Percent Voltage Drop

%Vd=100 VdVs\%V_{d} = \frac{100 \, V_{d}}{V_{s}}

Worked example: 6.9 V lost on 230 V → 3% — press Try an example to run it live, then adjust anything.

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Percent Voltage Drop explained

VdVs%Vd

Volts lost only mean something relative to volts supplied: 6.9 V is a trivial loss on 4160 V and a serious one on 230 V, where it is 3%. The familiar limits — the NEC's informational 3% on a branch circuit and 5% overall, and similar figures in IEC and CEC practice — are recommendations aimed at equipment performance, not safety minimums, but designers treat them as hard rules because the consequences are real.

Undervoltage is unkind to motors in particular: torque falls with the square of voltage, so a motor at 90% voltage makes only 81% of its torque and pulls extra current to compensate, running hotter for it. Incandescent lamps dim visibly at a few percent, and electronic supplies simply draw more amps as voltage sags, deepening the drop. Note this solver's percent fields also accept a plain fraction.

Percent Voltage Drop formula

%Vd=100 VdVs\%V_{d} = \frac{100 \, V_{d}}{V_{s}}
Where
  • %Vd\%V_{d}= Percent voltage drop (%)
  • VdV_{d}= Voltage drop (V)
  • VsV_{s}= Supply voltage (V)