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TDLR voltage drop practice

Voltage drop is the voltage you lose pushing current down a long run of wire. Too much and motors run hot and lights dim. This is a TDLR Calculations-part question type, now scored on its own. Here is the pattern, one worked example, and an original 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%

Try an original question

Sample question · original

A 240-volt single-phase circuit feeds a 20-amp load 100 feet away on 10 AWG copper. Copper's K is 12.9, and 10 AWG is 10,380 circular mils.

What is the approximate voltage drop?

  • A2.5 volts
  • 5.0 volts
  • C9.9 volts
  • D7.9 volts

Answer B. VD = (2 x K x I x L) / CM = (2 x 12.9 x 20 x 100) / 10,380 = 51,600 / 10,380, which is about 5.0 volts.

  • NEC 2023 Chapter 9, Table 8

Why the other answers tempt you

  • A: 2.5 volts forgets the 2 for the round trip. On single phase the current goes out and comes back, so the run counts twice.
  • C: 9.9 volts applies the round-trip factor twice. The 2 goes in once.
  • D: 7.9 volts uses 12 AWG circular mils (6,530) instead of 10 AWG (10,380). Smaller wire, bigger drop, wrong wire.

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