I have come across a few online voltage drop calculators and I believe I am misunderstanding the application or input parameters for these calculators because the voltage drop results I calculate by hand conflict with the calculator results. I have posted an example below.

__Assumptions__

Wire: 14 awg, CU Wire, I used 3.14 Ohms/Kf or 0.00314 Ohms/ft for a resistivity

__Parameters__

Voltage: 12 VDC (12 Vdc battery)

Rwire: 0.157 ohms (I am using 50' of wire, 0.157 was found by .00314*50')

Rload: 4 ohms

Schematic:

Now, according to typical voltage drop methods using the online calculators. Vdrop is found by using the load current multiplied by the wire resistance. I agree with this, however I am confused why they do not factor in the wire resistance in series with the load when determining load current. See example below.

__Typical Online Voltage Drop Calculator Method__

Load Current (IL) = Voltage/Rload = 12/4 = 3 AMPS

Vdrop = IL* (2*Rwire) = 3 * (0.314) = 0.942 Vdc

__My Calculations__

Rtotal = Rload + 2*Rwire = 4.314

Load Current (IL) = Voltage/Rtotal = 12/4.314 = 2.78 AMPS

Vdrop = IL* (2*Rwire) = 2.78 * (0.314) = 0.873 Vdc

I am just looking for any feedback for this discrepancy.

Thank you all for your time,

Chad