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Home / Technical Articles / Voltage Drop Calculations For Engineers – Beginners

Voltage drop formulas

Voltage drop calculations using the DC-resistance formula are not always accurate for AC circuits, especially for those with a less-than-unity power factor or for those that use conductors larger than 2 AWG.

Voltage Drop Calculations For Engineers - Beginners
Voltage Drop Calculations For Engineers - Beginners (on photo: Low voltage circuit breakers type NSX 250H, 600V)

Table 1 allows engineers to perform simple ac voltage drop calculations. Table 1 was compiled using the Neher–McGrath ac-resistance calculation method, and the values presented are both reliable and conservative. This table contains completed calculations of effective impedance (Z) for the average ac circuit with an 85 percent power factor (see Calculation Example 1).

If calculations with a different power factor are necessary, Table 1 also contains the appropriate values of inductive reactance and AC resistance (see Example 2).

The basic assumptions and the limitations of Table 1 are as follows:

  1. Capacitive reactance is ignored.
  2. There are three conductors in a raceway.
  3. The calculated voltage drop values are approximate.
  4. For circuits with other parameters, the Neher–McGrath ac-resistance calculation method is used.

Calculation Example #1

A feeder has a 100 A continuous load. The system source is 240 volts, 3 phase, and the supplying circuit breaker is 125 A. The feeder is in a trade size 1¼ aluminum conduit with three 1 AWG THHN copper conductors operating at their maximum temperature rating of 75°C. The circuit length is 150 ft, and the power factor is 85 percent.

Using Table 1 below, determine the approximate voltage drop of this circuit.

See the solution //

STEP-1 // Find the approximate line-to-neutral voltage drop.

Using the Table 1 column “Effective Zat 0.85 PF for Uncoated Copper Wires”, select aluminum conduit and size 1 AWG copper wire. Use the given value of 0.16 ohm per 1000 ft in the following formula:

Voltage drop example

STEP-2 // Find the line-to-line voltage drop:

Voltage drop example

STEP-3 // Find the voltage present at the load end of the circuit:

Voltage drop example

Calculation Example #2

A 270 A continuous load is present on a feeder. The circuit consists of a single 4-in. PVC conduit with three 600-kcmil XHHW/USE aluminum conductors fed from a 480 V, 3-phase, 3-wire source. The conductors are operating at their maximum rated temperature of 75°C.

If the power factor is 0.7 and the circuit length is 250 ft, is the voltage drop excessive?

See the solution //

STEP-1 // Using the Table 1 column “XL (Reactance) for All Wires”, select PVC conduit and the row for size 600 kcmil. A value of 0.039 ohm per 1000 ft is given as this XL. Next, using the column “Alternating-Current Resistance for Aluminum Wires”, select PVC conduit and the row for size 600 kcmil. A value of 0.036 ohm per 1000 ft is given as this R.

STEP-2 // Find the angle representing a power factor of 0.7.

Using a calculator with trigonometric functions or a trigonometric function table, find the arccosine (cos-1) θ of 0.7, which is 45.57 degrees. For this example, call this angle.

STEP-3 // Find the impedance (Z) corrected to 0.7 power factor (Zc):

Voltage drop example

STEP-4 // As in Calculation Example 1, find the approximate line-to-neutral voltage drop:

Voltage drop example

STEP-5 // Find the approximate line-to-line voltage drop:

Voltage drop example

STEP-6 // Find the approximate voltage drop expressed as a percentage of the circuit voltage:

Voltage drop example

STEP-7 // Find the voltage present at the load end of the circuit:

Voltage drop example

Conclusion // According to 210.19(A)(1), Informational Note No. 4, this voltage drop does not appear to be excessive.

TABLE 1 //

Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F)

Three Single Conductors in Conduit //
TABLE 1 - Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F) - Three Single Conductors in Conduit
TABLE 1 – Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F) – Three Single Conductors in Conduit

Reference // National Electrical Code Handbook – Mark W. Earley, P.E., Jeffrey S. Sargent, Christopher D. Coache and Richard J. Roux (National Fire Protection Association, Quincy, Massachusetts)

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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. deepak
    Mar 19, 2021

    Hi All,

    Does anyone know how to use this table for 90 deg C XLPE cables?

  2. Audu Eneojo
    Jul 31, 2019

    Hello Edverd,

    hwo about in a situation of new installation on site where,
    1. generator capacity is 15KVA – 3phase
    2. cable disctance is 400m from Gen-set to Load
    3. Cable size is 4core 35Sq.m copper

    what will be the voltage drop using the above parameters.
    please give detail calculations.

    • Deepak
      Mar 19, 2021

      Assuming Voltage as 400V, power factor as 0.8
      For 4C, 35mm2 cable, R=0.5171 ohm/km; X = 0.0830 ohm/km @ 20deg C.
      Voltage drop = Sqrt(3) * Amps * (R cos phi + X Sin phi) * length in km

      Voltage drop = 1.732 * 21.6 * (0.5171*0.8+0.0830*0.6) *0.4 = 6.93 V.

      Voltage drop % = 6.73 / 400 % = 1.68 %

  3. Salako Olakunle
    Jan 11, 2017

    Good work, it has been very educative. But please, at what percentage does the voltage drop becomes excessive and what is the probable effect on the circuit?

  4. manjunatha
    Nov 23, 2015

    how mccb trip on off wiring diagram send mi please

  5. Madhavanpillai
    Nov 21, 2015

    Highly informative articles

  6. Vishnu
    Oct 11, 2015

    A valuable site for Electrical Engineers

  7. Doug McIntire, PE
    Sep 25, 2015


    This is excellent and a good reminder to avoid “rule of thumb” calculations in engineering.

    I might point out that the last few steps are not needed. The L-L voltage drop is the same as L-N in percentage the beauty of per unit type calculation. See in step #4 of example #2 voltage drop is 3.584V. Divided into the line to neutral voltage of 277V you get the same answer of .0129 or 1.29%. That is all that is required to meet the code “noted” for feeders and branch circuits operating at reasonable efficiency.

    Be careful too as this defined as a feeder. I am led to assume that there are branch circuits following on to this feeder load and that must be included in the consideration. See NEC Article 100 definitions of branch circuit and feeder. I’m possibly splitting hairs but we are making finer distinctions. Thanks for doing this. Very useful.

  8. S.D.Buddhisagar
    Sep 25, 2015

    R/s : I am , 64 , try to read your articles , through which I have benefited a lot , I request you to guide me on how to do circuit analysis dependent sources, Ialways get confused.

  9. Ali C. Hussein
    Sep 21, 2015

    good work

    Sep 20, 2015

    its a very good articles for updating me. thank you.

  11. Khalid Ur Rehman Shah
    Sep 20, 2015

    Good work

  12. James M. Casallo, P.E.
    Sep 19, 2015

    Great article! I seem to always use the Line to Neutral value. Thanks for the update and exercise. Keep up the great work and sharing of the knowledge.

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