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    <p>I remember doing some measurements on a long wire run, and being
      surprised that the actual voltage drop was much less than the
      calculations.  That's when I found a small note at the bottom of
      the NEC Table 8 conductor properties used for official VD
      calculations.   The resistance is listed at 167 F (75C), so if you
      use the adjustment equation in Note 2, it is much less VD, and
      corresponds to reality.  For underground conductors, I might use
      25C.  Use common sense.  </p>
    <p>The formula is R2=R1 (1+ a(T2-75))</p>
    <p>I made a spread sheet about 15 years ago, that uses that formula
      for doing wire sizing.  Happy to share, if anyone is interested.</p>
    <p>Ray Walters<br>
      Remote Solar</p>
    <div class="moz-cite-prefix">On 3/21/2026 3:21 PM, Jay via
      RE-wrenches wrote:<br>
    </div>
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      cite="mid:6B74B822-29E3-4726-9208-6AB35CFF597B@gmail.com">
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      <div dir="ltr">Are you trying to run the whole house or just the
        burner?  </div>
      <div dir="ltr">I’m also curious about how you calculated the VD. I
        ran it on a few programs and it came up much less VD. I’d go
        direct or try the load and measure the VD. It won’t use the full
        power for long. Or at least when I use my induction top it takes
        so short to bring anything to a boil, after that it’s turned
        down. </div>
      <div dir="ltr">Lots of options. </div>
      <div dir="ltr"><br>
        <blockquote type="cite">On Mar 20, 2026, at 11:23 PM, William
          Miller via RE-wrenches
          <a class="moz-txt-link-rfc2396E" href="mailto:re-wrenches@lists.re-wrenches.org"><re-wrenches@lists.re-wrenches.org></a> wrote:<br>
          <br>
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            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">Friends:</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">A
                long-time client has an ADU 755 conduit feet from the
                main house.  They started off-grid but they are now grid
                connected.</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">In
                1999 we pulled a 120 volt feeder of #1 aluminum to the
                ADU.  This provides about 10 amps at 5% voltage drop.</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">This
                client wants to abandon a propane cook-top in the ADU in
                favor of an induction cook-top (indoor air pollution as
                a consideration).  A sample two burner induction
                cook-top is 1800 watts--too much power draw for this
                feeder.  </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">I
                was wondering if a portable power station rated at
                enough power to run the induction cook-top might solve
                this problem.  The 10 amp power feed can charge the
                power station at (hopefully) an acceptable amperage over
                an extended time period and the power station can
                provide enough surge power to run the cook-top
                intermittently.</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">I
                see the Bluetti AC500, for example, can be adjusted to
                draw no more than 10 amps when plugged into AC. 
                Unfortunately this charge rate, when running at the full
                10 amps, allows little overhead for other loads.  I
                presume this 10 amps draw is only 10 amps when the
                internal battery is fully discharged.  Maybe other units
                can charge at a lower rate.</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">Have
                any of you tried this?  Any input is appreciated.</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif"">William</span></p>
            <p class="MsoNormal"><span
style="font-size:14.0pt;font-family:"Arial","sans-serif""> </span></p>
            <p class="MsoNormal">William Miller</p>
            <p class="MsoNormal"><span
style="font-size:12.0pt;font-family:"Arial","sans-serif"">Miller
                Solar</span></p>
            <p class="MsoNormal"><span
style="font-size:12.0pt;font-family:"Arial","sans-serif""><a
                  href="http://www.millersolar.com/"
                  moz-do-not-send="true">www.millersolar.com</a></span></p>
            <p class="MsoNormal"><span
style="font-size:12.0pt;font-family:"Arial","sans-serif"">CA
                License C-10 77398</span></p>
            <p class="MsoNormal"> </p>
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      <pre wrap="" class="moz-quote-pre">_______________________________________________
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</pre>
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