## 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.

**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 E_{VD} = 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.

where:

**E**– Voltage drop, line-to-neutral, volts_{VD}**E**– Source voltage, line-to-neutral, volts_{s}**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.

**E _{R}** is the receiving end voltage.

#### 2. Exact Method #2

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

or

where:

**E**– Receiving line-line voltage in kV_{R}**E**– Sending line-line voltage in kV_{S}**MVA**– Receiving three-phase mVA_{R}**MVA**– Sending three-phase mVA_{S}**Z**– Impedance between and receiving ends**γ**– The angle of impedance Z**R**– Receiving end PF**S**– Sending end PF, positive when lagging

### 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.

### Calculations

**To calculate voltage drop:**

- 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.
- Divide by 100.
- 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)**.

**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:**

- Determine maximum desired i voltage drop, in volts.
- Divide voltage drop by ii (amperes x circuit feet).
- Multiply by 100.
**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.- 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.

#### 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.

**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**

**Reference //** Power Distribution Systems by EATON

Mohamed Fadhil AL-Mukhtar

Hi everybody,

In fact I have a confussion for which power factor I use in determining voltage drop of main cable (30m length) which supplies two motors (1st one 50HP , o.85 p.f, with 100m cable) and (2nd one 100HP , o.75 p.f,with 300m cable) .

Your answer is too much appreciated .

Thanks in advance

Chirantan Gupta

I saw the comments above and I guess what Gerald is referring to is the K factor method which is popularly followed by NEC. However, NEC also supports the IEEE 141-1993 formula. Refer to bottom of Table 9 Note 2 of NEC which clearly says: Multiplying current by effective impedance gives good approximation for line to neutral voltage drop. Thus, the approximate formula (Sqrt(3)*I*(Rcosphi+Xsinphi) can be followed. For all practical purposes, I have used this formula for last 25 years. The exact formula can also be used, and there is a vectorial method to get the solution.

Eng. Abdolgabar Ahmed Mahmood

Dear Engineers,

GERALD NEWTON AND TOMMY RICE.

Please tell me your methods of voltage drop calculations I will be grateful for you to let me know the best way of calculation because I am new in this field.

best regards

Eng. Abdolgabar Ahmed

Gerald Newton

Your voltage drop calculation article is off the wall. Sorry, but this is not how it is done. Furthermore, the State of Washington specifically states in a news letter how the inspectors want the calculation done. I have been in this business for over 40 years and have never, ever seen anything like your method. I suggest you do some research on how practical engineers and electricians do voltage drop calculations.

tommy rice

could you e-mail me and tell me how Washington state wants their calculations done. I use a different method for voltage drop calculations than edvard csanyi