When a single conductor isn't large enough to handle the required current or meet voltage drop limits, parallel conductors provide an elegant solution. This guide explains when parallel conductors make sense and how to properly apply them.

What Are Parallel Conductors?

Parallel conductors are multiple conductors of the same size connected together at both ends to share the current of a single circuit. Instead of one very large conductor per phase, you might use two, three, or four smaller conductors. The effective ampacity and voltage drop performance equals the sum of the individual conductors.

For example, two 350 kcmil conductors in parallel provide approximately the same performance as a single 700 kcmil conductor—but the 350s are much easier to pull, terminate, and work with.

NEC Requirements (310.10(G))

The NEC permits parallel conductors under specific conditions designed to ensure current divides equally among the parallel paths:

Minimum Size: 1/0 AWG or larger for each parallel conductor
Same Length: All parallel conductors must be identical length
Same Material: All copper OR all aluminum—no mixing
Same Size: All conductors per phase must be the same AWG/kcmil
Same Insulation: Identical insulation type and temperature rating
Same Termination: All must terminate the same way

When Parallel Conductors Make Sense

Good Applications

• Currents over 400A where single conductors are impractical
• Long feeder runs where massive single conductors are needed
• When conduit fill limits prevent larger single conductors
• Service entrance conductors to large facilities

Poor Applications

• Small branch circuits (under 1/0 AWG)
• Short runs where single conductors work fine
• When unequal lengths can't be avoided
• Cost savings aren't significant

Voltage Drop Calculation

For parallel conductors, simply divide the single conductor resistance by the number of parallel sets:

R_effective = R_single / N

Where N is the number of parallel conductors per phase

Example: 600A Feeder, 400ft

Option 1: Single 750 kcmil

R = 0.0171 Ω/kft

Vd = 2.85% (marginal)

Option 2: 2 × 350 kcmil

R = 0.0382/2 = 0.0191 Ω/kft

Vd = 3.18% (slightly higher)

Note: 3 × 300 kcmil would give R = 0.0135 Ω/kft effective, achieving 2.25% drop

Calculate Parallel Conductors

Our voltage drop calculator supports parallel conductor configurations. Enter your circuit parameters and number of parallel sets to get accurate results.

Parallel Conductor Calculator

When to Use Parallel Conductors for Voltage Drop