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Second year Lesson 12 of 13 About 16 min

Voltage drop

What you’ll learn

  • Why every conductor loses some voltage, and why long runs lose more.
  • The single-phase formula, VD = 2 × I × R × L, and the balanced three-phase form with √3.
  • How to turn the drop into a percentage and a voltage at the load.
  • Where the resistance figure must come from — and what to change when the drop is too big.

Every conductor has a little resistance, and every amp that flows through it loses a little voltage on the way — that loss is voltage drop. It is Ohm’s law, V = I × R, applied to the wire instead of the load. On a short run it hardly matters. On a long one it can leave a motor starving or a light dim, and it heats the wire for nothing. The arithmetic is easy once you see where the “× 2” comes from.

120 V source 0.24 Ω → −2.88 V 0.24 Ω → −2.88 V load 114.24 V 12 A, 150 ft (45.7 m) each way
Practice numbers. 150 ft (45.7 m) of conductor out and 150 ft back, each length with 0.24 Ω. At 12 A each conductor drops 12 × 0.24 = 2.88 V, so 5.76 V never reaches the load.

Why the × 2

Current goes out on one conductor and comes back on the other. Both have resistance, and the current flows through both, so both drop voltage. If the run is L long (one way), the circuit contains 2 × L of wire. Multiply that by the resistance per unit length and you have the loop resistance; multiply by the current and you have the drop.

  • VD = 2 × I × R × Lsingle-phase or DC
  • VD = √3 × I × R × Lbalanced three-phase
  • % = VD ÷ V × 100as a percentage

Here R is the conductor’s resistance per unit length — per foot or per metre — and L is the one-way length in the same unit. Tables usually give resistance per 1,000 ft or per km, so divide by 1,000 first.

Where R comes from

The resistance of a conductor depends on its size, whether it is copper or aluminium, and its temperature. Take it from your code book’s conductor table or the cable maker’s data for the exact conductor you are using. The figures on this page are practice numbers, chosen to make the arithmetic clear — they are not values for any real conductor.

Worked examples

1. A 120 V branch circuit, in feet (practice resistance)

  1. 12 A load, 150 ft one way, 120 V. Practice resistance: 1.6 Ω per 1,000 ft (from “your book”).
  2. R per foot = 1.6 ÷ 1,000 = 0.0016 Ω. One conductor: 0.0016 × 150 = 0.24 Ω.
  3. VD = 2 × 12 × 0.0016 × 150 = 5.76 V.
  4. As a percentage: 5.76 ÷ 120 × 100 = 4.8%. Voltage at the load: 120 − 5.76 = 114.24 V.
  5. Heat wasted in the wire: I² × R = 12² × 0.48 = 69 W, all day, for nothing.
The same run in metres

150 ft = 45.72 m, and 1.6 Ω per 1,000 ft = 1.6 × 3.2808 = 5.25 Ω per km (0.00525 Ω/m). VD = 2 × 12 × 0.00525 × 45.72 = 5.76 V — the same answer, as it must be.

2. A 240 V feeder, in metres (practice resistance)

  1. 30 A, 60 m one way (196.9 ft), 240 V. Practice resistance: 0.8 Ω per km (0.244 Ω per 1,000 ft).
  2. R per metre = 0.8 ÷ 1,000 = 0.0008 Ω. One conductor: 0.0008 × 60 = 0.048 Ω.
  3. VD = 2 × 30 × 0.0008 × 60 = 2.88 V, which is 2.88 ÷ 240 × 100 = 1.2%.

3. A balanced three-phase load (practice resistance)

  1. 40 A per line, 100 ft (30.5 m) one way, 208 V three-phase. Practice resistance: 0.5 Ω per 1,000 ft (1.64 Ω per km).
  2. VD = √3 × 40 × 0.0005 × 100 = 1.732 × 2 = 3.46 V line to line.
  3. 3.46 ÷ 208 × 100 = 1.67%.

When the drop is too big

Compare your answer with the limit in your code book. If it is over, look at the formula: every term you can shrink cuts the drop in proportion. A larger conductor has less resistance per foot — the usual fix. A shorter route cuts L. Feeding the load at a higher voltage carries the same power on less current, and the percentage is of a bigger number. Splitting the load across two circuits halves I on each.

Try one: example 1 with the run doubled to 300 ft (91.4 m)

VD = 2 × 12 × 0.0016 × 300 = 11.52 V, or 9.6% of 120 V. Double the length, double the drop.

On the job

Tips from the job

  • Voltage drop shows up as dim lights at the far end, motors that run hot, and equipment that resets when something big starts. Measure at the load under load, not with it switched off.
  • Measure the run, do not guess it. Walk the route the cable or conduit will actually take, up and down included.
  • When the drop is too high, the usual fix is a larger conductor. Size it up from your code book’s tables, then re-check the drop.
  • Work the drop in the voltage drop calculator; it takes the resistance you type from your book, in feet or metres.

Common mistakes

  • Using the round-trip length for L and then multiplying by 2 again — the drop comes out double.
  • Using a resistance for the wrong size, the wrong metal, or per metre when the length is in feet.
  • Forgetting the × 2 (single-phase) or using 2 instead of √3 on a balanced three-phase circuit.
  • Quoting a “maximum percent” from memory. The limits and where they apply are in your code book.
  • Measuring voltage at the load with the load off: no current, no drop. Measure while it runs.

Check yourself

8 questions. Tap an answer — it is marked straight away and the reason opens under it. Get 80% or better and the lesson is ticked off on the year page.

Not started yet.

  1. Why does the single-phase voltage-drop formula multiply by 2?

    Answer and why

    B. Both the conductor going out and the one coming back have resistance and both lose voltage. L is the one-way length, so the wire in the circuit is 2 × L.

  2. Practice figures: 10 A, 100 ft one way, a conductor of 2 Ω per 1,000 ft, 120 V single-phase. What is the drop?

    Answer and why

    B. Resistance per foot = 2 ÷ 1,000 = 0.002 Ω. VD = 2 × 10 × 0.002 × 100 = 4 V.

  3. Same circuit, same load, but the run is twice as long. The drop…

    Answer and why

    C. VD = 2 × I × R × L. Double L and the drop doubles — it is a straight line, not a square.

  4. A 240 V circuit loses 6 V along its run. What percentage is that?

    Answer and why

    B. % = VD ÷ V × 100 = 6 ÷ 240 × 100 = 2.5%.

  5. For a balanced three-phase load, the multiplier in VD = k × I × R × L becomes…

    Answer and why

    C. In a balanced three-phase circuit the currents in the three lines are 120° apart, and the line-to-line drop works out to √3 × I × R × L using the one-way length.

  6. Where should the resistance per length for your calculation come from?

    Answer and why

    B. Resistance depends on the conductor size, copper or aluminium, and temperature. Take it from the table or data sheet every time — and the limit on how much drop is acceptable is also in your code book.

  7. An induction motor at the end of a long run sees low voltage. What happens?

    Answer and why

    B. To deliver the same mechanical power at a lower voltage, the motor pulls more current, which heats its windings and adds more drop on the run. Low voltage shortens motor life.

  8. Which change does NOT reduce voltage drop?

    Answer and why

    C. More current means more drop: VD rises with I. The other three all cut the drop — a bigger conductor or shorter run lowers R × L, and a higher voltage carries the same power with less current.

Questions people ask

What is the voltage drop formula?

For a single-phase (or DC) circuit: VD = 2 × I × R × L, where I is the current, R is the conductor resistance per unit length and L is the one-way length. For a balanced three-phase circuit, replace the 2 with √3 (1.732). This is the resistance-only form; long runs of large conductors on AC also have reactance, which the conductor tables in your code book account for.

How much voltage drop is allowed?

That is set by your code book (and sometimes by the job specification), and it differs between codes and between feeders and branch circuits. Look it up for your jurisdiction rather than using a number from memory. This page teaches how to calculate the drop you then compare against it.

Is voltage drop the same as ampacity?

No. Ampacity is about heat: how much current a conductor can carry without overheating its insulation. Voltage drop is about the voltage lost along the length. A long run can be fine on ampacity and still lose too much voltage, which is why long runs often need a larger conductor than the current alone would call for.

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