Date:         Sun, 22 Dec 2013 00:12:18 -0500
Reply-To:     James <jk_eaton@HOTMAIL.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         James <jk_eaton@HOTMAIL.COM>
Subject:      Re: Van Dies while in motion
Comments: To: Stuart <ve3smf@yahoo.com>
In-Reply-To:  <1387673013.13895.YahooMailNeo@web162704.mail.bf1.yahoo.com>
Content-Type: text/plain; charset="iso-8859-1"

Stuart, you're starting with an erroneous assumption - that power is constant.

Remember, in electricity, Ohm's law, V=I*R, is the most important law.  And since resistance is constant, as voltage drops, so does current.  In fact, since P=Vsquared/R, power will drop faster than voltage does.

Take a regular lightbulb.  At 12V, it's a particular brightness.  Now, change that to 9V.  Is it still as bright?  No.  It gets dimmer.  So power cannot be constant, or it would be the same brightness. 

James 
Ottawa, ON  

> Date: Sat, 21 Dec 2013 16:43:33 -0800
> From: ve3smf@YAHOO.COM
> Subject: Re: Van Dies while in motion
> To: vanagon@GERRY.VANAGON.COM
> 
> OK, I am willing to go out on the limb for this one.....
>  
> First some background.  Any device such as headlights, heater fans, blowers, etc will draw a fixed amount or power.  P=I*E where Power in watts = Current times Voltage.  The problem is, when voltage drops, the device still draws the same power, and the amount of current required (ampre) will go up.
> ref: http://www.sengpielaudio.com/calculator-ohm.htm