Field Electrical Quick Calc Electrical
The bread-and-butter electrical math you double-check on a service call, in one place. Ohm’s Law & Power is the wheel — type any two of V / I / R / P and it solves the other two. AC Power ties line voltage, amps, kW, kVA, kVAR, and power factor together for single- and three-phase. Motor HP / kW / FLA converts horsepower to kilowatts and estimates full-load amps. Voltage Imbalance runs the NEMA check on three line-to-line readings.
Input — any two of V, I, R, P
Enter exactly two — the other two are solved and shown in blue. DC or resistive loads (PF = 1); for motors and reactive loads use the AC Power tab.
Output
The Ohm’s-law wheel
| Solve for | From the other two |
|---|---|
| V (volts) | I·R · P/I · √(P·R) |
| I (amps) | V/R · P/V · √(P/R) |
| R (ohms) | V/I · V²/P · P/I² |
| P (watts) | V·I · I²·R · V²/R |
Worked example: a relay coil measures R = 4 Ω and the panel feeds it V = 12 V.
I = V/R = 12 ÷ 4 = 3 AP = V²/R = 144 ÷ 4 = 36 W
So the coil draws 3 A and burns 36 W — sanity-check that against the coil’s wattage rating before you trust the circuit.
Input
Line voltage is line-to-line — 208 / 240 / 480, not 120 / 277. Three-phase carries the √3 factor; pick 1φ for a line-to-neutral branch.
Output
Q is a magnitude; lagging (inductive) is assumed. Size conductors and breakers on amps (kVA-derived), not on kW.
Power factor → phase angle
| PF | Angle θ | kW / kVA |
|---|---|---|
| 1.00 | 0.0° | 1.00 |
| 0.95 | 18.2° | 0.95 |
| 0.90 | 25.8° | 0.90 |
| 0.85 | 31.8° | 0.85 |
| 0.80 | 36.9° | 0.80 |
Worked example (3φ): a 460 V supply fan pulls 30 A at PF 0.85, three-phase.
S = √3 × 460 × 30 = 23.90 kVA(apparent)P = S × 0.85 = 20.32 kW(real)Q = √(S² − P²) = 12.59 kVAR(reactive), θ = acos 0.85 = 31.8°
Line-to-line and line-to-neutral differ by √3 (VLL = √3 × VLN) — a 480 V system reads 277 V to neutral. Enter the L-L value here.
Input — nameplate
PF and η are your assumptions — edit them to the nameplate when you have it. Efficiency takes 0.90 or 90.
Output — estimate
Common motor full-load current (NEC)
| Motor HP | 3φ @ 460 V (A) | 1φ @ 230 V (A) |
|---|---|---|
| 1 | 2.1 | 8 |
| 2 | 3.4 | 12 |
| 3 | 4.8 | 17 |
| 5 | 7.6 | 28 |
| 7.5 | 11 | 40 |
| 10 | 14 | 50 |
| 15 | 21 | — |
| 20 | 27 | — |
| 25 | 34 | — |
| 30 | 40 | — |
| 50 | 65 | — |
Worked example: a 10 HP, 460 V, three-phase motor at PF 0.88 and η 0.90.
FLA = (10 × 746) / (√3 × 460 × 0.88 × 0.90) = 7460 ÷ 631 = 11.8 A— the estimate.- NEC 430.250 lists 14 A for this motor — about 16 % higher.
That gap is the whole point: the table builds in a representative low PF and efficiency, so it always runs higher than your stated-PF estimate. Size off the 14 A table value, never the 11.8 A estimate.
Input — three line-to-line readings
Measure all three phase-to-phase legs at the motor terminals with the load running. NEMA defines unbalance as the largest deviation from the average, over the average.
Output
NEMA MG-1 voltage-unbalance limits
| Unbalance | What it means |
|---|---|
| ≤ 1 % | Acceptable — no derate. |
| 1 – 5 % | Derate the motor per the NEMA curve; find and fix the cause. |
| > 5 % | Do not operate — correct the supply first. |
Worked example: a motor reads 460 / 455 / 450 V across its three legs.
avg = (460 + 455 + 450) ÷ 3 = 455 Vmax deviation = |460 − 455| = 5 Vunbalance = 5 ÷ 455 × 100 = 1.10 %— just into derate territory.
Voltage unbalance is brutal at the motor: the current unbalance runs roughly 6–10× the voltage unbalance, and the extra current is pure heat in the windings. Chase a 1 % reading to a loose lug, an unbalanced single-phase load on the feeder, or the utility.