morgans
[Top] [All Lists]

RE: Halogen headlamps

To: "'tjsouz@epix.net'" <tjsouz@epix.net>, Gary Kneisley <kneisley@ohio.net>
Subject: RE: Halogen headlamps
From: "Willburn, Gerry" <Gerry.Willburn@trw.com>
Date: Tue, 15 Dec 1998 12:17:53 -0800
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

<Prev in Thread] Current Thread [Next in Thread>