Re: [Shop-talk] ohms law help

[DAVID MASSEY <dave1massey@cs.com>]

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 The real question is what are you powering?=C2=A0 A 175 watt inverter powe=
ring a 100W load will draw 100W/12V =3D 8.3A.=C2=A0 Just because the invert=
er has a 175 capacity it will draw only the power it needs to run the load.=
=C2=A0 So if you have a specific load in mind start there.=C2=A0 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% ran=
ge.=C2=A0 So your 14.5 amps could be 17 amps.=C2=A0=20

Also consider that a 12 gauge wire has a resistance of 2.5 Ohms per 1000ft.=
=C2=A0 A 10 foot run there an back is a total of 20 ft.=C2=A0 So=20
2.5*20/1000 =3D .05 Ohms.=C2=A0 At 17 amps you will get a voltage drop of .=
85 volts.=C2=A0 Not an issue since an automotive battery typically produces=
 13 to 13.5 and a functioning alternator runs at 14 volts or more.=C2=A0 Bu=
t 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.=C2=A0 Per UL, CSA, IEC, etc standard thes=
e devices will have a name plate indicating the power in (worst case) and t=
he power out (max ratings).
Other than that it's pretty simple.
=20
Dave=20

=20
=20
-----Original Message-----
From: john niolon <jniolon@att.net>
To: shop-talk <shop-talk@autox.team.net>
Sent: Sat, Sep 26, 2020 3:22 pm
Subject: [Shop-talk] ohms law help

#yiv4470871034 body {margin:0.7em;}#yiv4470871034 body.yiv4470871034OECFntD=
ef, #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=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0175 watt ac/dc convert=
er12 volt dc=C2=A0looking for amp draw off battery=C2=A0power/voltage =3D a=
mps=C2=A0=C2=A0175/12 =3D 14.5 amps=C2=A0=C2=A0=C2=A0 right ??=C2=A0=C2=A0=
=C2=A0 and 10' of 12 gauge wire should do it=C2=A0 right ??=C2=A0thanksjohn=
=C2=A0_______________________________________________

Shop-talk@autox.team.net
Archive: http://www.team.net/pipermail/shop-talk http://autox.team.net/arch=
ive

ssey@cs.com


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<div style="color:black;font: 12pt Arial, Helvetica, sans-serif;">
<div> <font size="3">The real question is what are you powering?&nbsp; A 175 
watt inverter powering a 100W load will draw 100W/12V = 8.3A.&nbsp; Just 
because the inverter has a 175 capacity it will draw only the power it needs to 
run the load.&nbsp; So if you have a specific load in mind start there.&nbsp; 
If this is for generic use then go with the capacity.<br>
</font></div>

<div><font size="3"><br>
</font></div>

<div><font size="3">Then factor in efficiency which on any modern inverter is 
in the 85-95% range.&nbsp; So your 14.5 amps could be 17 amps.&nbsp; <br>
</font></div>

<div><font size="3"><br>
</font></div>

<div><font size="3">Also consider that a 12 gauge wire has a resistance of 2.5 
Ohms per 1000ft.&nbsp; A 10 foot run there an back is a total of 20 ft.&nbsp; 
So <br>
</font></div>

<div><font size="3">2.5*20/1000 = .05 Ohms.&nbsp; At 17 amps you will get a 
voltage drop of .85 volts.&nbsp; Not an issue since an automotive battery 
typically produces 13 to 13.5 and a functioning alternator runs at 14 volts or 
more.&nbsp; But if you were making a longer run the voltage drop could be more 
of a wire size determinant than the wire's capacity.</font></div>

<div><font size="3"><br>
</font></div>

<div><font size="3">Lastly, check the nameplate data.&nbsp; 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).</font></div>

<div><font size="3"><br>
</font></div>

<div><font size="3">Other than that it's pretty simple.</font><br>
</div>

<div> <br>
</div>

<div style="clear:both"><font size="4">Dave </font><br>
<br>
</div>

<div> <br>
</div>

<div> <br>
</div>

<div 
style="font-family:arial,helvetica;font-size:10pt;color:black">-----Original 
Message-----<br>
From: john niolon &lt;jniolon@att.net&gt;<br>
To: shop-talk &lt;shop-talk@autox.team.net&gt;<br>
Sent: Sat, Sep 26, 2020 3:22 pm<br>
Subject: [Shop-talk] ohms law help<br>
<br>

<div id="yiv4470871034">
<style type="text/css">#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;}</style>

 

<div>

<div>been a long time since I used this&nbsp;&nbsp;&nbsp;&nbsp;</div>


<div>&nbsp;</div>


<div>175 watt ac/dc converter</div>


<div>12 volt dc</div>


<div>&nbsp;</div>


<div>looking for amp draw off battery</div>


<div>&nbsp;</div>


<div>power/voltage = amps</div>


<div>&nbsp;</div>


<div>&nbsp;</div>


<div>175/12 = 14.5 amps&nbsp;&nbsp;&nbsp; right ??&nbsp;&nbsp;&nbsp; and 10' of 
12 gauge wire should do it&nbsp; right ??</div>


<div>&nbsp;</div>


<div>thanks</div>


<div>john</div>


<div>&nbsp;</div>
</div>
</div>
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