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Lights, Volts and Amps
> > =09So, if you have 55w driving lights, and your battery puts out=20
> > 12v, the conformulation would be: 110w / 12v =3D 9.17amps.
>
> I have found from actual current measurements that the wattage ratings=20
> are not very accurate, so you would be best to use a good Fluke (or=20
> similar meter) and actually measure the voltage and the current, to=20
> determine the wattage.
>
> The above formula also illustrates why relays work so well. They get you=
> =20
> more volts at the light, which reduces the current. The higher voltage=20
>
> ***WRONG***
Well, OK, I have to state that given the wattage as a constant, then the
above would be true. An error, yes, but a full blown "***WRONG***" ? I
guess that isn't as bad as a "Bbbzzzzzzzzztttttt!" :-)
> Current is directly proportional to voltage (if voltage goes up 10%, then
> the current flow will go up 10%), and the power dissipated goes up as the
> square of the voltage (or the current) change accordingly.
I thought that P=VI, (power = voltage * current)
Correct
and power = (Current)^2 * Resistance
Correct. Also, power = (voltage)^2 / Resistance
From P=VI, that means that power is proportional to voltage and also
proportional to current. Keeping power constant, if you increase
^^^^^^^^^^^^^^^^^^^^^^
$$$$$$$$$$$$$$$$$$$$$$
See below!
voltage, then current has to go down. I suppose it is possible that with
higher voltage at the bulbs, that the power rating of the bulb goes up a
little, which would allow a little more current.
<<Und>>
... only one minor quibble ... You are assuming that the resistance is con-
stant. Most lamps have a characteristic where the resistance goes up with
temperature, so you may not notice that purely resistive current increase
with voltage ...
True. For the range 12.6V to 14V, the lamps' resistance is reasonably con-
stant (close enuf for my main point here, anyways). See further below.
The rest of your post is dead-on ... the only addition that I would make
from recent experience is to make sure that you clean up the grounds at
the same time you run that nice plus-twelve lead ...
Actually, I would go so far as to replace all those pesky corroded
wire ends . . . and then solder the f,er,funny connectors onto the
wire after a nice crimp is made.
<<Und>>
>***WRONG***
>
>Current is directly proportional to voltage (if voltage goes up 10%, then
>the current flow will go up 10%), and the power dissipated goes up as the
>square of the voltage (or the current) change accordingly.
>
Sorry, big fella... You hook up an ampmeter up and vary that voltage, you'll
see that it is INVERSELY proportional. This is a DC system, and the power
draw is due to combined resistances - nun uh that funny RL/RC, Z stuff
(purely resistive load).
***BZZZZTTTTTTTZZZTZTTZTZTZTTTZ *** WRONG **** BBAAYWEIDKJHAMAUETT****
E = I X R, P = I X E Vterm = Egen - losses
Voltage is equivalent to pressure. Amperage is equivalent to flow.
If I reduce the I squared R losses (heat), and reduce the total resistance
by providing better connections, I actually lower the total power consumed
by that circuit and raise the terminal voltage, and lower the current draw.
***************************************************************************
* WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG *
***************************************************************************
* WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG *
***************************************************************************
* WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG *
***************************************************************************
What you have in an automotive system is, for all practical intents and
purposes is a ***==> CONSTANT VOLTAGE SOURCE <==***
That system voltage is not gonna [significantly] change no matter what
load you put on it, for all normal automotive loads. [Yes, yes, you can
drop the idle voltage from 13.6Vnominal to 12.3Vnominal by running the
lights/radiatorfan/freshairfan/allwindowmotors/elecseatheaters&servos/
defrostelement/wipers/horn/killertriampedstereo@bleedingearslevels, and
for good measure, the radar detector. Hell, you can drop the system
voltage to 6 or 8 volts just by engaging the starter motor on a cold
morning. *BUT* for purposes of this discussion, I limit the considerations
to running engine, up-to-par electrics [well, these are Audis, but I'll
let that pass...], and JUST THE HEADLIGHTs. OK? No Shorting of the
battery to ground with 000000000ga superconducting bus bars. Just
normal real life expectable and reproducible conditions - engine run-
ning, 13.6V nominal system voltage, and headlights.]
Now, given a constant voltage source, the current draw will be deter-
minded by the load impedance. In this case, the electric lights. And,
to a painfully-real degree, the resistance of the wiring/switches/re-
lays/etal that HerrDoktorAudi sees fit to put in the way of the lights
and the voltage source.
In particular, you can model this "circuit" as a voltage source or
generator with two "load"s (resistances or impedances) in series:
+---------------+
| |
| Load A
+ | |
Source |
- | |
| Load B
| |
+---------------+
|
--- (Ground)
///
Call "Load A" all the wiring/et al resistance.
Call "Load B" the light.
Source is a constant 13.6 volts -- it is a voltage source that is capable
of maintaining 13.6 volts at all current levels drawn by all combinations
of Load A & B values...normally realized in TheRealWorld(tm).
Load A & B form a voltage divider, and their sum (of resistance) will
determine the current flow through (and thus the power dissipated by)
the circuit -- FOR A CONSTANT SOURCE VOLTAGE. The voltage "developed
across" each of Load A and Load B sums to Source (13.6).
If you *DECREASE* Load A (i.e., you bypass all of the SFWEA [Shitty
Factory Wiring Et Al] with low-resistance/heavy-gauge wiring and relays),
call it Load A', then the total Load [resistance] seen by Source will be
smaller -- (A + B) > (A' + B); the current flow will then be LARGER,
for constant [yeah yeah yeah, see below] Load B, as per:
(current) = (voltage) / (resistance)
or
I = 13.6 / (A + B)
and
I' = 13.6 / (A' + B)
and in particular, I' > I for A' < A
Further, since the voltage "divides" directly-proportionally across
Load A and Load B, if A' < A, then less of the constant 13.6 volts will
develop across A' than A, while MORE voltage will develop across B.
Thus the power dissipated by Load B (the lights), as given by, variously:
(power) = (voltage) * (current)
= (voltage)^2 / (resistance)
= (current)^2 * (resistance)
will ***INCREASE*** for Load A' < Load A.
<<Und>>
Whoa... power isn't constant, it's going to depend on the voltage across the
bulb (V*V/R where R is the resistance of the bulb, which I'm assuming to
be fairly constant over the couple of volts range we are talking about).
So, going from 10V at the bulb to 12V at the bulb is going to increase
the power dissipation by a factor of 12*12/10*10, 1.44, a 44% increase!
Yup.
<<Und>>
With incandescent lamps, that is not strictly true. The resistance of
the filament increases with temperature, and ( off the cuff..) a 10%
increase in voltage will only use a 5% increase in current, but give a
whopping 22% increase in light output for which you will pay for by
a reduction in lamp life of nearly 40%. WELL WORTH IT.
True, my "Load B" above is not truly constant, but is a function of the
power dissipated by (how "bright == "hot" it gets) the filament; thus
"Load B" is an active impedance with a dynamic value depending on the
voltage developed across it, and thus dependent on the value of Load A
(which in itself can be a dynamic value depending on how corroded
connectors are jiggled by the smooth/rough idling engine, how hot/cold/
damp/wet the connectors are, whether the switch&relay contacts found
clean metal-to-metal, or jaggeddirtyhighresistancepit-to-metal, etc.).
For the "real life automotive case", my "analysis" above still holds:
Lower SFWEA ==> higher voltage across the light ==> more current through
the light ==> more power dissipated by the light.
However, for halogens, the "reduction in lamp life" is itself a very
dynamic variable, and in fact lamp life has an optimum voltage, and
will DECREASE for voltages both above and BELOW that optimum voltage
(although the decrease for higher voltages is drastically greater than
for lower voltages). [Assuming we're talking about useful light-pro-
ducing voltages; at 0.01V, the life of any "12v" light is essentially
infinite...and at 120V, the life of any "12v" light is essentially
nil...I get the feeling I need to qualify myself carefully in this
thread...]
I had posted actual numbers over a year ago, directly from the Sylvania
manufacturers data book, had outputs of 9004s 9007s etc. The file
is gone, sorry, ( may still be at my previous employer...)
what, no 13-redundant-copies backup tapes? You fool, you!
-RDH