for example, if 1.5 V with a current discharge of 500mA is it going to last longer than a 1.5V with a current discharge of 1 A? or does the voltage determine the battery?
2007-09-14 07:58:46 · 5 answers · asked by Lee Hal 2 in Science & Mathematics ➔ Engineering
The 500mA is going to last longer than the 1A. These are measures of the number of amps a battery puts out per hour with the 500mA being less that the 1A. The voltage has nothing to do with its battery life.
2007-09-14 08:11:28 · answer #1 · answered by Twizard113 5 · 0⤊ 0⤋
The problem is the internal resistance of the battery. This rises as the battery dies. A totally bad battery still reads full voltage, as the voltmeter has a much higher resistance than the battery's internal resistance.
A battery will last for years on slow discharge, but quickly on high current requirements. Voltage determines nothing. It is the chemical properties of the battery that determine its current capacity and discharge rates, as well as shelf life.
2007-09-16 07:51:53 · answer #2 · answered by Warren W- a Mormon engineer 6 · 0⤊ 0⤋
If you discharge less current more life, more current less life .you are not mentioned what type of Battery you are asking? .Ambient temperature also determine the life of battery ,if the ambient temp < 25'c or less will give double life.and also charging discharge cycle to be followed.battery should not be kept idle in discharge condition it should be charged immediately after discharging .If you are mentioning about dry cell it is comparatively less life with Rechargeble batteries
2007-09-14 08:39:39 · answer #3 · answered by ricky414 5 · 0⤊ 0⤋
The materials used in the battery and the construction of the battery determine its lifetime. Battery lifetime or capacity is specified as ampere-hours which is the product of the load in amperes multiplied by time in hours.
2007-09-14 11:05:31 · answer #4 · answered by Matt D 6 · 0⤊ 0⤋
CAPACITY
eg 40 AH
40 amps for 1 hour
or 4 Amps for 10 hours
2007-09-14 11:26:09 · answer #5 · answered by Anonymous · 0⤊ 0⤋