# Voltage Drop Calculation

**is done by Ohm’s law.**

*voltage drop calculation*## Voltage Drop in Direct Current Circuits

In direct current circuits, the reason for voltage drop is the resistance. For understanding the voltage drop in DC circuit, we can take an example. Assume a circuit which consist of DC source, 2 resistors which are connected in series and a load.Here; every elements of the circuit will have certain resistance, they receive and lose energy to some value. But the deciding factor of the value of energy is the physical features of the elements. When we measure the voltage across the DC supply and first resistor, we can see that it will be less than the supply voltage. The energy used by each resistance can be known by measuring the voltage across individual resistors. While the current flows through the wire starting from the DC supply to first resistor, some energy that is given by the source get dissipated owing to the conductor resistance. To verify the voltage drop, Ohm’s law and Kirchhoff’s circuit law are used which are briefed below.
Ohm's law is represented by
V → Voltage Drop (V)
R → Electrical Resistance (Ω)
I → Electrical Current (A)
For DC closed circuits, Kirchhoff’s circuit law is also used for **voltage drop calculation**. It is as follows:
Supply Voltage = Sum of voltage drop across each component of the circuit.

### Voltage Drop Calculation of a DC Power Line

Here, we are taking an example of 100 ft power line. So; for 2 lines, 2 × 100 ft. Let Electrical resistance be 1.02Ω/1000 ft and current be 10 A.## Voltage Drop in Alternating Current Circuits

In AC circuits; in addition to Resistance (R), there will be a second opposition for the flow of current – Reactance (X) which comprises of X_{C}and X

_{L}. Both X and R will oppose the current flow also the sum of the two is termed as Impedance (Z). X

_{C}→ Capacitive reactance X

_{L}→ Inductive reactance The amount of Z depends on the factors such as magnetic permeability, electrical isolating elements and the frequency of AC. Similar to Ohm's law in DC circuits, here it is given as E → Voltage Drop (V) Z → Electrical Impedance (Ω) I → Electrical Current (A) I

_{B}→ Full load current (A) R → Resistance of the cable conductor (Ω/1000ft) L → Length of the cable (one side ) (Kft) X → Inductive Reactance (Ω/1000f) V

_{n}→ Phase to neutral voltage U

_{n}→ Phase to phase voltage Φ → Phase angle of load

### Circular Mils and Voltage Drop Calculation

Circular mil is really a unit of area. It is used for referring the circular cross sectional area of the wire or conductor. The voltage drop using mils is given by L → Wire length (ft) K → Specific Resistivity (Ω-circular mils/foot). P → Phase constant = 2 meant for single phase ” = 1.732 meant for three phase I → Area of the wire (circular mils)### Voltage Drop Calculation of Copper Conductor from Table

The voltage drop of the copper wire (conductor) can be found out as follows: f is the factor we get from the standard table below.AWG | MM^{2} | SINGLE PHASE | THREE-PHASE |

14 | 2.08 | 0.476 | 0.42 |

12 | 3.31 | 0.313 | 0.26 |

10 | 5.26 | 0.196 | 0.17 |

8 | 8.37 | 0.125 | 0.11 |

6 | 13.3 | 0.0833 | 0.071 |

4 | 21.2 | 0.0538 | 0.046 |

3 | 0.0431 | 0.038 | |

2 | 33.6 | 0.0323 | 0.028 |

1 | 42.4 | 0.0323 | 0.028 |

1/0 | 53.5 | 0.0269 | 0.023 |

2/0 | 67.4 | 0.0222 | 0.020 |

3/0 | 85.0 | 0.019 | 0.016 |

4/0 | 107.2 | 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 |