If the electric potential is 220V and some device needs 1500watts then how does it suck exactly that amount of electrical energy from outlet?
$\begingroup$
$\endgroup$
8
-
$\begingroup$ Humor is allowed in the comments if it is reasonably related to the topic and projects a professional image. Unprofessional remarks deleted. $\endgroup$– dmckee --- ex-moderator kittenJan 22, 2012 at 21:36
-
$\begingroup$ you get payed for sitting on .sx? I dont get payed so I dont get to be professional, a valid reason would be that my remark was utter b_llshit $\endgroup$– propagandaJan 22, 2012 at 21:53
-
6$\begingroup$ @propaganda - this isn't hackernews or 4chan. We want SO to be a respected source of technical answers. This isn't going to happen if schools/businesses start to block it because of naughty words. $\endgroup$– Martin BeckettJan 22, 2012 at 22:20
-
$\begingroup$ who holds the copyright to questions and answers? if I dont get paid for my content users should not be enforced to do sx's homework for them... also sucking and pushing and shoving electric energy must rub off an image of being a respectable resource of technical answers $\endgroup$– propagandaJan 22, 2012 at 23:54
-
$\begingroup$ CC commons license question, Are moderators paid to moderate? $\endgroup$– dmckee --- ex-moderator kittenJan 23, 2012 at 0:27
|
Show 3 more comments
5 Answers
$\begingroup$
$\endgroup$
2
It is not the most straightforward viewpoint to say that a device "needs 1500 watts". This is more a consequence than a condition. What happens is that you create an electric circuit by plugging in a device into the outlet. That circuit follows Ohm's law: $$V = I R$$ So for a given voltage and resistance a certain current $I$ will flow. The power is simply $$P = V I = V^{2}/R = I^2 R$$The device does not know anything, it just has a property, it's resistance $R$.
-
$\begingroup$ Tiny correction: the question refers, most probably, to an AC case, where a factor of $\sqrt{2}$ has to be added. $\endgroup$– yohBSJan 22, 2012 at 18:22
-
$\begingroup$ @yohBS: I don't think any factor is missing here. For a purely resistive device the description is correct and does not change for AC. The 220V is not the amplitude but rms. $\endgroup$ Jan 22, 2012 at 19:21
$\begingroup$
$\endgroup$
If it's a light bulb or heater, it's just a resistor.
First, forget that it's alternating current, just to simplify things. Think of the power source as a really big 220 volt battery.
If it's drawing 1500 watts, divide that by 220, and that will tell you the current I in Amperes. (That just measures how many electrons per second are flowing. An Ampere is about 6x10^23 electrons per second.)
To get the resistance R in Ohms, just divide the voltage V (220 volts) by the current I that you got above. (An Ohm is just the number of volts it takes to push one Ampere through the resistor.)
I hope you can see that the smaller the resistance is, the bigger the current is, and when you multiply that by the voltage, you get the power.
So the way you make a bigger heater or light bulb is by giving it less electrical resistance.
If you want to go back to alternating current (AC) the power is a time-average, and it swings up and down at twice the AC frequency. I'll let you figure out why, if you want.
answered Jan 22, 2012 at 16:48
$\begingroup$
$\endgroup$
6
Probably easier to think of it in terms of pressure and 'pushing'
Electrical potential (volts) are an analogue of pressure - there is a certain amount of electrical pressure trying to push current through the circuit. The resistance is what impedes the flow of current. So a high power device, like a 3500W kettle, has a low resistance and so the 220V pressure can push a lot of current through it.
If you then take the same device to a country with wussy little 110V of electrical pressure then it will only be able to push half as much current through the same resistance and so only give 1/4 as much power. Which is why they generally don't use electrical kettles in north America.
answered Jan 22, 2012 at 17:13
-
1$\begingroup$ P = IV = V^2/R? Want to upvote since I think this is the best answer so far, but please fix this. $\endgroup$– user2963Jan 22, 2012 at 20:11
-
-
$\begingroup$ @zephyr - sorry lack of coffee, still waiting for the pathetic 110V kettle to boil! $\endgroup$ Jan 22, 2012 at 21:10
-
$\begingroup$ but then again you are wrong: if U_{NA}=220V=I*R and U'_{JEU}=110V=I'*R then I' = I/2, and P'=U'*I'=P/4! so jewropean outlets can
push4 times more energy through the same kettle resistor than in Northern States of America $\endgroup$ Jan 22, 2012 at 21:13 -
1$\begingroup$ @propaganda - brain-needs-coffeeeee....... ;-) $\endgroup$ Jan 22, 2012 at 21:30
$\begingroup$
$\endgroup$
1
Circuits in devices are designed to regulate their own power consumption: when they need more power they decrease their resistance, and when they need less they increase their resistance, often breaking the circuit open when they stop needing energy at all
-
$\begingroup$ LOL you remind me of Calvin's dad. You made me laugh $\endgroup$ Jan 25, 2012 at 20:56
$\begingroup$
$\endgroup$
The easy devices have been covered already. More complex devices, such as PCs use smart power supplies. These power supplies need to create low DC voltages (e.g. 12V) from the input voltage (110V or 220V AC).
A simplified model of such power supply has a small battery at 12V. This battery is continuously discharged to provide the device with the power it needs. The power supply checks the battery charge level, and when it drops too low the battery is quickly charged from the rectified (= AC turned into DC) input voltage. When the device draws more power, the power supply will recharge the battery more often. But if the battery is too small, the power supply can't keep up. Therefore power supplies for PC's come in different Watt ratings.
