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Home / Technical Articles / Voltage drop calculation methods with examples explained in details

Voltage drop considerations

The first consideration for voltage drop is that under the steady-state conditions of normal load, the voltage at the utilization equipment must be adequate.

Voltage drop calculation methods with examples explained in details
Voltage drop calculation methods with examples explained in details
Fine-print notes in the NEC recommend sizing feeders and branch circuits so that the maximum voltage drop in either does not exceed 3%, with the total voltage drop for feeders and branch circuits not to exceed 5%, for efficiency of operation.

In addition to steady-state conditions, voltage drop under transient conditions, with sudden high-current, short-time loads, must be considered.

The most common loads of this type are motor inrush currents during starting. These loads cause a voltage dip on the system as a result of the voltage drop in conductors, transformers and generators under the high current. This voltage dip can have numerous adverse effects on equipment in the system, and equipment and conductors must be designed and sized to minimize these problems.

In many cases, reduced-voltage starting of motors to reduce inrush current will be necessary.

Voltage drop formulas

Let’s see two most common methods for calculation of voltage drop – approximate and exact methods:

1. Approximate method

Voltage drop EVD = IR cosθ + IX sinθ where abbreviations are same as below “Exact Method”.

2. Exact method #1

If sending end voltage and load PF are known.

Voltage drop calculation - Exact method 1


  • EVD – Voltage drop, line-to-neutral, volts
  • Es – Source voltage, line-to-neutral, volts
  • I – Line (Load) current, amperes
  • R – Circuit (branch, feeder) resistance, ohms
  • X – Circuit (branch, feeder) reactance, ohms
  • cosθ – Power factor of load, decimal
  • sinθ – Reactive factor of load, decimal

If the receiving end voltage, load current and power factor (PF) are known.

Voltage drop - Exact method If the receiving end voltage, load current and power factor (PF) are known

ER is the receiving end voltage.

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2. Exact Method #2

If receiving or sending mVA and its power factor are known at a known sending or receiving voltage.

Voltage drop - Exact method 2


Voltage drop - Exact method 2


  • ER – Receiving line-line voltage in kV
  • ES – Sending line-line voltage in kV
  • MVAR – Receiving three-phase mVA
  • MVAS – Sending three-phase mVA
  • Z – Impedance between and receiving ends
  • γ – The angle of impedance Z
  • R – Receiving end PF
  • S – Sending end PF, positive when lagging

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Voltage drop tables

Tables for calculating voltage drop for copper and aluminum conductors, in either magnetic (steel) or nonmagnetic (aluminum or non-metallic) conduit, are shown below. These tables give voltage drop per ampere per 100 ft (30 m) of circuit length.

The circuit length is from the beginning point to the end point of the circuit regardless of the number of conductors.

Tables are based on the following conditions:

Condition #1

Three or four single conductors in a conduit, random lay. For three-conductor cable, actual voltage drop will be approximately the same for small conductor sizes and high power factors. Actual voltage drop will be from 10 to 15% lower for larger conductor sizes and lower power factors.

Condition #2

Voltage drops are phase-to-phase, for three-phase, three-wire or three-phase, four-wire 60 Hz circuits. For other circuits, multiply voltage drop given in the tables by the following correction factors:

Correction factors table:

Three-phase, four-wire, phase-to-neutral × 0.577
Single-phase, two-wire × 1.155
Single-phase, three-wire, phase-to-phase × 1.155
Single-phase, three-wire, phase-to-neutral × 0.577

Condition #3

Voltage drops are for a conductor temperature of 75 °C. They may be used for conductor temperatures between 60 °C and 90 °C with reasonable accuracy (within ±5%). However, correction factors in Table 1 can be applied if desired. The values in the table are in percent of total voltage drop.

  • For conductor temperature of 60 °C – SUBTRACT the percentage from Table 1.
  • For conductor temperature of 90 °C – ADD the percentage from Table 1.
Table 1 - Temperature correction factors for voltage drop
Table 1 – Temperature correction factors for voltage drop

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To calculate voltage drop:

  1. Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.
  2. Divide by 100.
  3. Multiply by proper voltage drop value in tables. Result is voltage drop.

Example #1

A 460 V, 100 hp motor, running at 80% PF, draws 124 A full-load current. It is fed by three 2/0 copper conductors in steel conduit. The feeder length is 150 ft (46 m).

What is the voltage drop in the feeder? What is the percentage voltage drop?
  • 124 A × 150ft (46m) = 18,600 A-ft
  • Divided by 100 = 186
  • Table: 2/0 copper, magnetic conduit,
    80% PF = 0.0187
    186 x 0.0187 = 3.48 V drop
    3.48/460 x 100 = 0.76% drop

Conclusion: 0.76% voltage drop is very acceptable. (See NEC Article 215, which suggests that a voltage drop of 3% or less on a feeder is acceptable.)

To select minimum conductor size:

  1. Determine maximum desired i voltage drop, in volts.
  2. Divide voltage drop by ii (amperes x circuit feet).
  3. Multiply by 100.
  4. Find nearest lower voltage drop value in tables, in correct column for type of conductor, conduit and power factor. Read conductor size for that value.
  5. Where this results in an oversized cable, verify cable lug sizes for molded case circuit breakers and fusible 4 switches. Where lug size available is exceeded, go to next higher rating.

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Example #2

A three-phase, four-wire lighting feeder on a 208 V circuit is 250 ft (76.2 m) long. The load is 175 A at 90% PF. It is desired to use aluminum 7 conductors in aluminum conduit.

What size conductor is required to limit the voltage drop to 2% phase-to-phase?
  • VD = 2/100 × 208  = 4.16 V
  • 4.16 / (175 × 250) = 0.0000951
  • 0.0000951 × 100= 0.00951
  • In table, under aluminum conductors, nonmagnetic conduit, 90% PF, the nearest lower value is 0.0091. Conductor required is 12 500 kcmil.
    (Size 4/0 THW would have adequate ampacity, but the voltage drop would be excessive.)

Table 2 – Voltage Drop—Volts per Ampere per 100 Feet (30 m); Three-Phase, Phase-to-Phase

Voltage drop - Volts per ampere per 100 Feet (30 m); Three-phase, phase-to-phase
Table 2 – Voltage drop – Volts per ampere per 100 Feet (30 m); Three-phase, phase-to-phase

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Reference // Power Distribution Systems by EATON

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Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.


  1. ely barruel
    Dec 02, 2021

    Im confused with correction factor table in condition #2. Should this item be “Three-phase, three wire, phase-to-phase x1.155” instead of “Single-phase, three wire, phase-to-phase x1.155”

    Nov 20, 2021




  3. Sigurd Nydal
    Aug 31, 2021

    If found it stranges how dificult it seem to calculate pressure drop.

    The resistance in conductors = Lengh x 0,0175/Divide by Cross sectional area.

    The voltage drop 20 meter conduct , 10 Ampere ,1,5mm2, =4,7 Volt. That is for 2 wire cable

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