Science
Electrical Wiring

Some places have 120 Volts and others 240 Volts What is the advantage of one over the other?

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2011-01-15 04:07:59
2011-01-15 04:07:59

The answer given below refers only to USA, Canada and other countries using the same type of 60 Hz, 240 Volts "balanced around ground" system for standard mains power supplies to homes, offices, etc. Basically, it comes down to reducing the amount of lost power due to resistance of the wires in the walls of your home. Higher voltages can deliver more power to a load, with less power lost in the transfer. The downside is that higher voltages are more dangerous, and require more insulation to keep the wires safe to touch.

The long explanation

The reason behind why it can transfer more power, is a bit complicated. It all comes down to Joule's laws. Power used (P) is equal to Voltage (V) times Current (I). (P=I*V) So twice the voltage will transfer twice the power. So why not just use double the current, and avoid the safety issues with higher voltages?

Wires conduct electricity, but they also have some resistance. The more the resistance, the more power is lost due to heat. The power loss in wires can be calculated by using the equation P(loss)=I^2*R (or current times current times resistance). We'll ignore resistance for now, and assume it's constant. If you double the amount of current going through a wire, you quadruple the power losses and heat generated by 4 (2 * 2 = 4). So by using twice the current, you have transferred twice the power, but are now losing 4 times more power in the wires. The amount lost is small, but not trivial.

So let's say we want to cut our power losses by a factor of 4, while still transmitting the same amount of power. Remember the equations P(loss)=I^2*R and P(used)=I*V.

We can double V and halve the I and still get the same amount of power:

1/2*I * 2*V = I*V = P(used)

This will cut our power losses by 4:

1/2*I * 1/2*I * R = 1/4*P(loss)

Or, we could cut the resistance by 4 and leave I and V alone:

I*I * 1/4*R = 1/4*P(loss)

To do this though, we would need to increase the size of our wires which can get expensive, and is much more difficult to install.

There are other much more complicated reasons as well, dealing with mutual inductance and power factor correction, but they can be ignored as they're way outside the scope of this answer.

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