Voltage Drop Calculator

Electrical circuit analysis • 2026 standards

Voltage Drop Formulas:

Show the calculator

\( \text{Voltage Drop} = \frac{2 \times K \times L \times I}{CM} \)

\( \text{Percentage Drop} = \frac{\text{Voltage Drop}}{\text{Source Voltage}} \times 100 \)

\( \text{Wire Resistance} = \frac{\rho \times L}{A} \)

\( \text{Power Loss} = I^2 \times R \)

\( \text{Acceptable Drop} = \text{Source Voltage} \times 0.03 \text{ (3%)} \)

These formulas calculate voltage drop in electrical circuits. K is the resistivity constant (12.9 for copper, 21.2 for aluminum), L is the length in feet, I is the current in amperes, and CM is the circular mils of the conductor.

Example: For a 120V circuit with 15A load, 100 ft wire run, 12 AWG copper wire:

CM = 6,530 (12 AWG)

Voltage Drop = (2 × 12.9 × 100 × 15) ÷ 6,530 = 5.94V

Percentage Drop = (5.94 ÷ 120) × 100 = 4.95%

Since 4.95% > 3% recommended limit, a larger wire gauge is needed.

Circuit Specifications

Advanced Options

Results

5.94 V
Voltage Drop
4.95%
Percentage Drop
89.10 W
Power Loss
3.60 V
Acceptable Drop (3%)
Wire Resistance
0.0396 Ω
Wire CM
6,530
Conductor Type
Copper
Circuit Type
Single Phase

Voltage Drop Fundamentals

What is Voltage Drop?

Voltage drop is the reduction in voltage along a conductor due to resistance. It occurs when current flows through a wire and is influenced by wire length, gauge, material, and current. Excessive voltage drop can cause equipment malfunction and energy loss.

Calculation Methods

Standard voltage drop formulas:

  • Single phase: VD = (2 × K × L × I) ÷ CM
  • Three phase: VD = (√3 × K × L × I) ÷ CM
  • Power loss: P = I² × R
Key Rules:
  • Keep drop under 3% for feeders
  • Keep drop under 5% for branch circuits
  • Use proper wire gauge
  • Consider circuit length
  • Account for temperature

Wire Sizing

AWG Standards

American Wire Gauge (AWG) specifies wire diameter and current carrying capacity. Larger numbers represent smaller wires. Proper wire sizing prevents overheating and ensures safety.

Sizing Factors

Key factors in wire selection:

  1. Circuit current
  2. Voltage drop
  3. Temperature rating
  4. Conduit fill
Sizing Guidelines:
  • 14 AWG: 15A circuits
  • 12 AWG: 20A circuits
  • 10 AWG: 30A circuits
  • 8 AWG: 40-50A circuits
  • 6 AWG: 55-65A circuits

Electrical Engineering Learning Quiz

Question 1: Multiple Choice - Voltage Drop Formula

What is the voltage drop formula for a single-phase circuit?

Solution:

The correct answer is A) VD = (2 × K × L × I) ÷ CM. This is the standard formula for single-phase voltage drop calculations, where K is the resistivity constant, L is length, I is current, and CM is circular mils.

Pedagogical Explanation:

The factor of 2 in the single-phase formula accounts for the round-trip path of current through both the hot and neutral conductors. The voltage drop occurs in both directions, so the total drop is doubled.

Key Definitions:

Voltage Drop: Reduction in voltage due to conductor resistance

Circular Mils (CM): Unit of cross-sectional area for wires

Resistivity Constant (K): Material-specific resistance factor

Important Rules:

• Single phase: Factor of 2

• Three phase: Factor of √3

• K = 12.9 for copper

Tips & Tricks:

• Remember: 2 for single, √3 for three phase

• Copper: K = 12.9, Aluminum: K = 21.2

• Always check NEC requirements

Common Mistakes:

• Forgetting the factor of 2 for single phase

• Using wrong K value for material

• Confusing single and three phase formulas

Question 2: Voltage Drop Calculation

Calculate the voltage drop for a 120V, 20A single-phase circuit with 100 feet of 12 AWG copper wire. (K = 12.9, CM = 6,530)

Solution:

Step 1: Apply the single-phase voltage drop formula

VD = (2 × K × L × I) ÷ CM

Step 2: Substitute values

VD = (2 × 12.9 × 100 × 20) ÷ 6,530

Step 3: Calculate

VD = 51,600 ÷ 6,530 = 7.90V

Step 4: Calculate percentage drop

Percentage = (7.90 ÷ 120) × 100 = 6.58%

The voltage drop is 7.90V or 6.58%.

Pedagogical Explanation:

This calculation demonstrates how to apply the voltage drop formula with actual values. The result shows that this setup exceeds the recommended 3% limit, indicating that a larger wire gauge would be needed for proper operation.

Key Definitions:

Single-Phase Circuit: Circuit with one alternating current path

Wire Gauge: Measurement of wire diameter

Current Carrying Capacity: Maximum safe current for a wire

Important Rules:

• Keep drop under 3% for feeders

• Keep drop under 5% for branch circuits

• 12 AWG = 20A maximum

Tips & Tricks:

• 12 AWG good for 20A circuits

• For 6.58% drop, use 10 AWG

• Always verify with NEC tables

Common Mistakes:

• Using wrong AWG for current

• Forgetting to check percentage

• Not accounting for length

Question 3: Word Problem - Wire Sizing

A 240V, 30A circuit needs to supply power 150 feet away. What is the minimum wire gauge needed to keep voltage drop under 3%? (Use copper wire, K = 12.9)

Solution:

Step 1: Calculate maximum allowable voltage drop

Max VD = 240V × 0.03 = 7.2V

Step 2: Rearrange voltage drop formula to solve for CM

CM = (2 × K × L × I) ÷ VD

Step 3: Substitute values

CM = (2 × 12.9 × 150 × 30) ÷ 7.2

Step 4: Calculate

CM = 116,100 ÷ 7.2 = 16,125 CM

Step 5: Find wire gauge

Looking at AWG tables, 8 AWG has 16,510 CM, which is sufficient.

8 AWG wire is needed for this circuit.

Pedagogical Explanation:

This problem demonstrates how to work backwards from a voltage drop requirement to determine the necessary wire size. The rearrangement of the voltage drop formula allows us to solve for the required cross-sectional area.

Key Definitions:

Wire Sizing: Selecting appropriate wire gauge for circuit requirements

Cross-Sectional Area: Area of wire perpendicular to current flow

Circuit Requirements: Voltage, current, and distance needs

Important Rules:

• Rearrange formula to solve for unknown

• Use next larger gauge if between sizes

• Always check current capacity too

Tips & Tricks:

• Round up to next standard gauge

• Check current capacity also

• Consider temperature derating

Common Mistakes:

• Not solving for CM first

• Using wrong formula arrangement

• Forgetting current capacity check

Question 4: Application-Based Problem - Power Loss

Calculate the power loss in watts for a circuit with 15A current and 0.05 ohms resistance.

Solution:

Step 1: Apply the power loss formula

Power Loss = I² × R

Step 2: Substitute values

Power Loss = (15)² × 0.05

Step 3: Calculate

Power Loss = 225 × 0.05 = 11.25W

The power loss in the circuit is 11.25 watts.

Pedagogical Explanation:

Power loss represents energy converted to heat in the conductor due to resistance. This is why undersized wires can overheat and potentially cause fires. The quadratic relationship with current means that doubling the current quadruples the power loss.

Key Definitions:

Power Loss: Energy lost as heat in electrical conductors

I²R Losses: Resistive losses proportional to current squared

Heat Generation: Thermal energy produced by electrical resistance

Important Rules:

• Power Loss = I² × R

• Loss increases with current squared

• Larger wires have less resistance

Tips & Tricks:

• Minimize resistance to reduce losses

• Proper wire sizing prevents overheating

• Consider ambient temperature effects

Common Mistakes:

• Forgetting the squared relationship

• Not considering thermal effects

• Ignoring cumulative losses

Question 5: Multiple Choice - Temperature Effects

How does temperature affect wire resistance?

Solution:

The correct answer is B) Resistance increases with temperature. As temperature increases, atomic vibrations in the conductor increase, causing more collisions with electrons and thus higher resistance.

Pedagogical Explanation:

This temperature coefficient effect is important in electrical design because warm wires have higher resistance, leading to greater voltage drops and power losses. This is why wire ampacity ratings are based on specific temperature conditions.

Key Definitions:

Temperature Coefficient: How resistance changes with temperature

Ampacity: Maximum safe current for a conductor

Thermal Effects: Temperature-related changes in electrical properties

Important Rules:

• Copper: 0.00393/°C temperature coefficient

• Resistance increases with temperature

• Derate for high temperatures

Tips & Tricks:

• Use derating factors for high temperatures

• Consider ambient conditions

• Account for heat buildup in conduits

Common Mistakes:

• Ignoring temperature effects

• Not derating for high temps

• Forgetting conduit fill effects

Voltage Drop Calculator

FAQ

Q: How do I calculate voltage drop for a three-phase circuit?

A: For three-phase circuits, use this formula:

$$ \text{VD} = \frac{\sqrt{3} \times K \times L \times I}{CM} $$

Where √3 ≈ 1.732. The factor √3 accounts for the phase relationship in three-phase systems.

For example, with 208V, 30A, 100 ft, 10 AWG copper:

CM = 10,380 (10 AWG)

VD = (1.732 × 12.9 × 100 × 30) ÷ 10,380 = 6.48V

Percentage = (6.48 ÷ 208) × 100 = 3.11%

Q: What's the difference between copper and aluminum wire?

A: The main differences relate to resistivity and cost:

  • Copper: K = 12.9, better conductivity, more expensive
  • Aluminum: K = 21.2, higher resistance, less expensive

Mathematically, for the same voltage drop:

Aluminum requires larger gauge than copper

For example, if 12 AWG copper is needed, 10 AWG aluminum would be required for similar performance.

Aluminum is commonly used for service entrances and large feeders due to cost savings.

About

Electrical Team
This calculator was created
This calculator was created by our General & Utility Calculators Team , may make errors. Consider checking important information. Updated: April 2026.