# Why is electricity generated as alternating current and not as DC?

Jun 9, 2017

This is not true

#### Explanation:

We have both AC and DC generators which generate AC and DC current , both run with same mechanism but differ by just , that DC generator uses split rings but ac uses slip rings . for more details you can see a proper demonstration on YouTube .

Jun 9, 2017

The main reason is that the voltage of Alternating Current can be "stepped up" for long distance distribution and then "stepped down" for local distribution, using transformers.

#### Explanation:

I believe that you are asking; "Why is Alternating Current commonly used by Electric Companies to provide electric energy to their customers?"

Here is an example:

Let's assume that you are a power company and you have chosen to use DC:

1 mega-Watt industrial load ($440$ Volts DC @ ~2300 Amperes)

10 kilometers away from the power generator and the resistance of the power lines are $0.1 \frac{\Omega}{\text{km}}$

The loss of the power by the 2 power lines (one for + and the other for the -) is:

P_"loss" = 2(2300"A")^2(10"km")(0.1Omega/"km")

${P}_{\text{loss}} \approx 10$ mega-Watts

The voltage drop on the power 2 power lines $4600$ Volts

This means that your generator voltage is $5040$ Volts and it must generate $2300$ Amperes. You generate 11 mega-Watts to deliver 1 mega-Watt.

Consider the same problem with AC and you can step-up to 250,000 volts for distribution and step-down at the load:

1 mega-Watt industrial load ($440$ Volts AC @ ~2300 Amperes)

We step up the voltage to 250,000 volts for distribution:

$P = \left(440 \text{VAC")(2300"A") = (250000"VAC") (x"A}\right)$

$x = \frac{\left(440\right) \left(2300\right)}{250000}$

x ~~ 4"A"color(red)(larr "This is the current in the distribution lines")

10 kilometers away from the power generator and the resistance of the power lines are $0.1 \frac{\Omega}{\text{km}}$

The loss of the power by the 2 power lines (one for + and the other for the -) is:

P_"loss" = 2(4"A")^2(10"km")(0.1Omega/"km")

${P}_{\text{loss}} \approx 32$ Watts

The choice between losing 10 million Watts to deliver 1 million Watts or losing 32 Watts to deliver 1 million Watts to a customer is simple. That is why power companies use AC.