# Voltage Drop

## Ohm's law and calculating voltage drop

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Voltage drop can be calculated using Ohm's law like

E = R I(1)

where

E = voltage drop (volts, V)

R = electrical resistance (ohms, Ω)

I = current (amps, A)

### Example - Voltage Drop

Voltage drop in a 100 ft power line:

*2 x 100 ft**#10 copper**electrical resistance 1.02**Ω*/1000 ft*current 10 amps*

can be calculated as

E = 2 (0.00102Ω/foot) (100 ft) (10 A)

= 2.04 V

### Circular mils and Voltage drop

Voltage drop can also be calculated using mils like

E = K P L I / A (2)

where

K = specific resistivity (Ω- circular mils/foot)

P = phase constant = 2 (for single phase) = 1.732 (for three phase)

L = wire length (ft)

A = wire area (circular mils)

Specific resistivity various wires

*Solid Copper, K = 11 (temp 77*^{o}F - 121^{o}F), K = 12 (temp 122^{o}F - 167^{o}F)*Solid Aluminum, K = 18 (temp 77*^{o}F - 121^{o}F), K = 20 (temp 122^{o}F - 167^{o}F)*Stranded Copper, K = 11 (temp 77*^{o}F - 121^{o}F), K = 12 (temp 122^{o}F - 167^{o}F)*Stranded Aluminum, K = 19 (temp 77*^{o}F - 121^{o}F), K = 20 (temp 122^{o}F - 167^{o}F)

### Example - Specific resistivity and Voltage drop

With values from the example above the voltage drop can be calculated as

E = (11ohm- circular mils/foot) 2 (100 ft) (10 A) / (10400 mils)

= 2.11 V

### Copper Conductor - Voltage Drop Table

The voltage drop in copper conductors can be estimated with

*E = f I L (3)*

*where *

*f = factor from table below*

*I = current (amps)*

*L = conductor length (ft)*

AWG | Factor | |

Single-phase | 3-phase | |

14 | 0.476 | 0.42 |

12 | 0.313 | 0.26 |

10 | 0.196 | 0.17 |

8 | 0.125 | 0.11 |

6 | 0.0833 | 0.071 |

4 | 0.0538 | 0.046 |

3 | 0.0431 | 0.038 |

2 | 0.0323 | 0.028 |

1 | 0.0323 | 0.028 |

1/0 | 0.0269 | 0.023 |

2/0 | 0.0222 | 0.020 |

3/0 | 0.019 | 0.016 |

4/0 | 0.0161 | 0.014 |

250 | 0.0147 | 0.013 |

300 | 0.0131 | 0.011 |

350 | 0.0121 | 0.011 |

400 | 0.0115 | 0.009 |

500 | 0.0101 | 0.009 |

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