ELEC-210 · Lesson 10 of 12 · 30 min
Voltage drop in AC circuits
ELEC-110 introduced voltage drop as the product of current and conductor resistance. That is accurate for DC and a good approximation for small branch circuits, but in AC feeders, especially large ones, the conductor's reactance and the load's power factor change the result, sometimes substantially. This lesson develops the AC voltage drop calculation: the resistance and reactance of conductors and what affects them, the effective impedance method, single-phase and three-phase formulas, the code limits and tables, sizing conductors for voltage drop, and special cases such as motor starting and long runs to outbuildings, EV chargers and pumps.
Conductor impedance in AC circuits
An AC conductor has both resistance and inductive reactance:
- AC resistance is slightly higher than DC resistance, because of skin effect (alternating current crowds toward the surface of the conductor, using less of its cross-section) and proximity effect (currents in nearby conductors distort the current distribution). The difference is negligible for small conductors and a few percent for large ones such as 500 kcmil. Resistance also rises with temperature, by about 0.4% per °C for copper; tables are usually given at 75 °C.
- Inductive reactance arises from the magnetic field around the conductors and depends on their spacing: conductors close together (in a cable or tight conduit) have less reactance than widely spaced ones. Steel conduit increases reactance compared with PVC or aluminum conduit, because the steel concentrates the magnetic field. Reactance changes little with conductor size, typically 0.04 to 0.06 Ω per 1,000 ft (0.13 to 0.2 Ω/km) for building wiring.
For small conductors, resistance dominates and reactance can be ignored. For large conductors, the resistance becomes small and reactance becomes comparable or even larger, so ignoring it underestimates voltage drop.
| Copper conductor | AC resistance (Ω/km, PVC conduit, 75 °C) | Reactance (Ω/km) | Ratio X/R |
|---|---|---|---|
| 12 AWG | 6.6 | 0.18 | 0.03 |
| 6 AWG | 1.6 | 0.17 | 0.1 |
| 1/0 AWG | 0.39 | 0.14 | 0.36 |
| 4/0 AWG | 0.20 | 0.13 | 0.65 |
| 500 kcmil | 0.089 | 0.13 | 1.4 |
The effective impedance method
The voltage drop along a conductor is the phasor \(I(R + jX)\). What matters for the load is how much this reduces the magnitude of the voltage, which, for the small drops of practical circuits, is closely approximated by the component of the drop in phase with the source voltage:
where \(\theta\) is the load's power factor angle (\(\cos\theta\) is the power factor). At unity power factor, only resistance matters. At lagging power factors, reactance contributes; for large conductors with X/R near or above 1, the effective impedance at 0.85 power factor can be more than the resistance alone, while for small conductors it is less (because \(\cos\theta < 1\) reduces the resistive part).
The voltage drop formulas are then:
where \(L\) is the one-way length and \(Z_{\text{eff}}\) the effective impedance per unit length. The factor 2 accounts for the outgoing and return conductors in a single-phase circuit; \(\sqrt3\) converts the drop in one line to the change in line-to-line voltage for a balanced three-phase circuit. For a three-phase line-to-neutral drop, use \(I\,L\,Z_{\text{eff}}\).
Worked problem
Feeder voltage drop, with and without reactance
A three-phase 600 V feeder of 4/0 AWG copper in PVC conduit runs 120 m and carries 200 A at 0.80 power factor lagging. Calculate the voltage drop using resistance only and using effective impedance.
Show solution
- Resistance per km 0.20 Ω, reactance 0.13 Ω; for 120 m: \(R = 0.024\ \Omega\), \(X = 0.0156\ \Omega\).
- Resistance only: \(\sqrt3 \times 200 \times 0.024 = 8.3\) V (1.4%).
- Effective: \(Z_{\text{eff}} = 0.024 \times 0.8 + 0.0156 \times 0.6 = 0.0286\ \Omega\); \(V_d = \sqrt3 \times 200 \times 0.0286 = 9.9\) V (1.65%).
Answer: About 9.9 V, or 1.65%, rather than the 8.3 V (1.4%) given by resistance alone: ignoring reactance underestimates this drop by about 20%. For a 500 kcmil feeder the error would be larger.
Code requirements and tables
CEC Rule 8-102 limits voltage drop in an installation: it must not exceed 5% from the supply side of the consumer's service (or the transformer secondary of a separately derived system) to the point of utilization, and must not exceed 3% in a feeder or branch circuit. The drop is calculated on the basis of the demand load of a feeder or, for a branch circuit, the connected load if known, or 80% of the rating of the overcurrent device protecting the circuit. Appendix D of the CEC includes tables (such as Table D3) giving circuit lengths for a 1% voltage drop for each conductor size and current, with correction factors for system voltage and type, which make routine checks quick.
Voltage drop matters for more than compliance. A motor running at 90% voltage draws more current, runs hotter and produces about 81% of its rated torque (torque varies with voltage squared); LED drivers and electronic equipment may shut down; resistive heaters lose output (with the square of voltage); and every volt lost is energy wasted as heat in the conductors.
Sizing conductors for voltage drop
To find the conductor needed for a given maximum drop, rearrange the formula for the allowable effective impedance per unit length:
and choose the smallest conductor whose effective impedance does not exceed it, then check that it also meets the ampacity requirements. For small conductors where reactance is negligible, the equivalent shortcut uses resistance only, or the circular-mil formula familiar from ELEC-110, \(\text{cmil} = k\,K\,I\,L/V_d\), with \(K\) about 12.9 Ω·cmil/ft for copper at 75 °C.
Worked problem
Sizing a long branch circuit
A 240 V single-phase circuit supplies a 32 A EV charger 55 m from the panel. What copper conductor keeps the drop within 3%? (Use resistance only; R per km: 10 AWG 4.0 Ω, 8 AWG 2.6 Ω, 6 AWG 1.6 Ω.)
Show solution
- Allowable drop: \(0.03 \times 240 = 7.2\) V.
- Allowable R per km: \(7.2/(2 \times 32 \times 0.055) = 2.05\ \Omega\)/km.
- 8 AWG (2.6 Ω/km) is too high; 6 AWG (1.6 Ω/km) qualifies, giving \(2 \times 32 \times 0.055 \times 1.6 = 5.6\) V (2.3%).
Answer: Use 6 AWG copper, although 8 AWG would carry the current (the charger is a continuous load needing a 40 A circuit). When conductors are upsized for voltage drop, check whether the code requires the equipment bonding conductor to be increased in proportion, and that the terminals of the breaker and charger accept the larger conductor.
When conductors are increased in size to limit voltage drop, the bonding conductor is generally increased in proportion to the increase in circular mil area of the circuit conductors, so that fault current remains high enough to operate the protective device (Lesson 11).
Voltage drop with many loads and in distribution systems
A feeder supplying several loads along its length carries decreasing current toward its far end. Its total drop is the sum of the drops in each section, each calculated with the current flowing in that section. A convenient equivalent is to treat the total load as concentrated at the load centre, the current-weighted average distance.
Worked problem
Distributed loads on a feeder
A single-phase 240 V circuit (resistance 1.6 Ω/km, both conductors counted by the factor 2) supplies 10 A at 20 m, 10 A at 40 m and 10 A at 60 m. Find the voltage drop at the far end.
Show solution
- Section currents: 0-20 m carries 30 A; 20-40 m carries 20 A; 40-60 m carries 10 A.
- Drop \(= 2 \times 1.6 \times (30 \times 0.020 + 20 \times 0.020 + 10 \times 0.020) = 3.2 \times 1.2 = 3.84\) V.
- Load centre: \((10 \times 20 + 10 \times 40 + 10 \times 60)/30 = 40\) m; \(2 \times 1.6 \times 30 \times 0.040 = 3.84\) V, the same.
Answer: 3.84 V (1.6%). Treating all 30 A as drawn at the far end would have given 5.8 V, overstating the drop by half.
In a complete installation, the voltage at a load is the transformer secondary voltage (set by taps) minus the transformer's own regulation under load (Lesson 8), minus the feeder drop, minus the branch circuit drop. Planning distribution so that large loads are near their sources, using higher voltages for long feeders (a 600 V or 480 V feeder carries the same power with a third or less of the current of a 208 V feeder, and the drop as a fraction of voltage falls with the square of the voltage ratio), and locating transformers near load centres are the most effective ways to control voltage drop.
Motor starting
Induction motors draw five to seven times their full-load current when starting, at low power factor (0.2 to 0.4). Voltage drop during starting can be large enough to stall the motor, drop out contactors (which may release below about 70% to 80% voltage) and flicker lights. For starting, the effective impedance uses the starting power factor, at which reactance dominates.
Worked problem
Starting dip on a motor feeder
A 600 V three-phase motor has a full-load current of 40 A and a locked-rotor current of 240 A at 0.35 power factor. Its feeder has \(R = 0.050\ \Omega\) and \(X = 0.012\ \Omega\) per conductor. Find the running drop (at 0.85 PF) and the starting drop.
Show solution
- Running: \(Z_{\text{eff}} = 0.050 \times 0.85 + 0.012 \times 0.527 = 0.0488\ \Omega\).
- Running drop: \(\sqrt3 \times 40 \times 0.0488 = 3.4\) V (0.6%).
- Starting: \(Z_{\text{eff}} = 0.050 \times 0.35 + 0.012 \times 0.937 = 0.0287\ \Omega\); drop \(= \sqrt3 \times 240 \times 0.0287 = 11.9\) V (2.0%).
Answer: The feeder drop during starting is about three to four times the running drop. The total dip at the motor also includes the drop in the transformer and supply (Lesson 9), usually larger than the feeder's, which together determine whether the motor can start acceptably or needs a reduced-voltage starter (ELEC-310).
Measuring voltage drop in the field
Calculations predict voltage drop; measurements confirm it. The most direct method is to measure the voltage at the source (panel) and at the load at the same time, under the same load, with two meters or a meter with remote probes, because the supply voltage itself varies minute to minute. Alternatively, measure the voltage at the load with the load off and then on: the difference is the drop in the whole circuit, including the supply transformer, so it overstates the circuit drop slightly. A large drop on one conductor, measured end to end under load, points to a loose or corroded connection rather than an undersized conductor, especially if one conductor of a circuit shows much more drop than the other. Recording voltage at a critical load over a week with a power quality analyzer (Lesson 5) shows whether low voltage coincides with particular loads starting, which separates voltage drop in the building from low supply voltage.
Common mistakes
MistakeIgnoring reactance on large feeders.
InsteadUse effective impedance \(R\cos\theta + X\sin\theta\), especially above 1/0 AWG.
MistakeUsing the factor 2 for three-phase circuits.
InsteadUse \(\sqrt3\) for line-to-line drop in balanced three-phase circuits.
MistakeTreating distributed loads as all at the far end.
InsteadSum section drops or use the load centre.
MistakeUpsizing conductors for voltage drop but not the bonding conductor.
InsteadIncrease the bonding conductor proportionally as the code requires.
MistakeChecking only running conditions for motor feeders.
InsteadCheck starting dip at the starting power factor and current.
Summary and key terms
- AC conductors have resistance, slightly increased by skin and proximity effects, and reactance depending on spacing and raceway.
- Effective impedance \(R\cos\theta + X\sin\theta\) gives the drop in voltage magnitude.
- Single-phase drop is \(2ILZ_{\text{eff}}\); three-phase line-to-line drop is \(\sqrt3\,ILZ_{\text{eff}}\).
- Codes and tables limit or recommend 3% for feeders or branch circuits and 5% overall.
- Conductors are sized for the allowable impedance, with proportional increases in bonding conductors.
- Distributed loads use load centres; motor starting drops are calculated at locked-rotor current and power factor.
- Skin effect
- Concentration of alternating current near a conductor's surface.
- Proximity effect
- Distortion of current distribution caused by nearby conductors.
- Effective impedance
- The impedance component producing voltage magnitude drop, \(R\cos\theta + X\sin\theta\).
- Voltage drop
- The reduction in voltage between source and load caused by conductor impedance.
- Load centre
- The current-weighted average distance of distributed loads.
- Locked-rotor current
- The current a motor draws at standstill when first energized.
- Regulation
- The change in transformer output voltage from no load to full load.
- Circular mil
- A unit of conductor area equal to the area of a circle 0.001 in in diameter.
- X/R ratio
- Reactance divided by resistance of a conductor or circuit.
- Point of utilization
- The location where a load is connected.
Practice problems
Work each problem on paper first, then reveal the answer to check.
Effective impedance for \(R = 0.10\), \(X = 0.05\) at 0.8 PF?
Show answer
\(0.08 + 0.03 = 0.11\ \Omega\).
Single-phase drop: 20 A, 30 m, 5.2 Ω/km (resistive load)?
Show answer
\(2 \times 20 \times 0.03 \times 5.2 = 6.2\) V.
Three-phase drop: 100 A, \(Z_{\text{eff}}\) total 0.05 Ω per conductor?
Show answer
8.7 V.
Maximum drop for a 3% limit on a 208 V circuit?
Show answer
6.24 V.
Why does reactance matter more for large conductors?
Show answer
Their resistance is small, so reactance becomes comparable.
Torque of a motor at 90% voltage?
Show answer
About 81% of rated.
Load centre of 20 A at 10 m and 10 A at 40 m?
Show answer
20 m.
Ratio of current for the same power at 600 V versus 208 V (three-phase)?
Show answer
About 0.35.
Why is the starting voltage drop calculated with a different \(Z_{\text{eff}}\)?
Show answer
The starting power factor is much lower, so reactance dominates.
What else must be checked when upsizing conductors for voltage drop?
Show answer
Bonding conductor size and terminal suitability.
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