Re: Re(2?): Lights, Volts and Amps (Long geeky thread)

[Don Hoefer <hoefer@eng.pko.dec.com>]

Dave Head wrote: (and quoted from RDH?)

  (much deleted for brevity)

> >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.
> >

> NO **** You just decreased your total circuit load - Which means your total
> current DRAW will decrease ****
> 
> Ptotal = lowered value due to fixing connections + Loadlight
> 

I think that Mr. Head is confused between power regulation and voltage
regulation.  Most car radios have power regulation, which is why they
still work well when you start the car (V ~= 8V instead of 12V) but
the headlights dim, as they depend upon voltage regulation for proper
power consumption.  If you reduce the resistive load on a constant
voltage, the current will increase, therefore the power will increase
also.  V = I * R (derived from the electro-magnetic equations) in the
limited engineering universe.  If everything else is constant (which
is why some physicists and many mathemeticians don't like engineeers,
we're always taking shortcuts) and V is held constant and R is reduced,
I will increase.  There is nothing in a light bulb to regulate current or power
save it's own resistance and the voltage applied across it.  In the example
written about here, resistive load in series is removed from the circuit,
therefore the resistance of the total load has gone down, current and power
will increase, and the light bulb will get a higher percentage of the power
dissipated in the circuit due to increased voltage and current.  This was
amply demonstrated in the equations presented in the originally quoted letter.
 
> ****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?
> 

Total RESISTIVE LOAD was changed on a constant voltage circuit, therefore
power changes as a result.

> >
> >   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!
> >

The Key.

> >
> >   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.
> >

> 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.

POWER REGULATED loads only, plain light bulbs and toaster ovens will fry as
the current (and power) go up.

We assume that current, power, and voltage are proportional in our
simplified equations of V = I * R, and P = I * V = I^2 * R = V^2 / R.

Voltage regulated (apples), power regulated (oranges).

> 
> 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.

Resistance in the total load has been lowered,  power will go up (given constant
voltage).  The location of some of the power consumption has been changed to where
we want it.

> 
> 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.
> 

Power loads add up in parallel.  Power in a particular load may be a series
problem.  (Ever wire your toaster and a light bulb in series?  Power in
each and total power will go way down.)

> 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.
> 

Yes, same for transmission lines, higher voltage means less current for the
same POWER.  Big power distribution systems assume given loads, a macro view.
Our little Audis require a micro view where little details count.

> 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.
> 

Which is why it takes Engineers and Technicians to build real things that work.
You need to remember details like these.

Don Hoefer
'82 Coupe