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<p class="MsoNormal"><span style="font-size: 12pt; font-family: "Times New Roman";">Hello Jason,<br>
<br>
The correct voltage-drop calculation requires round-trip distance. Two
conductors, each the same gauge and carrying the same current will have equal
voltage drop.<br>
<br>
For example:<br>
<br>
Conductor # 1<br>
-----------------------------------------------------------------<br>
|
|<br>
Power
Source
Load<br>
|
|<br>
-----------------------------------------------------------------<br>
Conductor # 2<br>
<br>
<br>
The equivalent resistive circuit:<br>
Resistor # 1<br>
----------------------------/\/\/\/\/\/\------------------------<br>
|
|<br>
Power
Source
Load<br>
|
|<br>
----------------------------/\/\/\/\/\/\------------------------<br>
Resistor # 2<br>
<br>
Rather than get into the esoteric terminology (Thevenin's Theorem) .. suffice
to say the total of the voltage dropped across R1 + Load + R2 is equal to the
Power Source voltage, as is easily seen above.<br>
<br>
Voltage is applied to the entire circuit. It doesn't flow just "one
way". The same is true of current, which also flows in the entire
loop.<br>
<br>
Overall, "power" is generated by the Power Source, and
"consumed" in the Load, which makes us want to think of energy
flowing "one way". Some of the power output by the Power Source
is lost in Resistors #1 and #2. Even so .. the total energy dissipated by
R1 + Load + R2 must equal the energy delivered by the Power Source.<br>
<br>
Hope this helps...<br>
<br>
Dan<br>
Sr. Engineer<br>
Exeltech<br>
<br>
<br>
<br>
--- On <b>Mon, 12/8/08, Jason Lombard <i><openhandsolar@gmail.com></i></b>
wrote:<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-bottom: 12pt;"><span style="font-size: 12pt; font-family: "Times New Roman";">From: Jason Lombard
<openhandsolar@gmail.com><br>
Subject: Re: [RE-wrenches] voltag drop<br>
To: "RE-wrenches" <re-wrenches@lists.re-wrenches.org><br>
Date: Monday, December 8, 2008, 4:44 PM<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-bottom: 12pt;"><span style="font-size: 12pt; font-family: "Times New Roman";">Marco,<br>
<br>
One way it the appropriate answer, you mostly have voltage only going in that
direction is my thought.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size: 12pt; font-family: "Times New Roman";">On
Mon, Dec 8, 2008 at 3:12 PM, Marco <<a href="mailto:marco@pvthawaii.com" target="_blank">marco@pvthawaii.com</a>> wrote:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="">I feel kinda
foolish in asking this.....but when calculating voltage drop in a conductor
over distance, do I use the one-way or round-trip distance?</span><span style="font-size: 12pt; font-family: "Times New Roman";"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size: 12pt; font-family: "Times New Roman";"> <o:p></o:p></span></p>
<p class="MsoNormal"><span style="">thanks,</span><span style="font-size: 12pt; font-family: "Times New Roman";"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="">marco</span><span style="font-size: 12pt; font-family: "Times New Roman";"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size: 12pt; font-family: "Times New Roman";"> <o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size: 12pt; font-family: "Times New Roman";"><!--[if gte vml 1]><v:shapetype
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<p class="MsoNormal" style=""><span style="">Marco Mangelsdorf, Ph.D.</span><span style="font-size: 12pt; font-family: "Times New Roman";">, </span><span style="">President</span><span style="font-size: 12pt; font-family: "Times New Roman";"> <br>
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</span><span style="">69 Railroad Avenue, Suite A-7</span><span style="font-size: 12pt; font-family: "Times New Roman";"> <br>
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</span><span style="">(808) 969-3281, fax 934-7462</span><span style="font-size: 12pt; font-family: "Times New Roman";"><o:p></o:p></span></p>
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<p class="MsoNormal"><span style=""><o:p> </o:p></span></p>
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