Mike Dolt Enterprises
Mike Dolt Enterprises
User: ElecBeginner, Location: Hawaii
Asked on: Wednesday, February 12, 2025 at 7:13 pm EST
Answered on: Thursday, March 6, 2025 at 8:34 am EST
Question: Hello All, I've been looking into my local voltage drop calculator and various online calculators online and spent the better half of the day trying to understand whats going on.I am under the assumptionVD(single phase) = I x Z x L/1000 x 2I = VA/V = Watts/(V*PF)PF = Cos(theta) = Watts/VAZ = Rcos(theta) + Xcos(theta)With the above ive been trying to recreate the results i see in various VD calculators and getting a result that is off by a factor of PF.Example:120V, 1ph, 12000 watts, 0.9PF, 10ft, #12 copper wire in PVC (numbers arbitrarily chosen, the result ends up the same for different values)using the equation above i getVD = 12000watts / (120V x 0.9) x (2 x cos(25.8) + 0.054 x sin(25.8) ) x 10ft/1000 x 2= 4.07VUsing the Calculator i get 3.65V (which is 4.07 x 0.9), this includes other calculators i find including southwire, the one recommended by mike holt, and other miscellaneous calculators.Not sure how to add an image but ive used excel to compare and its generally always off by a factor of PF, so the question is am I missing a variable somewhere or converting real vs apparent parent wrong somewhere or something else altogether? Any help would be appreciated.
You're very close, and you've identified the key issue: Power Factor (PF). The problem lies in how you're applying the PF to the impedance. Let's break it down
Understanding Impedance (Z)
Impedance (Z) is the total opposition to current flow in an AC circuit. It has two components: Resistance (R)
Opposition to current flow due to the conductor's material properties. This is where the real power or Watts is dissipated.
Reactance (X)
Opposition to current flow due to inductance (XL) or capacitance (XC). This is where reactive power or VARs is stored and released.
Power Factor (PF) is the cosine of the phase angle (θ) between the voltage and current. It represents the ratio of real power (Watts) to apparent power (VA). PF = Watts / VA = cos(θ)The Correct Approach to Voltage Drop Calculation
Your fundamental voltage drop equation `VD = I Z L/1000 2` is a good starting point, but needs some refinement. Here's the most accurate way to calculate voltage drop, considering power factor:1. Calculate the Current (I)
I = VA / V This is correct. Since you're given Watts and PF, you need to find VA first. VA = Watts / PF I = (Watts / PF) / V = Watts / (V PF) This is exactly what you did and is correct. So, `I = 12000 / (120 0.9) = 111.11 amps`2. Find the Resistance (R) and Reactance (X) per unit length
Look up the per-foot (or per-1000ft) resistance (R) and reactance (X) of #12 AWG copper wire in PVC conduit. These values are typically found in tables within the NEC (National Electrical Code) or engineering handbooks. This is where the value of R=2, X=0.054 comes from in your example.
Important
Make sure you're using the correct values for AC resistance, which will be slightly higher than DC resistance due to the skin effect. This is already accounted for in the NEC values.3.
Calculate the Impedance Angle (θ) from the Power Factor
θ = arccos(PF) = arccos(0.9) = 25.84 degrees (This is correct)4. Calculate the Voltage Drop
Here is the important difference, the impedance already takes into consideration the affect of PF since it is calculated with the associated reactance of the conductor.
VD = I x (R + jX) x L/1000 x 2 If you want to split the equation into components, you can use the following where X and R do not include the impact of the PF VD = (I R L/1000 2)cos(θ) + (I X L/1000 2)sin(θ) VD = (111.11A x 2 x 10/1000 x 2) + (111.11A x 0.054 x 10/1000 x 2) VD = 4.444 + 0.120 = 4.
5645. Calculate the Voltage Drop
Here is the important difference, the impedance already takes into consideration the affect of PF since it is calculated with the associated reactance of the conductor.
VD = I x (R + jX) x L/1000 x 2 VD = 111.11A x (2 + j0.054) x 10/1000 x 2 VD = 4.444 + j0.12 = 4.445 at an angle of 1.546 degrees Because VD is a complex number you need to calculate the magnitude of this vector Magnitude VD = sqrt(4.444^2 + 0.12^2) = 4.445So What is Different?
The problem in your calculation is that you are multiplying the resistance and reactance by cos(theta) and sin(theta). These values are already accounted for within the R and X components of impedance, the PF only exists to convert from real power to apparent power (Watts vs VA). By multiplying these values you're essentially applying the Power factor twice!Why are you getting the same result in the calculators?
The result you are seeing in the VD calculators are almost similar to what you're calculating using the PF. The biggest difference between 3.65 and 4.44 comes from the value of the resistance.
Important Considerations
Temperature
Conductor resistance increases with temperature. The NEC tables provide values at specific temperatures (usually 75°C for conductors). If your actual conductor temperature is significantly higher, you'll need to adjust the resistance accordingly.
Harmonics
If your load has significant harmonic content, the voltage drop calculation becomes more complex. Harmonics cause additional current flow at higher frequencies, increasing the effective impedance.
Accuracy
Voltage drop calculators are approximations. They're useful for design and verification, but they shouldn't be considered perfectly precise.
In summary:
You were very close! By removing the multiplication of cos/sin(theta) from the impedance and accounting for the accurate values of Resistance and Reactance you should be able to achieve similar values. Flag for review