Electrical puzzle

[Ivy]

Just been to the web site, all very interesting, it's a good job I was sitting down when I got to the bottom of the page, I nearly had a heart attack.

RS do some led cluster units that might fit into a cabin light, this would save some battery power but they are still around £30.00.

I will stay with my 25w tri-col, etc.

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[VicS]

Tom, while I think your figure of 8 ohms for a series resistance is wrong (about 3.3 according to my arithmetic) you are right about the total load only being reduced to about 16 watts. But you will find that if a 25w bulb is under-run to this extent the light output will be reduced by very much more than you anticipate.

During a dicussion a few weeks ago I did some simple experiments and found that if the volage applied to a 21w bulb was reduced by 2v the light output was reduced by about 50% even though the electrical power being dissipated in the bulb was reduced by only about 25% and the saving in current drawn from the battery was reduced by only about 14%. Reducing the light output of a 25w bulb to the equivalent of a 10 w bulb is not very different so one would expect to reduce the current consumption by only a little more than say 15%, adding even more weight to the argument that reducing power consumption by adding a series resistance is pretty inefficient.

However this all deviates considerably from the original question but I have no idea as to the answer to that. Perhaps all the bulbs tried really came from the same source but were 'badged' differently.

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[tome]

VicS

The arithmetic is simple. A 25W bulb will draw around 2.1A at 12V and the filament can be assumed to be a simple resistor of ~5.75 ohms in this application. A series resistor will provide a voltage divider and since W=IV there is a linear relationship between applied voltage and power. You need to drop the voltage across the bulb by 15 watts (ie to 10W) so this gives 15/25x12 =7.2V across the series resistor. To get your ratio, you need the series resistance to be 7.2/4.8 x 5.75 = 8.64 ohms. Your 3.3 ohms would only reduce the power to 16W.

I agree about the light output being non-linear but the point of the thread was to reduce the power to 10W without regard for the light output. As I said, I'll be sticking to 25W.

Regards
Tom

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[ccscott49]

Keep the 25 watt bulbs and dont HATE starting your engine occasionally or get a bigger battery.

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[andyball]

Nooo !

If you add an 8ohm resistor in series with a 5.75 ohm bulb, you get 13.75 ohms.

Stick 12V across 13.75 ohms & I=V/R gives you a current flowing of 12/13.75= 0.87 Amps.

Power = I (current) squared x R so the power in the bulb would now be 0.87 x 0.87 x 5.75 = 4.35 Watts :not the 10 W wanted.


So:

P =I squared x R ....P/R = I squared .... we need 10W in a 5.75 ohm bulb .... 10/5.75= 1.74 . That's I squared, so the current needed is sq root of 1.74= 1.32Amps.

The total resistance incl bulb can be found from V/I =R : 12/1.32 = 9.09 ohms.

take 5.75 (the bulb) away from 9.09 (the total resistance) and you need a 3.34 ohms series resistor; the power in which would be (from P= I squared x R) 1.32 x 1.32 x 3.34 = 5.82 Watts

A long winded explanation imo, which is why I'd make a poor teacher.

otoh...if you want a total load (bulb resistor ) of 10Watts, then an 8 ohm is the thing, but your earlier post mentioned a 16W total.



Whatever....it isn't a good idea, we're all agreed on that.

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[tome]

Quite right Andy, I stand corrected (sorry VicS). It's been a very slow day for me after a self-indulgent evening, and I'm definitely not operating any machinery.

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[peterb]

Noooo!

The trouble is that a tungsten filament bulb is not a simple resistor. As its temperature increases so does its resistance. Try measuring the cold resistance of a bulb; you'll find that it's nothing like the value that you would expect from its wattage rating. The result is that calculations based on Ohm's Law just don't work.

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[andyball]

yes, we know that....Tom "assumed" it was a simple resistor in an earlier post to show the maths.

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[VicS]

Yes Peter, but I decided to leave that bit out to avoid further complicating matters. It would have been nice if you had quantified it though but has you didn't I went back to my 'laboratory' with my trusty 21w bulb, a length of reistance wire and a couple of multimeters and this is what I found:

Volts Amps Watts Ohms

4.64 1.06 4.92 4.38
5.44 1.15 6.26 4.72
6.33 1.24 7.85 5.10
7.20 1.33 9.58 5.41
8.20 1.42 11.6 5.77
9.13 1.51 13.8 6.05
10.10 1.59 16.1 6.35
11.08 1.67 18.5 6.63
11.95 1.74 20.8 6.81
12.30 1.78 21.9 6.91


The resistance ot the bulb cold was 3.1ohms.

I doesn't serve any useful purpose to the original theme of the thread but it's all jolly interesting if youv'e got nothing better to do and SWMBO has charge of the television remotes. Pity though that I can't set the results table out properly.

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