electrical question

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hi all, i thought this would be a good place to find the answer to my question.

I tinker with electrics a bit, normally car related stuff, audio and other bits. im no expert. theres a question thats been bugging me.

I have a set of spotlights. if I put bulbs in than are higher than 55 watts, will those bulbs then demand the extra power from the battery?

I guess my question is what is the relationship between the load and the power source?

I had radiator fan (15ah draw )hooked up to a small 7ah battery, it ran but obviously not at full whack. does the load just deal with what its given... its a hard one to phrase...

will me putting higher watt bulbs in my spotlights increase theyre brightness? obviously ill use chunkier wiring.

cheers for reading my ramblings,

ned .
 
Hi,

100W bulbs draw twice as much current as 50W bulbs for the same
supply voltage rating. A 100W 12V bulb will be a 50W bulb if its run
on 6V instead of 12V. a 10W 12V bulb will draw 1/10 the current,
about an amp (0.83) whilst 100W about 8 amps.

A 100W bulb that isn't brighter than a 50W bulb is pretty poor, however a
10W halide bulb is equivalent to a 30W to 50W normal bulb in brightness.

rgds, sreten.

P=VxI, P=I(squared)xR, P=V(squared)/R, R=V/I.
 
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Ah is capacity. Think of it as a bucket of water. A battery with a higher amp-hour rating is like a larger bucket.

The fan has an amp rating, not an amp-hour rating. The amperage that the fan draws doesn't depend on the battery's ah rating. A fan with a higher amp rating means that the fan draws water from the bucket faster than one with a lower amp rating.

The 7Ah battery can likely deliver far more than 15 amps of current for short periods of time. In a car, the Ah rating of the battery becomes a non-issue when the engine is running because all power comes from the alternator in most vehicles.

You cannot always use replacement bulbs that are of a higher rating because the socket may not be able to handle the added stress. Sometimes the added heat melts the fixture. Sometimes the added heat and the increased current draw causes the socket contacts/wiring to burn.
 
A 100W 12V bulb will be a 50W bulb if its run
on 6V instead of 12V.

Sorry to insist on details, but look at the formula you posted: A 100W 12V connected to a 6V source will draw half the current. So half the voltage, half the current will lead to a factor of 0.25 in power.

A 100W 12V bulb on 6V will dissipate approx. 25W not 50W. (Different resistance due to different filament temperature neglected.)

Greetings,
Andreas
 
Sorry to insist on details, but look at the formula you posted:
A 100W 12V connected to a 6V source will draw half the current.
So half the voltage, half the current will lead to a factor of 0.25 in power.

A 100W 12V bulb on 6V will dissipate approx. 25W not 50W.
(Different resistance due to different filament temperature neglected.)

Greetings,
Andreas

Hi,

Oops .... you are quite right, a schoolboy error on my part. Not only
will it be a 1/4, as you say, the colour spectrum will be much redder
due to lower operating temperature, the "colour" of a filament bulb
is a good clue as to how hard its being run, and how long it will last.

For half power you'd need to run a 12V bulb at 8.5V.

rgds, sreten.
 
Sorry to insist on details, but look at the formula you posted: A 100W 12V connected to a 6V source will draw half the current. So half the voltage, half the current will lead to a factor of 0.25 in power.

A 100W 12V bulb on 6V will dissipate approx. 25W not 50W. (Different resistance due to different filament temperature neglected.)

Greetings,
Andreas
don't want to be picky, either, but a 12v bulb on 6v will consume a great deal more current than you think. nominally, it should be pulling 7-8a current, however, it will take/pass more than 2x the current cold, and the thermal resistance will drop the consumption,once lit. so, on 6v, it will barely light, meaning the nominal amp consumption should be over 10a, which would be 60watts+. this is real-world occurrence/experience, and tested. i used to re-wind small-engine stators, and trying to get the highest amp output i can, however, as part, i needed to have several different test bulbs to drag them down. are you working on something that does not have a power generation source? whenever adding compionents that draw power, you never want to be running more than 70% max load of the generator nominally. sure you can rely on the battery to handle spikes and surges of power (at lower voltage), but continuously, anything over 70% and efficiency really drops bad, strain is excessive, and you really don't gain. as mentioned, fixture power ratings are there to tell you the limit it can handle, before heat, or contact areas exceeds what it can safely handle before failure.
 
don't want to be picky, either, but a 12v bulb on 6v will consume a great deal more current than you think [...] so, on 6v, it will barely light, meaning the nominal amp consumption should be over 10a, which would be 60watts+.

Basically you're right. The cold resistance is significantly lower, so the inrush current into a cold bulb could be up to approx. 4-5 times the nominal current. But you overestimate the effect in this example. A 12V bulb will start glowing red at roughly 1amp, it lights up from 3V up. So at 6V, the consumption will be higher than the 'theoretical' value of 25watts, but surely not 60+watts. As I stated above, the 25watts were calculated neglecting the effects of temperature-dependent filament resistance.

this is real-world occurrence/experience, and tested.

Well, never trust calculations if you can measure. So your 60+watts against my bet of slightly above 25watts :)

attachment.php


The data above were taken in my lab using a 12V/100W halogen bulb. You can see nicely that the resistance (black) is not constant due to the changing filament temperature. This leads to a dissipation that does not follow the theoretical square law, especially at low voltages.

But how large is the effect:

At 6V applied, the theoretical expectation for a constant resistance would be 25watts, real world says 33watts due to lower cold resistance.

The 60+watts dissipation you estimated require approx. 9V to be applied.

25watts vs. 33watts = 8w underestimation
33watts vs. 60+watts = 27+watts overestimation

Rundmaus won :D:D:D

Best regards,
Andreas
 

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The power source is typically rated to have a specific voltage (12v, for example). There is also a current limit for the source. If the device is well protected or if it has significant internal resistance, the voltage will not hold at the design voltage (12v in this example) and the voltage will drop. For a well protected supply, it may survive having a high current load connected to it. If it's not well protected and more current is drawn from it than it was designed to deliver, the supply could be damaged.

In a car, the source is the battery and/or charging system plus the wiring. It's not likely that you'll draw more than the battery or charging system can supply but the wires feeding the device (lamps here) need to be large enough and must be fused properly.
 
ahhh, cheers once again. so both devices are, im grasping at words here, - relative.

I used the example of the car lights, but it was a question floating around in my head for a while now. thanks for the info.

might i ask, what would be an example of a well protected supply? if you point one out I can read further on the matter.
 
Apart from current limits and power consumption issues, keep in mind the following, important detail when talking about lights/lamps:

Conventional (or halogen) incandescent bulbs dissipate roughly 95% of the supplied electrical power as heat. Though a higher wattage bulb may physically fit into your spotlight, the construction (housing, reflector, socket) may not be able to survive the significantly higher heat load!

Andreas
 
There are a lot of examples of a well protected supply. The simplest is the battery/wire with the proper fuse. If the circuit is overloaded, the fuse blows and no damage is done.

In more complex supplies, there may be several types of protection. A voltage regulator like THIS one has both thermal and over-current protection. A simple regulator can be produced with just a few parts. If you look at the internal schematic diagram for this regulator, you can see that it's much more complex.

Other supplies shut down or significantly limit their output until the load is removed or reduced to a safe level.

For better examples, you'd need to be very specific about the type of supply.
 
One option that you should look into is LED. The newer LEDs can easily produce as much light as a 55w incandescent. They will generally use less current/power (depending on how many you use). They will, however, cost more for an equal amount of light output. You won't be able to use bare LEDs. High power LEDs need heatsinks. To get the best beam, you'll need a properly designed reflector.
 
sorry, those were fma figures. i didn't bother with graphs and such, just measuring current and voltage and used a small collection of mr16, and h3 bulbs in 20, 30, 50, and 55, for differnt arrangements. the biggest goal was to get 110w @ 13v+ for me. only really bothered checking anything below 8.xv.... not trying to pick a fight, but was just trying to illustrate the effect. perhaps greater with a pair of bulbs, but i don't really care. kind of dropped off on re-winding stators years ago, when my back got too bad to ride much.

Anyways, it will help a great deal to know exactly what you are working with. what are the fixtures? fog-lights? (description of housing size helps) h3 bulbs? mr16? 6y 6.35, or other base? home 12v accent lights? color temperature an issue? i've been dealing with just about all the different lighting out there for the past couple years, and led's can, by far, out-perform incandescent bulbs in efficiency. the good ones are going to cost you, though. you have to look at the lumen output, and many have a mcd rating, so you have to do the calculation to convert. the color temperature makes a big differnce. i usually go after 4700k, or close/under, and the actual appearance is more like 6-7000k. if the fixture is good enough, you might be able to sneak some hid bulbs in there. hid's are more temperamental, though. they need an initial "burn-in" period, take a few seconds to light up full, and can have a problem with voltage fluctuation. they, too, create a great deal of heat/ir output, so the fixture needs to handle it. led's also do create their own heat, and the limitations should be printed on the package.
next, what kind of power source are you really using? is it on some sort of orv, trailer, stand-alone home/rec use? what power supply? are you planning on charging the battery, and then using the setup remotely, just from battery power, like a sort of "lantern"?
 
sorry, those were fma figures. i didn't bother with graphs and such,[...]... not trying to pick a fight, but was just trying to illustrate the effect.

:)

Didn't understand it as a fight, I just had a power supply, DMM and halogen bulb dummy load standing around in the lab, so why not measure it...

Could be that the effect is more exaggerated with a set of paralleled bulbs and such...

Andreas
 
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