Make use of our user-friendly Voltage-Drop Calculator to calculate the voltage drop or load current of a conductor easily. Just enter load current, conductor material, and initial voltage details in the input fields of the calculator and press the calculate button to obtain the result in a short span of time.

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more about the concept and check the solved questions

The following are the simple steps to evaluate the voltage drop. Use these guiding principles and get the output in a less amount of time.

- Check conductor length, load current, initial voltage and wire resistivity.
- Multiply the double of load current with the length and resistivity.
- Divide the product by the cross-sectional area of the wire to get the voltage drop.

The voltage drop is defined as the reduction of voltage. It occurs when the electric current moves through passive elements of the circuit.

The elements that affect the voltage drop of a conductor is conductor material type, length of the conductor, cross-sectional area of the wire, and load current.

The formula for voltage drop is along the lines:

**For DC or 1-phase AC, V = 2 * I * L * R / A / n**

**For 3-phase AC, V = √3 * I * L * R / A / n**

Where,

V is the voltage drop

I is the load current

L is the length of the wire

R is the resistivity of the wire

A is the cross-sectional area of the wire

n is the number of conductors that are connected in parallel

**Typical AWG wire sizes**

AWG is a wire guage system used predominantly in North America for the diameters of round, non-ferrous and electrically conducting wire. The list of typical AWG wires and their sizes are follows:

AWG | Diameter (inch) | Turns of Wire (per inch) | Area (kcmil) | Copper resistance (Ω/km) |
---|---|---|---|---|

0000 (4/0) | 0.4600 | 2.17 | 212 | 0.1608 |

000 (3/0) | 0.4096 | 2.44 | 168 | 0.2028 |

00 (2/0) | 0.3648 | 2.74 | 133 | 0.2557 |

0(1/0) | 0.3249 | 3.08 | 106 | 0.3224 |

1 | 0.2893 | 3.46 | 83.7 | 0.4066 |

2 | 0.2576 | 3.88 | 66.4 | 0.5127 |

3 | 0.2297 | 4.36 | 52.6 | 0.6465 |

4 | 0.2043 | 4.89 | 41.7 | 0.8152 |

5 | 0.1819 | 5.50 | 33.1 | 1.028 |

6 | 0.1620 | 6.17 | 26.3 | 1.296 |

7 | 0.1443 | 6.93 | 20.8 | 1.634 |

8 | 0.1285 | 7.78 | 16.5 | 2.061 |

9 | 0.1144 | 8.74 | 13.1 | 2.599 |

10 | 0.1019 | 9.81 | 10.4 | 3.277 |

11 | 0.0907 | 11.0 | 8.23 | 4.132 |

12 | 0.0808 | 12.4 | 6.53 | 5.211 |

13 | 0.0720 | 13.9 | 5.18 | 6.571 |

14 | 0.0641 | 15.6 | 4.11 | 8.286 |

15 | 0.0571 | 17.5 | 3.26 | 10.45 |

16 | 0.0508 | 19.7 | 2.58 | 13.17 |

17 | 0.0453 | 22.1 | 2.05 | 16.61 |

18 | 0.0403 | 24.8 | 1.62 | 20.95 |

19 | 0.0359 | 27.9 | 1.29 | 26.42 |

20 | 0.0320 | 31.3 | 1.02 | 33.31 |

21 | 0.0285 | 35.1 | 0.810 | 42 |

22 | 0.0253 | 39.5 | 0.642 | 52.96 |

23 | 0.0226 | 44.3 | 0.509 | 66.79 |

24 | 0.0201 | 49.7 | 0.404 | 84.22 |

25 | 0.0179 | 55.9 | 0.320 | 106.2 |

26 | 0.0159 | 62.7 | 0.254 | 133.9 |

27 | 0.0142 | 70.4 | 0.202 | 168.9 |

28 | 0.0126 | 79.1 | 0.160 | 212.9 |

29 | 0.0113 | 88.8 | 0.127 | 268.5 |

30 | 0.0100 | 99.7 | 0.101 | 338.6 |

31 | 0.00893 | 112 | 0.0797 | 426.9 |

32 | 0.00795 | 126 | 0.0632 | 538.3 |

33 | 0.00708 | 141 | 0.0501 | 678.8 |

34 | 0.00630 | 159 | 0.0398 | 856 |

35 | 0.00561 | 178 | 0.0315 | 1079 |

36 | 0.00500 | 200 | 0.0250 | 1361 |

37 | 0.00445 | 225 | 0.0198 | 1716 |

38 | 0.00397 | 252 | 0.0157 | 2164 |

39 | 0.00353 | 283 | 0.0125 | 2729 |

40 | 0.00314 | 318 | 0.00989 | 3441 |

**Example**

**Question: A current of 9 A flows through a circuit that carries
resistance of 10 Ω and the length of the wire is 5 m. The cross
sectional area of the wire is 24 sq m, find the voltage drop?**

**Solution:**

Given that

Load current I = 9 A

Resistance R = 10 Ω

Cross-sectional Area A = 24 sq m

Wire length L = 5 m

Voltage drop V = 2 * I * L * R / A

V = 2 x 9 x 5 x 10 / 24

= 900/24

= 37.5

Therefore, the voltage drop is 37.5 V.

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** 1. How to calculate the voltage drop?**

To find the voltage drop in a wire, get the product of load current, wire length, and wire resistance. Divide the product by the cross-sectional area of the wire and again divide the result by a number of conducts.

**2. What influences the magnitude of the voltage drop?**

The voltage drop occurs in a wire when the current has to travel along the wire. The factors affecting the voltage drop are wire material type, wire size, length of the wire and load current.

**3. What is the formula for voltage drop?**

The formula to calculate the voltage drop is V = 2 * I * L * R / A / n for DC or 1st phase AC and V = √3 * I * L * R / A / n for 3-phase AC.

**4. How can voltage drop be reduced?**

The simple steps to minimize the voltage drop is decreasing the temperature of the conductor, reducing the power load, increasing quality or increasing the conductor's size, decreasing the wire length.