[DML] Ohm's Law (Cigar lighter output)
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[DML] Ohm's Law (Cigar lighter output)

Something is confusing about your terminology, "How many amps is the cigar  
lighter -supposed- to put out??"
What do you mean my this?  As I see it the cigar lighhter socket, does not put 
out anything.  A circuit that you plug into it will "draw" current from  it as 
limited by Ohm's Law (Volts= Amps x Ohms) and the fuse.
The cigar light is protected by fuse # 17 which is supposed to be 20  amps.  
Using Ohm's Law that would mean that the lowest resistance device  you could 
plug in would be about 0.6 ohms.  Pretty darn low and drawing  alot of current 
(20 amperes).  A short is theoretically zero ohms, thus  would want to draw 
infinite current (amps) and thus it will blow the fuse.
You should be able to plug in devices that will use up to 20 amps of  current 
without blowing fuse #17.
If you plug something in the cigar lighter socket and it blows fuse #17  then 
it is "drawing" more than the 20 Amps  ...and probably has a  short.  Look 
closely inside the socket to make sure there is no foreign  material in there 
and DON'T clean it out with something metallic ... unless the  power is already 
If this does not answer your question then please repost more clearly or  
send it to me.

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