Chuck, There are to ways a battery's characteristics change with temperature, voltage and internal resistance.
Pb-acid batteries change their voltage -.01 V per degree C or something like that. A colder battery will have higher voltage than a warmer battery. Also, like others have said a warmer battery has less internal resistance. So if you charge with a constant voltage charger the colder battery will have a lower voltage between battery and charger plus it will have a high internal resistance. Hence lower current. If you want the math here goes. V(Voltage)=I(current)*R(resistance) --> I = V/R Charger is 15V Cold battery(0C) is 12V with .016 Ohms Warm battery(20C) is 11.8V with .012 Ohms Current in cold battery 15V - 12V = 3V I = 3V / .016 Ohms = 187.5 Amps Current in hot battery 15V - 11.8V = 3.2V I = 3.2V / .012 Ohms = 266.7 Amps The effect of the voltage being .2V different is only about 15 Amps. Most of the change is the internal resistance. The warmer battery is more efficient. If charge both batteries at 30 amps and the warmer battery only needs 14V where the cold battery needs 15V than the warmer batter is using less power to charge. 14V*30A=420W vs 15V*30A=450W. Regards, Jeremy Maus Belleville, MI [EMAIL PROTECTED] http://www.emidget.info |-----Original Message----- |From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On |Behalf Of Chuck Hursch |Sent: Friday, August 16, 2002 4:54 PM |To: [EMAIL PROTECTED] |Subject: battery temperature vs charging voltage and amps | | |Hello All, | |I'm not understanding why a warmer (Pb-acid) battery takes more |current to produce a given voltage under charge than a cooler |battery. Seems that lead-acid batteries are supposed to be more |efficient (less internal resistance) when warm than cold. So I |would expect the opposite of what occurs. Any explanation? | |Thanks, |Chuck Hursch |Larkspur, CA |www.geocities.com/nbeaa | 1979 |
