andro
Junior Member level 3
Hello Friends
i am confusing about using Sealed acid battery and i have some questions, but before i ask i would to put what i am understanding to find out if i am right or wrong
i have lead acid battery 12 v/ 7.5 AH i want to use it for lighting 12 leds Red color
red color led VF = 2 V and IF = 25 mA
if i connect them in parallel , then the total current consumed by them = 25 X 12 = 300 mA
to calculate resistor in series it should be = (12-2)/0.3
R = 33 Ohm
power dissipation by resistor = IV = 10*0.3 = 3 W , so i should use 5 watt right ???
since the battery 7500 mA and the led consume 300 mA , so the battery can power leds for about 7500/300 = 25 Hours
my first question
1- Is the current consumption by leds i.e. 300 mA for one hour?
2- If i use 3 batteries with the same specification, so it will serve for 3*25 = 75 Hours right??
that was the first section of my question
second section
i want to build simple charging Sealed lead acid battery for the same one that i have mentioned
i want to use DC power supply for example 18 DC, if i am using charging current for example 500 mA from supply so the resistance between supply and battery is
R = 18-12/0.5 = 12 Ohm
power dissipated in resistor = 12 * .5 = 6 W
so i should use resistor 8 watt right ??
3- If i use 500 mA charging current, then it will take
7500 / 500 = 15 Hours for complete charging right??
4- That was simple and direct charging , but some sites mentioned the term overcharging what is that ?? is it over current or over voltage ???
5- i want to use that type of batteries to power a device and that device consumes 200 Watt-Hour
my calculations is
power delivered by battery = 7.5 * 12 = 90 watt-Hour
no. of batteries = 200/90 = 2.222 then i use 3 batteries
so 3 batteries can be used for one hour right ??
6- can i use that battery to power up light bulb(60 W) directly with no need inverter ?
Best regards
i am confusing about using Sealed acid battery and i have some questions, but before i ask i would to put what i am understanding to find out if i am right or wrong
i have lead acid battery 12 v/ 7.5 AH i want to use it for lighting 12 leds Red color
red color led VF = 2 V and IF = 25 mA
if i connect them in parallel , then the total current consumed by them = 25 X 12 = 300 mA
to calculate resistor in series it should be = (12-2)/0.3
R = 33 Ohm
power dissipation by resistor = IV = 10*0.3 = 3 W , so i should use 5 watt right ???
since the battery 7500 mA and the led consume 300 mA , so the battery can power leds for about 7500/300 = 25 Hours
my first question
1- Is the current consumption by leds i.e. 300 mA for one hour?
2- If i use 3 batteries with the same specification, so it will serve for 3*25 = 75 Hours right??
that was the first section of my question
second section
i want to build simple charging Sealed lead acid battery for the same one that i have mentioned
i want to use DC power supply for example 18 DC, if i am using charging current for example 500 mA from supply so the resistance between supply and battery is
R = 18-12/0.5 = 12 Ohm
power dissipated in resistor = 12 * .5 = 6 W
so i should use resistor 8 watt right ??
3- If i use 500 mA charging current, then it will take
7500 / 500 = 15 Hours for complete charging right??
4- That was simple and direct charging , but some sites mentioned the term overcharging what is that ?? is it over current or over voltage ???
5- i want to use that type of batteries to power a device and that device consumes 200 Watt-Hour
my calculations is
power delivered by battery = 7.5 * 12 = 90 watt-Hour
no. of batteries = 200/90 = 2.222 then i use 3 batteries
so 3 batteries can be used for one hour right ??
6- can i use that battery to power up light bulb(60 W) directly with no need inverter ?
Best regards