>Mark et al., From the depths of my rusty store of electrical formulae,
>I seem to remember that watts divided by volts equals amps. Given that
>is so, if the voltage drops for a given wattage of lamp, that SHOULD
>mean that the amperage goes up. I don't know how much current that
>headlight circuit is fused for, but a drop of 4 volts in a 12 volt
>circuit means an increase of 50% in the amperage flowing through it.
>Somewhere, something has got to give; let's hope it doesn't require a
>fire extinguisher! :-)
>
Sorry Glen, but .... don't quit your day job.
The only fixed value in the circuit we're talking about is the resistance of
the filament in the lamp. Power (watts) in a purely resisitive circuit (like
a headlamp) is a function of two factors: The voltage applied and the
current drawn through the load. Since the load is fixed by the physical
properties of the lamp filament, a simple application of Ohm's law, P=I(2)
R, if the voltage across the load is halved then the current drawn by the
load will also be halved and the power output of the load (watts) will be
reduced by the square of the difference.
In your example a reduction of 4 volts in a 12 volt system would result in a
drop of 33% (1/3 = 4/12) in the current through the lamp and a corresponding
drop in power output (4/9 = 32/72).
ie 12 volts through 2 ohms = 6 amps
6 amps x 12 volts = 72 watts ( or 6 amps x 6 amps x 2 ohms)
8 volts through 2 ohms = 4 amps
4 amps x 8 volts = 32 watts (or 4 amps x 4 amps x 2 ohms)
So for a reduction of 33% (4/12) in applied voltage we see a reduction of
33% + 33%(2) = 44% (32/72) in power output and a logrythmicly extended lamp
life.
End of lecture........
___ \______ Ross MacPherson
/ __ \ __ / /------|) arm@unix.infoserve.net
/ (___)---------/ (___) Vancouver, BC, Canada
1947 MG-TC 3528 1966 MGB-GT
|