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Re(2?): Lights, Volts and Amps (Long geeky thread)
At 12:54 PM 6/6/96 -0400, you wrote:
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>* WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG *
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>* WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG *
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>* WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG WRONG *
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Sorry - don't feel the need to express myself so strongly - you are just wrong.
>
>What you have in an automotive system is, for all practical intents and
>purposes is a ***==> CONSTANT VOLTAGE SOURCE <==***
>
I agree - see my equation explained in your drawing.
>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:
*** Vterm (at light) = Egen (batt) - losses (load A) ***
>
> +---------------+
> | |
* | Load A (connections) = (losses)
> + | |
* Source (Egen) |
> - | |
> | Load B (light) = Vterminal
* | |
> +---------------+
> |
> --- (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:
NO **** You just decreased your total circuit load - Which means your total
current DRAW will decrease ****
Ptotal = lowered value due to fixing connections + Loadlight
>
> (current) = (voltage) / (resistance)
>
> or
>
> I = 13.6 / (A + B)
>
> and
>
> I' = 13.6 / (A' + B)
>
>and in particular, I' > I for A' < A
This is certainly true, if you're totally disregarding the fact that power
total has decreased.
>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=LOAD!)
But you've already raised the voltage to the light - and if you raise that
voltage, then the current must go down.
> = (current)^2 * (resistance)
>
>will ***INCREASE*** for Load A' < Load A.
****But loadtotal is now lower!**** If the circuit was consuming 65 watts
(10 watts losses, 55 watt light), and I remove 8 watts by fixing or
bypassing the losses, my total power goes to 57 watts, and my total current
goes down while my resistance goes down... And I raised the voltage at the
light because of the voltage divider you so conveniently mentioned. How do
you explain that?
>
><<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.
>
The current change is the result of the resistance (load) change.
My equation still holds true - we are talking apples and oranges here. IF
you believe that if you raise voltage you automatically raise current, you
are still wrong, and you will always be. Current will change as the result
of the load change in a power distribution system.
You can't say that voltage and current are directly proportional when P=IxE.
You're not arguing with me, you're arguing with physics.
Otherwise, we would not have any need for high voltage transmission lines
(lower current = less line losses). I could keep voltage low and keep
current low. Wow!
If I take a generator and raise voltage while keeping load constant, current
will go down.
You say you decrease the total resistance by decreasing the connection
voltage drop, that means current goes up. In reality, when you decrease the
connection voltage drop, YOU ARE ALSO DECREASING TOTAL POWER consumed by
that circuit. The only way current draw will increase above the original is
if the resistance change in the filament results in a greater total power
dissipation than before the connection problems were fixed.
BTW, in the RealWorld(tm) - it's all parallel circuit, and as you add load
resistance, you decrease total circuit resistance and increase current flow
from the constant voltage source. This includes Audis.
+--------+------+
| | |
| | Load A (connections) = (losses)
+ | |------+ Bypass using relay...
Source (Egen) |
- | |
| Load B (light) = Vterminal
| |
+---------------+
|
--- (Ground)
///
So what you do here by using the relay is short around funky high resistance
connections, reducing losses (part of load). Your current through load A
will be minimum, which means Isquared R losses (major load component) will
be minimum.
I may not fix Audis very well, but I've taught and lived electric power
generation for 20 years. In 68 class aircraft carriers, the distribution
system was designed for 4160V vice the industry standard 450V. The reasoning
was that since the current would be so low, motors could be built much
smaller and compact for a given power. This allowed them to have reactor
coolant pumps (kinda like a BIG Audi aux cooling pump) only 15 feet high
instead of 60 feet.
And, of course, in the RealWorld(tm), everything works only within a certain
band. If I raise voltage too high, then I get insulation breakdown and short
circuits. If I lower it too much, my heat losses (IsquaredR) go sky high and
I get hot spots and meltdowns.
********************************AUDI FAN******************************
EMCM(SW) Dave Head (nuclear grade electrician)
87 5KCSTQ 170K miles and not counting (damned VDO odometers)...
1.9 bar boost (charlie spring, no shim) - @ 1.3 the shuttle launches!
Maitland FL
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