How to figure the voltage drop at different points along a circuit?

  • #36
alan2023 said:
Looking at this original question from an electrical install perspective, my challange to someone would be how to build a spreadsheet to plug in the distances and any other necessary values (voltage, wire size/resistance, amps, etc.).
An estimate of the voltage drop can be made by assuming equal loads, distributed equally along the cable.

The number of loads is, n = 13.
The current drawn by each load is, i = 0.35 A.
The length of the cable is, d = 580 ft.
The resistance per unit length is, r = 1.6 milliohm/ft.

The average current is, i * n / 2.
The return resistance per unit length is, 2 * r.
The twos will cancel.

V drop = n * i * d * r = 4.2224 volt.
V far end = 120.0000 - 4.2224 = 115.7776 V.
 
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  • #37
alan2023 said:
Looking at this original question from an electrical install perspective, my challenge to someone would be how to build a spreadsheet to plug in the distances and any other necessary values (voltage, wire size/resistance, amps, etc.). As an electrician, I use custom calculators usually in a spreadsheet to determine voltage drop, conduit fill, cable tray fill. I first understand the code/math requirements, then build the sheet/formulas.
Here's how you could setup a spreadsheet using 7 columns, A,B,C,D,E,F,G. This setup assumes
a constant load current of .35 A per load.

XLoads.png


Think of the length of wiring between loads as constituting a "zone". There are 13 zones also counting the distance from the 120V feed to the first load as a zone.

Column A is the distance of each load from the 120 volt feed.
Column B is the distance between loads, including the distance from the feed to the first load.
Column C is just a reverse integer sequence starting with the number of loads.
Column D is .35 times column C (element by element). These are the currents in each zone.
Column E is .0032 (2 * the resistance per foot of the wire used; the approximate value used here is
the one given by Berkeman. For other than 12 ga, use 2 * the value for that wire) times column D.
This is the voltage drop per foot of wire in a zone.
Column F is column B times column E (element by element). At this point the elements of column F are the voltage drops within each zone; they are the decrease in voltage from one load to the next.

Column G is the voltage at each load. Creating column G requires the use of a procedure. Start with 120V and subtract the first element of column F; this becomes the first element of column G. Now subtract the second element of column F from the just calculated first element of column G and this will be the second element of column G. Then subtract the third element of column F from the second element of column G and this will be the third element of column G. Continue this to the end.
 
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  • #38
The Electrician said:
Here's how you could setup a spreadsheet using 7 columns, A,B,C,D,E,F,G. This setup assumes
a constant load current of .35 A per load.

View attachment 331825

Think of the length of wiring between loads as constituting a "zone". There are 13 zones also counting the distance from the 120V feed to the first load as a zone.

Column A is the distance of each load from the 120 volt feed.
Column B is the distance between loads, including the distance from the feed to the first load.
Column C is just a reverse integer sequence starting with the number of loads.
Column D is .35 times column C (element by element). These are the currents in each zone.
Column E is .0032 (2 * the resistance per foot of the wire used; the approximate value used here is
the one given by Berkeman. For other than 12 ga, use 2 * the value for that wire) times column D.
This is the voltage drop per foot of wire in a zone.
Column F is column B times column E (element by element). At this point the elements of column F are the voltage drops within each zone; they are the decrease in voltage from one load to the next.

Column G is the voltage at each load. Creating column G requires the use of a procedure. Start with 120V and subtract the first element of column F; this becomes the first element of column G. Now subtract the second element of column F from the just calculated first element of column G and this will be the second element of column G. Then subtract the third element of column F from the second element of column G and this will be the third element of column G. Continue this to the end.
What application is that in the snip?
 
  • #39
It's just a math app that I used to format the data in columns, but I didn't do the calculations with it; I did the actual calculations on my HP50g calculator. I don't have Excel on my computer so I couldn't actually give an example of the calculations done in Excel, but I wanted to show what it probably would look like in Excel.
 

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