Calculating voltage drop is very important in every electrical design especially when we are dealing with sensitive loads. Failure to calculate the voltage drop properly will result to under-voltage at the receiving end of the system. And undervoltage could lead into inefficient performance of our equipments.

The National Electrical Code (NEC) provide basic method in calculating voltage drop in the system. This code provides data of standard conductor properties that can be used in voltage drop calculation.

Voltage drop formula:

- Vd = ( 2 x Z x I x L )/ 1000 ---> for single phase system
- Vd = ( 1.73 x Z x I x L) / 1000 ---> for three phase system

where:

Vd = voltage drop

Z = impedance of the conductor per 1000 ft. (see NEC chapter 9 table 8 or 9)

I = load current in amperes

L = length in feet

1000 = constant to compensate with "per 1000 ft" in impedance value.

The National Electrical Code chapter 9 provides conductor properties based on 75 deg.C installations. Therefore for installations higher than above mentioned value we are required to consider special calculation procedure which is to be discuss in other article.

Table 8 refers to direct current system while table 9 refers to alternating current system. But there are important considerations in using this NEC table 8 and 9, viz:

- Use Table 8 for conductor sizes up to #4/0 since in this sizes R is approximately equal to Z
- Use Table 9 for conductor sizes higher that #4/0

NEC Chapter 9 Table 9 |

**Example:**

**A 10 HP motor is to be installed in a 220 V, 3-phase AC system. If the motor is located at 300 feet from the source, calculate its voltage drop?**

Solution:

There are three approach in getting the current drawn of any motor.

- Nameplate Rating
- NEC given value
- Conventional Calculation

For the sake of discussion we will take the third method by assuming power factor and efficiency of the motor to be 80%.

I = (20 HP x 745 W) / (1.73 x 220 x 0.80 x 0.80)

I = 61.25 Amperes ---> full load ampres (FLA)

NEC provides that the the conductor must not be lesser than 125% of the motor's FLA.This value, 76.6 amperes will the basis of our voltage drop calculation.

- Therefore: I = 56.0 x 1.25 = 76.6 Amperes
- NEC table of conductor ampacities provides that at 76.6 amperes wire size must be #4 copper based on NEC Art 310

*note: the voltage drop calculation should be taken after we get the 125% safety factor provided by the NEC.*

It follows that,

Vd= (1.73 x

**Z of #4**x 76.6 x 300 ft. )/

**1000**

Based on NEC Chapter 9 Table 9,

- Z = 0.321 ohm/ 1000 ft
- Therefore: Vd = (1.73 x 0.321 x 76.6 x 300) /1000
- Vd = 12.76 Volts

It follows that, %Vd = 12.76 V / 240 V = 5.3%

The NEC recommends that 5% voltage drop percentage is allowable in any circuit installation so if we strictly follow this recommendation then we will not accept #4 conductor but rather we will move up to the next higher size.

## 3 comments

Furthermore, by varying the current setting and the restraint on the movable element, Used Lab Equipment Distributors

...and the 10 HP motor in the problem statement somehow turns in to a 20HP motor. It don't hurt to revise your work before publishing

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