Voltage Drop Formula & Example Calculation

What is Voltage Drop?

Voltage drop is the decrease of electrical potential along the path of a current flowing in an electrical circuit. Or more simply, a “drop in voltage”. Voltage drops occur due to the internal resistance of the source, passive elements, across conductors, across contacts, and across connectors are undesirable because some of the energy supplied is dissipated.

The voltage drop across an electrical load is proportional to the power available to be converted in that load to some other useful form of energy. Voltage drop is calculated through Ohm’s law.

Voltage Drop in Direct Current Circuits

In direct current circuits, the reason for voltage drop is resistance. For understanding the voltage drop in the DC circuit, let’s look at an example. Assume a circuit that consists of a DC source, 2 resistors that are connected in series, and a load.

Here, every element of the circuit will have a 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.

We can calculate the energy consumed by each resistance by measuring the voltage across individual resistors. While the current flows through the wire starting from the DC supply to the first resistor, some energy given by the source gets dissipated due 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, we also use Kirchhoff’s circuit law for voltage drop calculation. It is as follows:
Supply Voltage = Sum of the voltage drop across each component of the circuit.

Voltage Drop Calculation of a DC Power Line

Here, we are taking an example of a 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 currentReactance (X), which comprises XC and XL. Both X and R will oppose the current flow also. The sum of the two is termed Impedance (Z).
XC → Capacitive reactance
XL → Inductive reactance

The amount of Z depends on the factors such as magnetic permeability, electrical isolating elements, and AC frequency.
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)
voltage drop in alternating current circuits
IB → 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)
Vn → Phase to neutral voltage
Un → Phase to phase voltage
Φ → Phase angle of the load

Circular Mils and Voltage Drop Calculation

A circular mil is really a unit of area. It is used for referring to 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.

SIZE OF COPPER CONDUCTORFACTOR, f
AWGmm2SINGLE-PHASETHREE-PHASE
142.080.4760.42
123.310.3130.26
105.260.1960.17
88.370.1250.11
613.30.08330.071
421.20.05380.046
3 0.04310.038
233.60.03230.028
142.40.03230.028
1/053.50.02690.023
2/067.40.02220.020
3/085.00.0190.016
4/0107.20.01610.014
250 0.01470.013
300 0.01310.011
350 0.01210.011
400 0.01150.009
500 0.01010.009
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