# AC vs. DC Electricity

1. May 11, 2005

### WarrickF

Hi All,

A friend of mine made a strange claim that I was very curious about.

He claims that power lines ( the type that run town to town ) run AC because it travels further distances without loss than DC.

Is this true, and if so why is there a difference?

Thanks
Warrick

2. May 11, 2005

### chroot

Staff Emeritus
This is potentially, but not necessarily true. The impedance of a transmission line can indeed be different at different frequencies. However, AC is used not because it suffers lower losses, but because it makes step-up and step-down conversions (with transformers) much easier. Such conversions are necessary, because transmission lines must use high voltages; line loss is dependent upon current. Larger voltages imply smaller currents, which lead to lower loss.

- Warren

3. May 11, 2005

### WarrickF

Thanks chroot, I guess I'll do some more reading before I bug you with the transformer questions. Why is there lower loss when there is smaller currents though?

4. May 11, 2005

### chroot

Staff Emeritus
The transmission line is effectively a large, distributed resistance. The power dissipated by a resistor depends on the current through it, not the voltage applied to it. (Consider that applying the same voltage to two different resistances results in two different amounts of power consumption; applying 12 V to an infinite resistance yields no power consumption; applying 12 V to a 1 ohm resistance yields 12 W of power consumption.)

- Warren

5. May 11, 2005

### WarrickF

That makes sense, thanks for you time chroot.

6. May 11, 2005

### jdavel

Are you sure you want to say that something in a resistor depends on current but not on voltage? Your argument works just as well the other way around. 12 amps through zero resistance yields no power consomption.

P.S. applying 12 volts to a 1 ohm resistance yields 144 watts, not 12. ;-)

7. May 11, 2005

### chroot

Staff Emeritus
Well, okay, voltage and current are related through resistance, so power consumption does depend on both. I didn't want to confuse him too much, but you're right, I misspoke.

The power loss through a resistance R is:

$$P = I^2 R = \frac{V^2}{R}$$

The voltage appears divided by the resistance, while the current appears multiplied by it. You don't want a large current for power transmission, you want a large voltage and a small current.

- Warren