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Understanding Voltage Drop in Electrical Systems

Learn how voltage drop occurs, why it matters for electrical design, and how to calculate it for single-phase and three-phase circuits.

Written by Marcus Vance, PE

Voltage drop is the decrease in electrical potential along the path of a current flowing in an electrical circuit. In power distribution systems, excessive voltage drop can lead to inefficient operation of equipment, overheating of cables, and potential failure of sensitive electronics.

Why Voltage Drop Occurs

Every conductor (usually copper or aluminum) has some amount of inherent electrical resistance ($R$) and reactance ($X$). According to Ohm’s Law, when current ($I$) flows through this impedance ($Z$), a voltage drop occurs:

$$V_{drop} = I \times Z$$

Where impedance $Z$ is calculated as:

$$Z = \sqrt{R^2 + X^2}$$

For relatively short cables or small wire sizes, the reactance $X$ is negligible, and we can simplify the impedance to just the resistance ($R$).

Standards and Compliance (NEC & IEC)

  • National Electrical Code (NEC): Recommends a maximum voltage drop of 3% for branch circuits, and a total of 5% combined for feeder and branch circuits.
  • IEC 60364-5-52: Recommends a maximum drop of 3% for lighting circuits and 5% for other uses (power/heating) in low-voltage installations.

Core Formulas

Single-Phase Circuits

For a single-phase AC or DC system, the current must travel to the load and return. Thus, the path length is doubled:

$$V_{drop} = \frac{2 \times L \times I \times R_{cond}}{1000}$$

Where:

  • $L$ = Length of the run (in meters or feet).
  • $I$ = Load current (in Amperes).
  • $R_{cond}$ = Resistance of the conductor (in $\Omega$ per 1000m or 1000ft).

Three-Phase Circuits

In a balanced three-phase AC system, the return currents cancel out, resulting in a factor of $\sqrt{3} \approx 1.732$ instead of $2$:

$$V_{drop} = \frac{\sqrt{3} \times L \times I \times R_{cond}}{1000}$$

Practical Example

Consider a 120V single-phase circuit running a 15A load over 150 feet using 12 AWG copper wire ($R \approx 1.93\ \Omega$ per 1000 feet).

  1. Double length run: $2 \times 150 = 300$ feet.
  2. Calculate drop: $$V_{drop} = \frac{300 \times 15 \times 1.93}{1000} = 8.685\text{ V}$$
  3. Percentage drop: $$%V_{drop} = \left(\frac{8.685}{120}\right) \times 100 = 7.24%$$

This exceeds the NEC recommendation of 3%, suggesting that a larger wire size (such as 10 AWG or 8 AWG) should be used.