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Voltage Drop Practice for the California Electrician Exam

Voltage drop is the voltage you lose pushing current down a long run of wire. Too much and motors run hot and lights dim. On the California exam, calculations are part of the Determination of electrical system requirements domain. Here is the pattern, one worked example, and a question to try.

Last reviewed June 2026

One worked example

A 120-volt garage circuit feeds a 16-amp battery charger 70 feet from the panel on 12 AWG copper. Is the voltage drop a problem?

  1. 1
    Name the problem

    This is a single-phase voltage drop. The current runs out to the load and back, so the distance counts twice. That is where the 2 comes from.

  2. 2
    Pick the formula

    K is the resistance constant for the metal. I is the load current. D is the one-way distance. CM is the conductor area in circular mils.

    VD = (2 × K × I × D) / CM

  3. 3
    Pull the numbers

    Copper K is 12.9. Load is 16 A. One-way run is 70 ft. 12 AWG is 6,530 circular mils.

  4. 4
    Run the arithmetic

    Multiply across the top, then divide by the circular mils.

    VD = (2 × 12.9 × 16 × 70) / 6,530

    VD = 28,896 / 6,530

    VD = 4.43 V

  5. 5
    Check it against the supply

    4.43 V is 3.69% of 120 V. NEC recommends staying under 3% on a branch circuit. Above the 3% target for branches. Borderline. Check the application.

    4.43 / 120 = 3.69%

Now try one

Your turn. A 240-volt, 24-amp load runs 150 feet on 10 AWG copper.

Run it. What's the voltage drop, in volts?

  1. 1
    Run the arithmetic

    Multiply across the top, then divide by the circular mils.

    VD = (2 × 12.9 × 24 × 150) / 10,380

    VD = 92,880 / 10,380

    VD = 8.95 V

  2. 2
    Check it against the supply

    8.95 V is 3.73% of 240 V. NEC recommends staying under 3% on a branch circuit. Above the 3% target for branches. Borderline. Check the application.

    8.95 / 240 = 3.73%

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