Journeyman refresher Refresher 10 of 10 · about 5 min
Journeyman formula card
What you’ll learn
- Every formula from the third-year, fourth-year and refresher material on one page.
- Which variable is which — line or phase, per winding or per line.
- A tool and a lesson for each group, when you need the working or the why.
Three-phase power
| kVA = √3 × V × I ÷ 1000 | V line-to-line, I line |
| kW = kVA × PF | real power |
| kVAR = √(kVA² − kW²) | reactive power |
| I = kW × 1000 ÷ (√3 × V × PF) | current from kW |
| Wye: V L = √3 × V ph, I L = I ph | Delta: V L = V ph, I L = √3 × I ph |
Transformers
| FLA = kVA × 1000 ÷ (√3 × V) | three-phase, each side |
| FLA = kVA × 1000 ÷ V | single-phase |
| Turns ratio = V winding pri ÷ V winding sec | per winding, not line to line |
| Open delta ≈ 57.7% of the three-can bank | 1 ÷ √3 |
Available fault current
| Isc = FLA × 100 ÷ %Z | infinite bus, at the terminals |
| Multiplier = 100 ÷ %Z | e.g. 5% → 20 × FLA |
Motors
| Ns = 120 × f ÷ P | synchronous speed, rpm |
| Slip % = (Ns − N) ÷ Ns × 100 | N = actual speed |
| f rotor = slip × f | rotor current frequency |
| I = HP × 746 ÷ (√3 × V × eff × PF) | three-phase current from HP (use the nameplate for real work) |
Drives
| V/Hz = V rated ÷ f base | held constant below base |
| N ≈ 120 × f ÷ P − slip rpm | shaft speed on a drive |
| DC bus ≈ 1.35 to 1.41 × V line | loaded to no-load |
Power factor correction
| kVAR = kW × (tan φ1 − tan φ2) | φ = arccos PF |
| C = Q ÷ (2π f V²) | per capacitor; Q in VAR, V across it |
Voltage unbalance
| % = max deviation from average ÷ average × 100 | three line-to-line readings |
4–20 mA
| PV = LRV + (mA − 4) ÷ 16 × span | current to value |
| mA = 4 + 16 × (PV − LRV) ÷ span | value to current |
| V = I × R | 4–20 mA × 250 Ω = 1–5 V |
Delta ↔ wye impedance
| Za = Zab × Zca ÷ (Zab + Zbc + Zca) | delta to wye |
| Zab = (ZaZb + ZbZc + ZcZa) ÷ Zc | wye to delta |
| Balanced: Z wye = Z delta ÷ 3 |
Worked check: three formulas on one motor
- 10 HP, 460 V, eff 0.90, PF 0.85: I = 10 × 746 ÷ (1.732 × 460 × 0.90 × 0.85) = 7,460 ÷ 609.5 = 12.2 A (a physics estimate — the nameplate FLA is what you use).
- Nameplate speed 1760 rpm on 60 Hz: 4-pole, Ns 1800, slip = 40 ÷ 1800 = 2.2%.
- On a drive at 50 Hz: shaft ≈ 120 × 50 ÷ 4 − 40 = 1460 rpm; output ≈ 460 ÷ 60 × 50 = 383 V.
On the job
Tips
- Print it double-sided and keep it with your code book tabs while you study.
- Before using a formula, say out loud what V and I it wants — line or phase.
- Work each answer back the other way; a formula that does not close was used wrong.
Common mistakes
- Using 3 instead of √3 with line values.
- Using the line ratio as the turns ratio on a delta–wye bank.
- Dividing by %Z as a whole number without the 100 (5 instead of 0.05).
- Using a calculated motor current where your code book tells you to use its own values.
Check yourself
-
1.What is the synchronous speed of a 6-pole motor on 60 Hz?
Ns = 120 × f ÷ P = 120 × 60 ÷ 6 = 1200 rpm.
-
2.A 300 kVA transformer with 5% impedance and a 208 V secondary. Infinite-bus fault current at its terminals?
FLA = 300,000 ÷ (1.732 × 208) = 832.7 A. Isc = FLA × 100 ÷ %Z = 832.7 × 20 = 16,654 A.
-
3.A 4–20 mA pressure transmitter is ranged 0–100 psi. What does 8 mA represent?
PV = LRV + (mA − 4) ÷ 16 × span = 0 + (8 − 4) ÷ 16 × 100 = 25 psi.
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4.kVAR to correct a 50 kW load from PF 0.80 to 0.95?
kVAR = kW × (tan φ1 − tan φ2) = 50 × (0.750 − 0.329) = 50 × 0.421 = 21.1 kVAR.
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5.Three equal 12 Ω impedances in delta. The equivalent wye has arms of…
Balanced: Z wye = Z delta ÷ 3 = 12 ÷ 3 = 4 Ω.
Related
- Single & three phase formula sheet — the first- and second-year set
- Ohm's law & power wheel
- Rusty on three-phase math?
- Journeyman practice exam
- Exam prep strategy
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