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 1st October 2012, 04:51 PM #1 NED 209   diyAudio Member     Join Date: Mar 2007 electrical question 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 .
 1st October 2012, 05:04 PM #2 sreten   diyAudio Member RIP   Join Date: Nov 2003 Location: Brighton UK 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. Last edited by sreten; 1st October 2012 at 05:12 PM.
 1st October 2012, 05:08 PM #3 Perry Babin   diyAudio Member   Join Date: Nov 2003 Location: Louisiana 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. __________________ Links >> Basic Car Audio Amp Repair * Basic Car Audio Electronics * New Site * Basic Switching Power Supply Design * Basic Computer Skills << Links
 1st October 2012, 06:07 PM #4 NED 209   diyAudio Member     Join Date: Mar 2007 Thank you Perry and Sreten for explaining that to me, its clearer now.
Rundmaus
diyAudio Member

Join Date: Aug 2005
Quote:
 Originally Posted by sreten 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

sreten
diyAudio Member RIP

Join Date: Nov 2003
Location: Brighton UK
Quote:
 Originally Posted by Rundmaus 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.

AKHeathen
diyAudio Member

Join Date: Apr 2011
Location: Anchorage AK
Quote:
 Originally Posted by Rundmaus 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.

Rundmaus
diyAudio Member

Join Date: Aug 2005
Quote:
 Originally Posted by AKHeathen 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.

Quote:
 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

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

Best regards,
Andreas
Attached Images
 Graph1.png (71.6 KB, 90 views)

Last edited by Rundmaus; 2nd October 2012 at 12:38 PM.

Rundmaus
diyAudio Member

Join Date: Aug 2005
Hey,

did a nicer version of the figure, with the theoretical square law in it.

Andreas
Attached Images
 haloneu.png (65.9 KB, 75 views)

Last edited by Rundmaus; 2nd October 2012 at 01:26 PM.

 2nd October 2012, 04:15 PM #10 NED 209   diyAudio Member     Join Date: Mar 2007 whoa. i still dont know, can the item demand power from the source? could a high current device damage a low current power source? or is it a one way street, with the load accepting and working with whatever it is fed from the power source

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