# Why less voltage drop with 240 wire vs 120 wire for the same length of run?

## Answer

###### Wiki User

###### 01/04/2011

### Answer for USA, Canada and countries running a 60 Hz supply service.

There is no such thing as 240 Volt or 120 Volt wire. All standard household electrical wiring is rated for up to 600 Volts. The voltage a wire carries is whatever you put into it. And anyway, considering the 120/240 Volt service as in a standard home, a single hot wire only has 120 Volts going through it at all times. You get 240 Volts by having two "hot" wires, each 120 Volts, fed from the Black and the Red "hot" legs of the incoming main supply.Voltage drop in wire is determined by the "gauge" of the wire, which is the thickness of it. A 12 gauge wire is thicker than a 14 gauge wire. The 14 gauge wire will have more voltage drop caused by the current passing through it because it is smaller and has more resistance.

Voltage drop will also be affected by the amount of voltage on a wire. It would be incorrect to assume that voltage drop is a percentage based on the length of the wire. The wire gauge affects the resistance of the wire, but assuming the same size (gauge) of wire, a higher voltage will have a greater force behind it than a lower voltage so essentially the resistance of the wire doesn't affect higher voltage as much as it does lower voltages.

If you were to compare electrical wiring with plumbing, voltage would be equivalent to water pressure, and amps would be equivalent to flow, or gallons per hour.

**Additional Comments and Answer for European Electrical
Installations**

The questioner probably *really* means "Why is there less
voltage drop for a cable operating at 240 V, *compared to the
same cable operating at 120 V".*

If this is the case, then the answer is simple. For a load of any given power, the higher the voltage, the lower the resulting current. For example, let's suppose we have a 1000-W load operating at 230 V -this load will draw a current of 4.35 A. The same load operating a 120 V will draw a current of 8.33 A. So, for any given conductor size, the voltage drop (Iload x Rconductor) will clearly be greater.

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