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I can't grasp the concept on why these two different voltage drop calculations yield different results. They should both arrive at the same answer.

In our intro EE classes when we were doing any power calculations for any standard circuit we learned that if we sum up all the series/parallel impedances into one equivalent impedance where there is just a voltage source and one equivalent impedance, we would get the same end-result for our calculations than if we were to solve the same circuit in it's original expanded form.

Here's the link to the full original PDF: http://contechnical.com/wp-content/uploads/Voltage-Drop-Calc.pdf

Thanks for any help, it's been keeping me up at night.

EOL Circuit

EOL Calculation

Point-To-Point Circuit

Point-To-Point Calculation

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The end of line resistor calculation shown at the start is a worst case approximation, assuming that all of the current is flowing through a single device \$30.\overline 3\overline0\$ rods away rather than several devices at varying distances. The second calculation is more accurate and quantifies the whole network but more time consuming.

schematic

simulate this circuit – Schematic created using CircuitLab

This shows the oversimplification being made. The top diagram is the actual circuit, the bottom diagram is what the end of line calculation represents.

Those are awful, weird overlapping diagrams you have there by the way.

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  • \$\begingroup\$ I agree with the overlapping diagrams haha. I still don't understand why the answer is different. It should be the exact same answer. As I said in the beginning of my question: "In our intro EE classes when we were doing any power calculations for any standard circuit we learned that if we sum up all the series/parallel impedances into one equivalent impedance where there is just a voltage source and one equivalent impedance, we would get the same end-result for our calculations than if we were to solve the same circuit in it's original expanded form." \$\endgroup\$ – HappyTurtle Sep 23 '18 at 4:22
  • \$\begingroup\$ @HappyTurtle I added a diagram for you to illustrate the simplification being made. \$\endgroup\$ – K H Sep 23 '18 at 4:44
  • \$\begingroup\$ Oh so you're saying that in an EOL Voltage Drop Calc, the impedance of the wires is improperly added together. Adding them all together as if they were in series with one another with no parallel loads in-between them will clearly give you a larger total equivalent impedance. This is why EOL Voltage Drop calcs are always more conservative than Point-To-Point Vdrop calcs, correct? \$\endgroup\$ – HappyTurtle Sep 23 '18 at 8:47
  • \$\begingroup\$ That's right. Not really improper but less precise. In practical applications of engineering you will see a lot of fudge factors, places where error is permitted in the name of simplicity so long as the potential error is within safe boundaries, especially, as in this case, if the error(larger wire if anything) favors function of the system. The more expensive the wire, the more minimization is justified, the more costly an undervoltage failure, the more maximization is. \$\endgroup\$ – K H Sep 23 '18 at 8:59
  • \$\begingroup\$ Perfectly explained, and thank you for the diagram. Very well done! \$\endgroup\$ – HappyTurtle Sep 23 '18 at 18:55
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The computation would be something like the following:

$$\begin{align*} V_\text{DROP}=&\frac{200'\cdot 2}{1000'}\cdot 1.98\:\Omega\cdot \left(75\:\text{mA}+125\:\text{mA}+200\:\text{mA}+100\:\text{mA}\right)\\\\ &+\frac{50'\cdot 2}{1000'}\cdot 1.98\:\Omega\cdot \left(125\:\text{mA}+200\:\text{mA}+100\:\text{mA}\right)\\\\ &+\frac{150'\cdot 2}{1000'}\cdot 1.98\:\Omega\cdot \left(200\:\text{mA}+100\:\text{mA}\right)\\\\ &+\frac{100'\cdot 2}{1000'}\cdot 1.98\:\Omega\cdot 100\:\text{mA}\\\\ =\:&697.95\:\text{mV} \end{align*}$$

Note that this is different from the answers you were given.

The result is \$19.1\:\text{V}-697.95\:\text{mV}=18.40205\:\text{V}\$.

Their error is in step 2, where they incorrectly calculate that \$100'\cdot .00198\frac{\Omega}{ft} = .0198 \:\Omega\$. Instead, they should have calculated that \$100'\cdot .00198\frac{\Omega}{ft} = .198 \:\Omega\$. Note the error?? This means their voltage drop is already off by a factor of 10 for this step.

Also, in step 2, not only did they make the above mistake but they also made another mistake. They computed that \$18.704-.008415= 18.691585\$ when that simple subtraction should have yielded \$18.695585\$, instead. Of course, they weren't computing the subtrahend correctly, regardless. But they failed to perform the subtraction correctly, as well.

Together, this makes for an error of \$-0.004\$ (because they couldn't subtract correctly) as well as the incorrect value to subtract, which is yet another error of \$+0.075735\$. Combined, this means they were off by \$+0.071735\$. You will then find that \$626.215\:\text{mA}+71.735\:\text{mV}=697.95\:\text{mV}\$, which is the value I computed near the top, above.

So that's all there is. They screwed up. Oh, well.

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  • \$\begingroup\$ While I certainly do appreciate you careful attention to detail. It still didn't really get to the heart of the question, being why point-to-point voltage drop calculations and end of line voltage drop calculation yield different results. Thank you for cleaning up the math though, much appreciated! \$\endgroup\$ – HappyTurtle Sep 23 '18 at 8:44

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