Gary,
Not only is P=IE but E=IR so if
260 = I x 12, R = 0.554 Ohms
Then is E = 10V, I = 18 Amps.
Nuf said.
Gerry
> -----Original Message-----
> From: tjsouz@epix.net [SMTP:tjsouz@epix.net]
> Sent: Tuesday, December 15, 1998 9:37 AM
> To: Gary Kneisley
> Cc: morgans@autox.team.net
> Subject: Re: Halogen headlamps
>
> Gary Kneisley wrote:
> > A neat way to remember the formula is PIE.
> > P=IE
> > Power = I (amps) times Electromotive force (volts)
> > thus
> > 260 watts = 21.66 amps times 12 volts
> > or
> > 260watts/12volts=21.66amps
> > Look what happens if your voltage is low....
> > 260watts/10volts = 26 amps
> >
> > Gary
> > Grafton, OH
> > 1991 +8
> >
> Hi Gary et al,
> When the voltage drops the headlamps no longer consume 260 watts. The
> resistance of the bulbs stays the same (relatively, there is a change in
> resistance with temperature of the filament), so when the voltage drops
> the current decreases. A depleted battery (read low voltage) produces
> dim headlights. A fully charged battery drives the current through the
> headlamps that gives bright illumination. The highest current occurs
> when the voltage impressed on the circuit is the highest (Ohm's Law).
>
> Regards, Tony
|