U=R*I vs. Lightbulb with 40W pulls 40W no matter how high the voltage is

Obviously I am not professionally into electronics as u might tell :wink:
But I want to learn and a thing I don't understand is how the power consumption of electronic devices obeys Ohm's law. If I have a lightbulb in my house with 230V and 40W it gets 0.17A but in my caravan with 24V it pulls 1.6A to get to 40W, which is why my caravan needs thicker cables than my home. But how does that comply with Ohm's law? Did the bulb "magically" change its resistance?

Maybe you can just tell me some electrical terms what I have to search for to get my answers.

Thank you in advance :slight_smile:

You simply can not use the same bulb on 230V and 24V and get the same power. A 24V on 230V will simply blow (because it will try to, according to Ohm, draw 230V x 40W / 24V2 = 16A! or 3680W!). Other way around, a 230V bulb may lit on 24V but will nowhere output 40W. That will be closer to 0,5W…

This will be slightly different because, although a load obeys Ohms law, a incandescent will change resistance with temperature. It’s a PTC. And that’s the biggest point, Ohm’s law works statically. Aka, you can only change a single parameter at a time. If you have a magic bulb (and it’s surely not an incandescent) that works on 230VAC and 24VDC and outputs 40W on both, that still obeys Ohms law only the load isn’t resistive anymore and you may not assume it to be fixed over the full voltage range :wink:

So is this website simply talking trash?

I know it is german, but they have a example calculation for a coffeemaker and they calculate it like this:
P=V*I
Amperage of Coffeemaker with 1000W in 12V
I=1000W/12V=83,3A
Amperage of same Coffeemaker with 230V
I=1000W/230V=4,3A

So this can only be a hypothetical example and is not true in reality?

No, it isn’t. But all formulas only work if the variables are fixed (and this might be fixed for a specific situation). For Ohms law (R = U / I), you can only calculate R is U and I are fixed (for that situation).

And your calculations are true, but it isn’t the same coffee maker* :wink:

Let’s make an analogy. You have a car and you know the average km/l (MPG). Now you’re going to make a journey and you know the length, you can calculate the fuel consumption. But that is, if you drive the same as you always do and the road is average aka if you may assume your km/l is valid. But if that road is 10% up and you drive twice as fast you will consume more fuel. But you can g up that hill with the same km/l as normal, but your average speed probably is a lot lower :wink:

Same goes for loads. Something is fixed. For a heating element (and roughly for an incandescent) R is fixed. Which means that only leaves one variable in Ohm’s law. Or you can define the voltage and calculate the current (I = U / R) or you can define the current and calculate the voltage drop you get (U = I x R). But more complex loads don’t have R fixed. The model can be A LOT more complex than just a resistor. But let’s assume we have a load with 100W @ 12V and 100W @ 230V. Which means the power now is fixed. And accrding to Ohm’s law and the power law, that means the other parts are not fixed. You can define the voltage which leads to a specified R (R = U2 / P) and I (I = P / U). Or you can define a current and calculate U (U = P / I) and R (R = P / I2). As you can see, in both cases all the right hand variables are or fixed or you defined them.

But it all depends on the load you have. Seeing every load as a resistor just isn’t true :slight_smile: And a resistor isn’t the only load with resistance, it’s simply a part/model with fixed resistance (unless otherwise specified like a PTC or LDR).

*At least not a normal one because they really are not fixed power but fixed resistance (roughly) :wink: If a coffee maker is designed for 1000W @ 230V it will not be

newby9000:
Did the bulb "magically" change its resistance?

No, that part you've missed is that it's a different 40W bulb. A 240V 40W bulb will be made with a much higher resistance than a 24V 40W bulb and you can't simply swap them or you will have either a bulb blown to bits (24V bulb run on 240V) or very very dim or no light (240V bulb on 24V).

Steve

Modern LED light bulbs (40W would be really bright!) can often handle a wide range of voltages, and some even can handle both DC and AC power. No matter what they'll shine as brightly thanks to built-in regulating electronics.

I see so the answer to my problem was a lot simpler than expected.
thank you for your responses :slight_smile:

I think I need a little electronics lab to check what I am actually doing when I build something

Another characteristic of an incandescent lamp is it changes resistance with temperature (of the filament, not ambient). So If I recall correctly a typical 120V bulb is approximately 1/6 th the resistance when "cold" (i.e. measured with an ohmmeter) that when it is at the normal 120V operating point.

So the resistance does change but not magical. You can also look up the temperature coefficient of resistance of tungsten to get an idea of the change.

What if instead of an incandescent bulb he had a resistive load like for am electric hot water heater.

Let's use 240 volts to make the math easy.

If the heating element is rated at 1500 watts at 240v currnet draw would be 6.25 a.
Now if he ran this off his car battery which is 24 volts current draw would be 62.5 a.

Would you agree?

With a resitive load, like in a hot water heater, would the thing heat according to 1500 watts following ohms law? In otherwords at a lower voltage will current increase to still draw 1500 watts? Or are there some other properties which come into play?

If the heating element is rated at 1500 watts at 240v currnet draw would be 6.25 a.
Now if he ran this off his car battery which is 24 volts current draw would be 62.5 a.

Would you agree?

No I would not agree. You are making the same error as the OP. A wattage rating is only valid for the voltage it was defined for. So 1500 watts at 240V can not be translated into 1500 watts at 24V.

So from the power rating the heater element has a resistance of 38.4 ohms.
Apply that with 24V and you get 0.625 Amps. That is a power of 15 watts.

Grumpy_Mike:
So from the power rating the heater element has a resistance of 38.4 ohms.
Apply that with 24V and you get 0.625 Amps. That is a power of 15 watts.

And that's exactly what we should expect with a fixed resistance load. If you only apply 1/10th of the voltage you get 1/10th of the current and since power still equals V x I you get 1/100th of the power (15W vs 1500W).

Steve

Thank you both, I agree. Resistance is fixed, so less V would result in less current. Less voltage and less current less wattage. Same reason a lamp ratted for 220v is dimly lit at 120 v.

Yeah..... basically.... you made an assumption that you knew what those value written on the box or bulb means. A set of rating values given by the manufacturer.... v i and power etc. typically mean... 'avoid exceeding ANY particular one of those values during operation'. It does not mean the device must necessarily be operated at those values.... even though there are many cases where a voltage rating happens to coincide with the recommended operating voltage etc.

Not exactly.... A guy who is into solar was telling me a water heater element designed for a 220 volt water heater could be used at 24 volts with solar pannels. He’s logic was the similar/same as the original poster. I had my doubts, but he tells me it works for him. Can’t argue with a believer. But then again over the years I’ve leaned the laws of Physice always seem to make an exception for some of his applications. Thank you for reaffirming my trust in science.

Laws of physics normally do not apply to sales folk. I've long learnt to distrust sales folk from especially chain shops (those kids tend to know nothing at all about whatever they sell).

I my have missed something in the earlier posts but I would approach the problem differently.

If I have a lightbulb in my house with 230V and 40W

If we assume for a moment the lightbulb resistance does not change.

Power = E2 / Resistance

40 = 230*230 / R

R = 1322.5 Ohms

If the same lightbulb was powered by 24 Volts:
Power = E2 / Resistance

P = 24*24 /1322.5

P = 0.4355 Watts

One must remember for a constant resistance power increases / decreases with the square of the voltage.

A guy who is into solar was telling me a water heater element designed for a 220 volt water heater could be used at 24 volts with solar pannels. He’s logic was the similar/same as the original poster. I had my doubts, but he tells me it works for him.

Yes it will work, but as noted before the power produced will be 1/100th of the power, but sure it will work.

Can’t argue with a believer.

Sure you can argue with him, but in general walls are a tad more receptive.

But then again over the years I’ve leaned the laws of Physice always seem to make an exception for some of his applications.

Sure yes. Thing is, that in the world of the blind the one eyed man is king, so all he needs is a few more brain cells than his moronic followers.

I agree. And they make up scientific facts on the spot.

I had a California Certified Solar Design Engineer tell me that during the day the solar panels on my roof get charged-up during the day and then at night discharge and supply electricity to my home. The guy then was trying to sell me four PowerWall battery banks. I asked him why I needed the batteries if all of the energy is stored in the solar panels?

He said you know your right. You don't need the batteries; all of the electricity is stored in the solar panels.

The other wonderful like I hear from the solar sales experts who just a few months ago were selling used cars is there is so much solar energy coming from the sun it would supply enough electricity to the world if only we had enough solar panels.

To which I reply, you know you are right, but you know, if all the suns energy striking the Earth were turned into electricity there wouldn't be any for planets, we would have no food, and it would be dark all of the item.

I hope you will join in my religion. I am a science fundamentalist. I M > HE It is the oldest religion there is. You will of course have to strictly obey the law of Physics and spread the good word of the knowledge of science to those who believe otherwise.

I hope you will join me and bring a friend.

Doug101:
I had a California Certified Solar Design Engineer tell me that during the day the solar panels on my roof get charged-up during the day and then at night discharge and supply electricity to my home. The guy then was trying to sell me four PowerWall battery banks. I asked him why I needed the batteries if all of the energy is stored in the solar panels? He said you know your right. You don't need the batteries; all of the electricity is stored in the solar panels.

So, basically you're saying that ---- the particular solar 'design engineer' wasn't an electronic engineer (or someone that understands basic electronics), and doesn't know the basic functionality of solar panels etc.

He had probably just been trained to know very well “red goes here, white goes there, and green connects to that”, for the electronics part, and a bit about how to safely install the panels themselves. That would mean he can install solar panels safely and properly, but has no clue on what’s going on. A bit of a stretch of the word “engineer”.