 | 105600 |

1 | 0.2893 | 83690 |

2 | 0.2576 | 66360 |

3 | 0.2294 | 52620 |

4 | 0.2043 | 41740 |

5 | 0.1819 | 33100 |

6 | 0.162 | 26240 |

7 | 0.1443 | 20820 |

8 | 0.1285 | 16510 |

9 | 0.1144 | 13090 |

10 | 0.1019 | 10380 |

11 | 0.09074 | 8234 |

12 | 0.08081 | 6530 |

13 | 0.07196 | 5178 |

14 | 0.06408 | 4110 |

15 | 0.05707 | 3257 |

16 | 0.05082 | 2583 |

17 | 0.04526 | 2048 |

18 | 0.0403 | 1624 |

19 | 0.03589 | 1288 |

20 | 0.03196 | 1022 |

21 | 0.02846 | 810.1 |

22 | 0.02535 | 642.5 |

23 | 0.02257 | 509.5 |

24 | 0.0201 | 404 |

25 | 0.0179 | 320.4 |

26 | 0.01594 | 254.1 |

27 | 0.0142 | 201.5 |

28 | 0.01264 | 159.8 |

29 | 0.01126 | 126.7 |

30 | 0.01003 | 100.5 |

31 | 0.008928 | 79.7 |

32 | 0.00795 | 63.21 |

33 | 0.00708 | 50.13 |

34 | 0.006305 | 39.75 |

35 | 0.005615 | 31.52 |

36 | 0.005 | 25 |

37 | 0.004453 | 19.83 |

38 | 0.003965 | 15.72 |

39 | 0.003531 | 12.47 |

40 | 0.003145 | 9.888 |

41 | 0.0028 | 7.842 |

42 | 0.002494 | 6.219 |

43 | 0.002221 | 4.932 |

44 | 0.001978 | 3.911 |

## Single phase voltage drop formula calculations

The single-phase voltage drop is defined as the product of 2K into current into L divided by circular mills of wire.

Example: Find the voltage drop of a 180-volt single phase # 6 copper conductor which provides 20 amp current to load. The length of wire is 200 feet. Take K = 12.9 as for 75° temp.

Solution: V_{d(single phase)} = [2K * L * I]/CM

= [2 * 12.9 * 200 * 20 ]/26240

= 3.93 V

Percentage voltage drop = 3.93/200 = 1.96%

As per standards the recomend drop is less than 3% .

## Three phase voltage drop calculations

The three-phase voltage drop calculations are similar to single phase with the difference of factor 1.73. Here the multiplication factor is 1.73 instead of 2. While all other calculations are same.

Example: Find the voltage drop of 1 kV three-phase system which uses #5 copper wire and delivers 30 amp load at a distance of 800 feet. Take K = 12.9 as for 75° temp.

Solution: V_{d(three phase)} = [1.73K * L * I]/CM

= [1.73 * 12.9 * 1000 * 200]/33100 = 16.18 V

%age V_{d(three phase)} = 16.18/1000 = 1.6%.

## Ohm’s law

Ohm’s law can also be used to calculate the voltage drop. Remember that Ohm’s law is only valid for single phase calculations.

Example: What is a voltage drop of two no 6 conductors when a current of 20 Amps flows through them. Use uncoated conductor at 75° for calculations.

Solution: R = 0.510/1000 feet. In our case:

R = 0.510/1000 * 400 = 0.204

V = IR

= 20 Amp * 0.102 = 4.08