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I'll put a set of ''right's''numbered based on my current assumptions, so if one of them is wrong, you can just point a couple out.

Let's say I have a device that is labeled ''660W/220V'', so It delivers 660W on a jack of 220V, right(1)? Then, let's move on: Based on Ohm's law: P = U²/R, applying the values, we get an INTERNAL RESISTANCE of the device of about 73,3 ohms, right(2)?

With U = R * i, the device draws a current of about 3A, right(3)?

So, if the internal resistance doesn't change, plugging the same device on a 110V jack now will only produce 165W and will only draw 1,5A right(4)?

Thing is, I've seen a video of a dude plugging 2x 60W lightbulbs on 2x outlets each, one in 110V and one in 220V. The one in the 110V drawed twice the amount of current compared with the one in 220V. That doesn't make sense.

Maybe the bulbs were different and designed to operate on the voltage he plugged? So in that case, a 60W/110V bulb have a smaller internal resistance, so it need to draw way more current to produce 60W, is that it?


Extreme stupid question: I don't understand the logic behind P = V * i, The amount of voltage in an outlet should only determine the capacity of current it can provide. More voltage should equal more current. More voltage equals less current makes sense in the formula, but not in my head lol.

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closed as unclear what you're asking by Andy aka, Daniel Grillo, placeholder, PeterJ, uint128_t Jun 22 '16 at 0:11

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ FYI, "tension" is not widely used in English. We more often use "potential difference" or just "voltage". \$\endgroup\$ – The Photon Jun 21 '16 at 16:43
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if the internal resistance doesn't change, plugging the same device on a 110V jack now will only produce 165W and will only draw 1,5A right(4)?

Correct. But there are very few loads that really act like pure unchanging resistors. Even an incandescent bulb's resistance changes value as the filament heats up.

Maybe the bulbs were different and designed to operate on the voltage he plugged? So in that case, a 60W/110V bulb have a smaller internal resistance, so it need to draw way more current to produce 60W, is that it?

Correct.

You could also see this when using a device powered by a switching power supply. The supply will adjust its current draw to supply the same power to its load. So on 110 V mains it will need to draw about twice as much current as on 220 V mains.

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  • \$\begingroup\$ Interesting on how everything is basically derived from the internal resistance. It's like they build the internal resistance first and then work the way up to decide how much voltage and current it needs? \$\endgroup\$ – Hiago Serpa Jun 21 '16 at 16:49
  • \$\begingroup\$ @HiagoSerpa, for a lightbulb mainly (but more likely you'd start knowing the mains voltage and the amount of light you want, then calculate what the filament resistance needs to be to achieve that). For other loads (computers, phone chargers, ...) a simple resistance isn't a good model for the way the device will load the mains. \$\endgroup\$ – The Photon Jun 21 '16 at 16:50
  • \$\begingroup\$ P = V I is the basic formula. I is how much charge is moved around per second. V is how much energy it takes to move each element of charge. Energy per second is power. \$\endgroup\$ – The Photon Jun 21 '16 at 17:23
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  1. Depends. That's an average or max. that shouldn't be exceeded. What device are we talking about? Because, if this is a power supply (for example), it can go from few Watts to 660W, yes. But also can go a bit higher too, if we put something such as motor on it or so.

  2. Internal resistance (as you call it) is called impedance. But that's just as long as it is a single device, not a power supply, which just converts (AC-AC, AC-DC, DC-AC or DC-DC voltage). In this case, it's impedance is much higher.

  3. Yes, about. During the time of work it probably falls down (at least a bit, but theoretically...yes).


The factor P (power) is actually just a energetics thing. Electronics doesn't care about it (unless seing for example what resistor to use, 1/4, 1/2, 1W), because we ALWAYS use determined voltage and determined current. Yes, they together combine power, but it's just not that simple. More voltage equals less current, why would wanted to take more current? It's the same effect. If you need to boil a jar of water, you can put it on stove, who is pre-heated on 100°C for 1 minute or to a preheated stove on 200°C for 30 seconds. The effect is the same, time is different. P=U*I is simmiliar.

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The resistance of the filament depends on the temperature. As long as the temperature is low the resistance is low as well. As soon as current flows the temperature rises and the resistance increases.

A certain power is required to heat up the filament.

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If we neglect that you are talking about AC currents and voltages (which make calculations a bit more complex for non-resistive loads), you are correct with (1), (2), (3) and (4).

The light bulbs seemed to be rated for the corresponding voltage, so in order for both to achieve 60W, the 110V version has to draw twice the current than the 220V bulb. Then both will deliver 60W in the voltage system they've been designed for.

Things would be as you expect only if the same type of 60W/220V bulb were to be used in a 110V and 220V system. Then, in the 110V system, the current would be half the current of the 220V system.

I also would like to add that often there is a small frequency difference between different mains systems (most commonly 50Hz vs. 60Hz), which adds small changes in current for inductive loads (motors), because the Impedance ("internal resistance") changes with frequency.

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Short:

Light bulbs are not well behaved resistors as they change resistance as the filament heats. This affects how they behave on different voltages, but in the video example that you cite there is almost certainly a simple explanation

The bulbs were almost certainly in series on 220 VAC and in parallel on 110 VAC. This would fully explain the current draws that were seen.

If they did not explain this then they were trying to trick you.

Longer:

DO NOT believe everything technical (or even most things) that you see in videos and/or on the internet. If you can provide a link to the video we may well be able to provide a better answer. Otherwise the best answer is -

"In a video ???????? !!!!". :-).

BUT do note that

  • Two 110V 60 W bulbs IN SERIES on 220V
    will draw about 120/200A = 0.545A.

  • AND the same two bulbs IN PARALLEL on 110V
    will draw 120W/110V = 1.09A
    or twice as much as on 220 VAC.

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For linear pure "ideal" resistors you can apply Ohms law and the related power formulae:

Ohms law is the first formula here but the three examples are simply rearrangements of the same formula.

  • Resistance = Voltage / Current
    R = V/I

  • Current = Voltage / Resistance
    I = V / R

  • Voltage = Current x Voltage
    V = I x R

Power dissipated in a resistance can be expressed by the following formula.
The three examples are simply the same expression rearranged with different variable substituted.

  • Power = Volts x Amps
    P = V x I

  • Power = Voltage drop squared / Resistance
    P = V^2/R
    as P
    = V x I
    = V x (V/R)
    = V^2/R

  • Power = Current squared x Resistance
    P = I^2 x R
    as P
    = V x I
    = (I x R) x I
    = I^2 x R

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  • \$\begingroup\$ @HiagoSerpa Your head needs adjusting :-). Water analogy: Voltage is equivalent to pumping pressure. Current = current. If you have a 14 psi pump and it delivers 1 gallon/second at that pressure it will fill a tank at about 32 feet head at 1 gal/sec (as 32 feet of head = 14 psi pressure in practice). If you had a pump that could deliver 3 gallons/sec at 14 psi it would fill the same tank 3 x as fast. ... \$\endgroup\$ – Russell McMahon Jun 21 '16 at 18:09
  • \$\begingroup\$ BUT if you had a pump that could deliver 1 gallon/second at 42 psi it would fill a tank at 96 feet head at the same rate as the first pump did at 32 foot head. Now, 3 x flow at same head OR 1 x flow at 3X head is 3x original power in each case. ie pressure of pumping x amount pumped equals power delivered. \$\endgroup\$ – Russell McMahon Jun 21 '16 at 18:09

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