Voltage Drop Calculations: The Formula You Need for the Exam

Voltage drop questions make up about 12% of the Calculations portion on the Texas journeyman electrician exam, which translates to roughly 3 questions. While that might not sound like many, these are often the questions that separate passing from failing because they require you to apply a formula, use NEC tables for conductor properties, and arrive at a precise numerical answer.

The good news is that voltage drop problems follow a consistent pattern. There are really only two formulas you need, and once you practice the process a few times, these become some of the most predictable questions on the exam.

What Is Voltage Drop?

When current flows through a conductor, some voltage is lost due to the resistance of the conductor. The longer the wire and the smaller its cross-sectional area, the more voltage is lost. By the time the current reaches the load at the end of the circuit, the voltage is lower than the voltage at the source.

This matters because electrical equipment is designed to operate within a certain voltage range. If the voltage at the outlet is too low, motors run inefficiently, lights dim, and sensitive equipment can malfunction or be damaged.

NEC Guidelines (Not Requirements)

The NEC does not mandate a maximum voltage drop. Instead, it provides recommendations through informational notes:

NEC 210.19(A) Informational Note No. 4: Branch circuit conductors should be sized so that the voltage drop does not exceed 3% at the farthest outlet.

NEC 215.2(A)(4) Informational Note No. 2: The combined voltage drop of feeder and branch circuit should not exceed 5%.

While these are recommendations rather than mandatory code requirements, the exam tests them as if they are the standard. If a question asks whether a circuit's voltage drop is acceptable, compare it against the 3% (branch) or 5% (total) guideline.

The Two Formulas

Single-Phase Voltage Drop

VD = (2 x K x I x D) / CM

Three-Phase Voltage Drop

VD = (1.732 x K x I x D) / CM

What Each Variable Means

| Variable | Meaning | Units | |---|---|---| | VD | Voltage drop | Volts | | K | Resistivity constant of the conductor material | Ohms per circular mil-foot | | I | Load current | Amps | | D | One-way distance from source to load | Feet | | CM | Cross-sectional area of the conductor | Circular mils |

The K Constant

Copper: K = 12.9 (at 75C)
Aluminum: K = 21.2 (at 75C)

Some references use slightly different K values depending on the temperature assumption. For the exam, use 12.9 for copper and 21.2 for aluminum unless the problem specifies otherwise.

Circular Mils (CM)

The circular mil area for each conductor size is found in NEC Chapter 9, Table 8. Here are the most common sizes you will need:

| Conductor Size | Circular Mils (CM) | |---|---| | #14 AWG | 4,110 | | #12 AWG | 6,530 | | #10 AWG | 10,380 | | #8 AWG | 16,510 | | #6 AWG | 26,240 | | #4 AWG | 41,740 | | #3 AWG | 52,620 | | #2 AWG | 66,360 | | #1 AWG | 83,690 | | 1/0 AWG | 105,600 | | 2/0 AWG | 133,100 | | 3/0 AWG | 167,800 | | 4/0 AWG | 211,600 | | 250 kcmil | 250,000 | | 300 kcmil | 300,000 | | 350 kcmil | 350,000 | | 500 kcmil | 500,000 |

Exam tip: Tab Chapter 9, Table 8 in your NEC codebook. You will need it for every voltage drop problem.

Worked Examples

Example 1: Basic Single-Phase

A 120V single-phase branch circuit supplies a 20A load through #10 AWG copper conductors. The one-way distance is 150 feet. What is the voltage drop and is it within the recommended 3% guideline?

Given:

Voltage = 120V
I = 20A
D = 150 ft
K = 12.9 (copper)
CM = 10,380 (#10 AWG from Table 8)

Calculate: VD = (2 x 12.9 x 20 x 150) / 10,380 VD = 77,400 / 10,380 VD = 7.46V

Percentage: (7.46 / 120) x 100 = 6.22%

This exceeds the 3% recommendation. The conductor needs to be upsized.

Example 2: Finding the Minimum Conductor Size

What is the minimum conductor size for a 120V single-phase circuit carrying 16A over 100 feet, if the voltage drop must not exceed 3%?

Step 1: Calculate the maximum allowable voltage drop. 3% of 120V = 3.6V

Step 2: Rearrange the formula to solve for CM. CM = (2 x K x I x D) / VD CM = (2 x 12.9 x 16 x 100) / 3.6 CM = 41,280 / 3.6 CM = 11,467 circular mils

Step 3: Find the conductor size with at least 11,467 CM from Table 8. #10 AWG = 10,380 CM (too small) #8 AWG = 16,510 CM (sufficient)

Minimum conductor size: #8 AWG copper

Example 3: Three-Phase Circuit

A 480V three-phase circuit supplies a 60A load through #6 AWG copper conductors over a one-way distance of 200 feet. What is the voltage drop?

Given:

Voltage = 480V
I = 60A
D = 200 ft
K = 12.9 (copper)
CM = 26,240 (#6 AWG)

Calculate: VD = (1.732 x 12.9 x 60 x 200) / 26,240 VD = 268,258 / 26,240 VD = 10.22V

Percentage: (10.22 / 480) x 100 = 2.13%

This is within the 3% recommendation for a branch circuit.

Example 4: Combined Feeder and Branch Circuit

A 240V single-phase feeder runs 100 feet from the panel to a subpanel using #4 AWG copper conductors, carrying 80A. From the subpanel, a branch circuit runs 50 feet using #10 AWG copper conductors, carrying 20A. What is the total voltage drop?

Feeder voltage drop: VD = (2 x 12.9 x 80 x 100) / 41,740 VD = 206,400 / 41,740 VD = 4.94V (2.06% of 240V)

Branch circuit voltage drop: VD = (2 x 12.9 x 20 x 50) / 10,380 VD = 25,800 / 10,380 VD = 2.49V (1.04% of 240V)

Total voltage drop: 4.94 + 2.49 = 7.43V (3.10% of 240V)

The branch circuit alone is within 3%, but the combined feeder plus branch circuit exceeds the 5% recommendation only slightly at 3.10%. This installation is within both guidelines.

Example 5: Aluminum Conductors

A 240V single-phase circuit uses #4 AWG aluminum conductors to supply a 40A load over 175 feet. What is the voltage drop?

Given:

K = 21.2 (aluminum)
I = 40A
D = 175 ft
CM = 41,740 (#4 AWG from Table 8)

Calculate: VD = (2 x 21.2 x 40 x 175) / 41,740 VD = 296,800 / 41,740 VD = 7.11V

Percentage: (7.11 / 240) x 100 = 2.96%

Just barely within the 3% guideline.

Alternative Method: Using Chapter 9, Table 9

For more precise calculations (particularly for AC circuits), you can use Chapter 9, Table 9, which provides AC resistance and reactance values for 600V cables in conduit. This method accounts for the AC resistance (which includes skin effect) rather than the DC resistance used in the K-factor method.

Single-phase using Table 9: VD = 2 x I x R x D / 1000

Three-phase using Table 9: VD = 1.732 x I x R x D / 1000

Where R is the effective impedance per 1,000 feet from Table 9 (for the conductor size and conduit type at the assumed power factor).

The K-factor method is simpler and is generally what the exam tests. However, if a problem specifically references Table 9, use that method instead.

Common Exam Pitfalls

Forgetting the factor of 2 (single-phase) or 1.732 (three-phase). The formulas account for the total circuit length. In a single-phase circuit, current travels to the load and back, so the total conductor length is twice the one-way distance. This is already built into the formula with the "2" multiplier. Do not double the distance AND use the "2" in the formula.

Using the wrong K value. K = 12.9 for copper, K = 21.2 for aluminum. Read the problem carefully to identify the conductor material.

Looking up CM from the wrong table. Circular mils are in Chapter 9, Table 8, not Table 5 (which gives conductor dimensions in square inches for conduit fill).

Confusing one-way distance with total distance. The "D" in the formula is always the one-way distance from source to load. If a problem gives you the total round-trip distance, divide by 2 before plugging it in.

Not converting kcmil to circular mils. 250 kcmil = 250,000 circular mils. The "kcmil" unit means thousands of circular mils. If the problem gives conductor size in kcmil, multiply by 1,000.

Quick Reference: The Voltage Drop Process

Identify: voltage, current, distance, conductor size, conductor material, single-phase or three-phase
Find K (12.9 copper / 21.2 aluminum)
Find CM from Chapter 9, Table 8
Apply the formula (use 2 for single-phase, 1.732 for three-phase)
Calculate percentage: (VD / Source Voltage) x 100
Compare to NEC guidelines: 3% for branch, 5% for feeder + branch combined

If the problem asks you to find the minimum conductor size, rearrange the formula to solve for CM, then find the conductor with at least that many circular mils in Table 8.

Practice voltage drop calculations with our free calculator

Sources:

National Fire Protection Association - NFPA 70: National Electrical Code, 2023 Edition
Texas Department of Licensing and Regulation (TDLR) - Electrician Exam Information
TDLR - Electrician Exam Statistics, Fiscal Year 2025
PSI Exams - TDLR Electrician Exam Candidate Information
International Association of Electrical Inspectors (IAEI) - NEC 2026 Significant Code Changes