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Q. What is voltage drop?

A. Voltage drop is the reduction of available voltage due to cables having resistance and/or impedance.

When a load is connected to supply, a cable is used to carry electrons from the supply to the load. Copper and aluminum are the two main conductors used in the electrical field. The problem with these two conductors is that they are not "pure" conductors. Due to their atomic structure, these conductors offer a small amount of opposition to current flow. The by-product of connecting these conductors in series with a load is that the voltage is reduced at the load end of the cable. The voltage drop will be relative to the amount of resistance that the cable has.

Why is voltage drop important ?

To many of us, voltage drop isn't a consideration. Cables have been selected on their ability to deliver current to a load and that is the end of the story.

Not only is voltage drop a regulatory issue, excessive voltage drop will affect the performance of loads and appliances.

AS/NZ 3000:2007 stipulated that all low voltage installations must not exceed a voltage drop of 5% of the nominal voltage.

This means for a 400V three phase systems, your maximum allowable voltage drop is 20V and for a 230V system it is 11.5V. This voltage is from the point of supply to anywhere in the installation.


To calculate voltage drop, a simple formula can be used.

Vd = L x I x Vc / 1000

L = length of cable, I = current and Vc is the losses associated with a cable (millivolts per amp per meter drop (mV/A.m)). Vc values can be obtained from AS/NZ 3008.1.1, tables 40 - 50.

AS3008 Tables 41 and 42 are the two most used tables for Vc (single insulated and multicore copper cables).

Tables 40 - 50 are all three phase mV/A.m figures. In order to used these figures on single phase circuits you must convert them by multiplying the mV/A.m by 1.155



A power circuit ran in 2.5mm V90 2c&e TPS is 45 meters long and will carry 16 amps. (T42 for TPS Vc figures)

Vd = 45 x 16 x (15.6 x 1.155) / 1000 = 12.9V

This example is a common circuit that electricians install. As you can see, voltage drop will be a consideration in this installation. BUT WAIT, we haven't considered the consumer mains in this calculation!!!

Remember we must keep our installation under 5%(11.5V) from the point of supply. This means the consumer mains must be calculated.



The consumer mains for the previous example are 16mm V90 building wires (230V) installed in a conduit. The length is 18 meters and the maximum demand for the house is 48 amps. (T41 for TPI Vc figures)

Vd = 18 x 48 x (2.55 x 1.155) / 1000 = 2.5V

Total voltage drop will be the sum of the consumer mains and final sub circuit voltage drop.

2.5V + 12.9V = 15.4V


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