[Shop-talk] ohms law help

[DAVID MASSEY]

 The real question is what are you powering?  A 175 watt inverter powering a 100W load will draw 100W/12V = 8.3A.  Just because the inverter has a 175 capacity it will draw only the power it needs to run the load.  So if you have a specific load in mind start there.  If this is for generic use then go with the capacity.

Then factor in efficiency which on any modern inverter is in the 85-95% range.  So your 14.5 amps could be 17 amps.  

Also consider that a 12 gauge wire has a resistance of 2.5 Ohms per 1000ft.  A 10 foot run there an back is a total of 20 ft.  So 
2.5*20/1000 = .05 Ohms.  At 17 amps you will get a voltage drop of .85 volts.  Not an issue since an automotive battery typically produces 13 to 13.5 and a functioning alternator runs at 14 volts or more.  But if you were making a longer run the voltage drop could be more of a wire size determinant than the wire's capacity.
Lastly, check the nameplate data.  Per UL, CSA, IEC, etc standard these devices will have a name plate indicating the power in (worst case) and the power out (max ratings).
Other than that it's pretty simple.
 
Dave 

 
 
-----Original Message-----
From: john niolon <jniolon at att.net>
To: shop-talk <shop-talk at autox.team.net>
Sent: Sat, Sep 26, 2020 3:22 pm
Subject: [Shop-talk] ohms law help

#yiv4470871034 body {margin:0.7em;}#yiv4470871034 body.yiv4470871034OECFntDef, #yiv4470871034 body.yiv4470871034OECFntDef div {font-family:"Segoe UI", Tahoma, Verdana, Arial, Helvetica, sans-serif;font-size:12pt;} been a long time since I used this     175 watt ac/dc converter12 volt dc looking for amp draw off battery power/voltage = amps  175/12 = 14.5 amps    right ??    and 10' of 12 gauge wire should do it  right ?? thanksjohn _______________________________________________

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