Question about battery
#1
Question about battery
alright guys. dont really know where to put this thread ut ill try here and see what happens. So a few months ago i got an ipod/iphone FM TRANSMITTER WHICH WORKS GREAT. when i unplug it from the cig lighter i have to go thorugh the radio stations on the transmitter to find a good one again becuase it resets. I was wondering if i just left it plugged in would it drain my battery when the car is off? because it would be a huge luxury to not have to find my perfect FM station every time i get in the car. thanks guys!
And Mods please move this if it is in the wrong section. thanks!
And Mods please move this if it is in the wrong section. thanks!
#2
It is very unlikely that your FM transmitter draws enough power to significantly drain the battery, even over a period of several days. Such devices typically draw only 50 to 100 mA, which added to the car's dark current (less than 50 mA per Ford's specs) would be a total draw of 100 to 150 mA.
So let's assume the car's dark current and the FM transmitter draw is 100 mA, the battery has a reserve capacity (RC) rating of 75 minutes (typical), and that the engine (worst case) is off for 24 hours a day. RC is the number of minutes the battery can support a 25 A draw, before the voltage drops to 10.5 V--so:
25 A * 75 minutes = 1875 Am (Amp minutes) = 31.25 Ah (Amp hours)
24 hours * .100 A per hour = 2.4 Ah per day
31.25 Ah / 2.4 Ah per day = 13 days before the battery would be drained to 10.5 V
However 100 mA is such a small load that the reserve capacity at that current would be much more than the rated 75 minutes at 25 A. It would probably take 3 to 4 weeks before the battery would no longer start the engine...
So let's assume the car's dark current and the FM transmitter draw is 100 mA, the battery has a reserve capacity (RC) rating of 75 minutes (typical), and that the engine (worst case) is off for 24 hours a day. RC is the number of minutes the battery can support a 25 A draw, before the voltage drops to 10.5 V--so:
25 A * 75 minutes = 1875 Am (Amp minutes) = 31.25 Ah (Amp hours)
24 hours * .100 A per hour = 2.4 Ah per day
31.25 Ah / 2.4 Ah per day = 13 days before the battery would be drained to 10.5 V
However 100 mA is such a small load that the reserve capacity at that current would be much more than the rated 75 minutes at 25 A. It would probably take 3 to 4 weeks before the battery would no longer start the engine...
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mrmrultimate
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09-10-2015 09:43 AM