Cabin light left on...
10-05-2007, 03:48 PM
#16
the formula is P=I x E
P is power in watts
I is current (in amps)
E is voltage
so:
#Watts=#Amps x Voltage
The interior bulb I believe is a 3 watt or 5 watt bulb
5 Watts=#Amps x Voltage which is 12
5 Watts=#Amps x 12V
Number of amps =5/12
So the current draw would be .42 amps for a 5 watt bulb
A 3 watt bulb would draw .25 amps at 12 Volts
Gee I hope my math is right
P is power in watts
I is current (in amps)
E is voltage
so:
#Watts=#Amps x Voltage
The interior bulb I believe is a 3 watt or 5 watt bulb
5 Watts=#Amps x Voltage which is 12
5 Watts=#Amps x 12V
Number of amps =5/12
So the current draw would be .42 amps for a 5 watt bulb
A 3 watt bulb would draw .25 amps at 12 Volts
Gee I hope my math is right
Sounds reasonable. Bet it's at least 5 watt, though.
I inadvertantly didn't completely close the trunk one time which left BOTH interior lights AND the trunk and engine courtesy lights on for about 2 hours on a not-so-strong battery. Had to jump it. Ordered a battery tender the next day.
Alan- How did this event work out for you?
