# Online Voltage Drop Calculators?

Joined Dec 11, 2019
3
Hi All,

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')
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
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,

#### OBW0549

Joined Mar 2, 2015
3,102
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.
Your calculations are correct. Forget the stupid, braindead online calculators and rely, as you did, on common sense and Ohm's Law.

• JohnInTX and dl324

#### SteveSh

Joined Nov 5, 2019
104
But when you used the on-line calculator, you used a current of 3 amps, through the load and wiring. When you did your hand calculation, you used 2.78 amps. If you scale the on-line calculator result by the difference in current, 0.93, you get (0.94 V)*0.93 = 0.87 A, just like your hand calc.

Last edited:

Joined Dec 11, 2019
3
Hi Steve,

Thank you for your feedback. I am sorry, but I am not following your comment. Where did you get 0.94 A?

In my original post I had mentioned I had a concern whether I understood the on-line calculator parameters. I assumed the on-line calculators asked for the load current for just the device. For example, if my device load was a solenoid rated at 36 watts at 12 volts then the load current I would enter would be 3 Amps regardless of what length the cable is which affects the overall resistance of the circuit. That how I obtained the 3 Amps and yes for that example I assumed this was the current through the wire as well which was not correct based on the resistance values I provided in the schematic.

Some on-line calculators specify its the current of the wire which needs to be entered in order to determine the voltage drop of the cable. To me this makes sense, but now the calculator seems useless at this point because we would need to physically measure the current in a circuit or perform circuit analysis like I did to enter the correct amperage (2.78 Amps) through the wire & device into the calculator to get the true voltage drop in the cable.

The only application I could see these calculators useful would be for sizing cables for branch circuits leaving a panel board. For example, if you had a branch circuit leaving a 30 amp breaker. You know the maximum load you will theoretically have on that circuit will be 30 amps. Thus, you can use 30 amps as a load current and play around with different cable types and lengths to estimate the voltage drop in the cable to be sure your voltage drop is within an acceptable percentage so the devices you are powering have enough voltage.

For my application I need to know the voltage drop based on what current (amps) is present at the time of operating this circuit. Not based on a current limiting device in my circuit. The application is not specified clearly on any of the websites for the on-line calculators so for now I have just been speculating what there true purpose is for.

Below are links for some of the online voltage drop calculators I have been checking out.

Thanks,

#### SteveSh

Joined Nov 5, 2019
104
Hi Steve,

Thank you for your feedback. I am sorry, but I am not following your comment. Where did you get 0.94 A?

Thanks,
Oops, I really hacked up that response. See edited version above, with changes in BOLD.

Joined Dec 11, 2019
3
Oops, I really hacked up that response. See edited version above, with changes in BOLD.
I am beginning to to make more sense of this, although 0.87 should be in volts not A. How did you figure the difference in current to be 0.93? Also, if 0.93 is measured in Amps and you multiply it by 0.94 Volts the product is a result in Watts, neither volts or amps.

"If you scale the on-line calculator result by the difference in current, 0.93, you get (0.94 V)*0.93 = 0.87 A, just like your hand calc."

#### SteveSh

Joined Nov 5, 2019
104
I am beginning to to make more sense of this, although 0.87 should be in volts not A. How did you figure the difference in current to be 0.93? Also, if 0.93 is measured in Amps and you multiply it by 0.94 Volts the product is a result in Watts, neither volts or amps.

"If you scale the on-line calculator result by the difference in current, 0.93, you get (0.94 V)*0.93 = 0.87 A, just like your hand calc."
Your right, the 0.87 should be amps, not volts.

The 0.93 is really a unitless factor - the ratio between two currents. Replace the word "difference" with "ratio".

Sorry again for the confusion.

#### leer7839

Joined Dec 15, 2019
1
Hi All,

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')
Schematic:
View attachment 194336
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
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,