2.2 Ohms Law, Power Factor, and Three-Phase Review
Key Takeaways
- Ohms law and power formulas must be matched to the circuit type, load type, and phase relationship shown in the problem.
- Power factor changes real power, apparent power, conductor current, transformer loading, and generator sizing decisions.
- Three-phase calculations require the square-root-of-three relationship for line quantities, but not every problem is a three-phase problem.
- Master-level review includes recognizing when nameplate data, code multipliers, or equipment articles replace simple theory formulas.
Theory as a calculation language
Electrical theory is the language behind many master-level code questions. The exam may not ask, What is Ohms law? It may give a motor load, a transformer, a long feeder, a nonunity power factor, or a three-phase panel schedule and expect you to select the right setup. The field expectation is similar. A master electrician must know whether a measured current is reasonable, whether a transformer is overloaded, whether a neutral is being misunderstood, and whether the voltage and phase listed on the plan match the equipment being installed.
Start with the basic relationships. Ohms law is E = I x R, I = E / R, and R = E / I. Power in a resistive direct-current or unity-power-factor single-phase situation can be written as P = E x I, P = I^2 x R, or P = E^2 / R. For alternating-current systems, the difference between real power, apparent power, and reactive power matters. Real power is watts or kilowatts. Apparent power is volt-amperes or kilovolt-amperes. Reactive power is vars or kilovars. Power factor is real power divided by apparent power.
Formula selection table
| Situation | Common formula | Notes |
|---|---|---|
| Single-phase apparent power | VA = V x I | Use line voltage and line current for the single-phase load. |
| Single-phase real power | W = V x I x PF | Applies where power factor is stated or implied by equipment. |
| Three-phase apparent power | VA = 1.732 x Vline x Iline | Use line-to-line voltage and line current. |
| Three-phase real power | W = 1.732 x Vline x Iline x PF | Add power factor only when calculating real power from current. |
| Current from three-phase kVA | I = kVA x 1000 / (1.732 x Vline) | Do not multiply by power factor when kVA is already apparent power. |
| Current from three-phase kW | I = kW x 1000 / (1.732 x Vline x PF) | Include power factor because kW is real power. |
A common exam trap is using power factor twice. If a transformer is rated 75 kVA, the rating is apparent power. Current on a 480 volt three-phase secondary is 75,000 / (1.732 x 480), or about 90 amps. Do not divide by power factor unless the problem gives kilowatts and asks for current. Another trap is forgetting the 1.732 factor. A 30 kVA three-phase load at 208 volts draws about 83 amps. The same apparent power on a single-phase 208 volt load would draw about 144 amps.
Phase, voltage, and neutral discipline
A master electrician should separate line-to-line, line-to-neutral, and phase winding quantities. In a 120/208 volt three-phase, 4-wire wye system, 208 volts is line-to-line and 120 volts is line-to-neutral. A three-phase load connected across all phases uses the 208 volt line-to-line value in the three-phase formula. A single-pole 120 volt load uses 120 volts in the single-phase formula. A 240/120 volt single-phase service is not a three-phase system just because it has two ungrounded conductors and a neutral.
Plan review often exposes voltage mistakes. A rooftop unit nameplate may require 208/230 volt single-phase or 460 volt three-phase. A panel schedule may show a two-pole breaker in a three-phase panel, but the load may still be single-phase. A sign transformer, welder, elevator controller, or kitchen appliance may carry nameplate data that directs branch-circuit sizing. The master-level habit is to read the nameplate basis before forcing the load into a generic formula.
Power factor and design judgment
Power factor affects conductor current and source loading. Low power factor means more apparent power is required for the same real power. A 50 kW three-phase load at 480 volts and 0.80 power factor draws about 75 amps: 50,000 / (1.732 x 480 x 0.80). At unity power factor it would draw about 60 amps. The real work is the same, but the current and apparent power are not.
That difference matters for transformers, generators, voltage drop, and utility metering. A generator must be evaluated for both kW and kVA capability. A transformer is normally rated in kVA, so loads with poor power factor can consume capacity even when the real power seems modest. Conductors are heated by current, not by the name of the load. Protective devices respond to current and time-current behavior, not simply to kW.
Code interaction
Theory formulas do not override code rules. Motor branch circuits, transformer secondaries, continuous loads, neutral calculations, appliance nameplate instructions, and special equipment can require article-specific sizing steps. The formula tells you a base current. The NEC article tells you whether to multiply, round, protect, derate, adjust, or select a standard ampere rating. On the exam, read whether the question asks for calculated load, conductor ampacity, overcurrent protection, transformer current, or voltage drop. Each answer may start with the same theory but end in a different code section.
A strong workflow is: list known values, identify phase, decide whether the value is W, kW, VA, or kVA, decide whether power factor applies, compute current or power, then apply the specific code rule requested. Write units on every step. Most wrong theory answers come from a correct formula used with the wrong unit, the wrong voltage, or the wrong load type.
A 75 kVA, 480 volt, three-phase transformer secondary is being checked for full-load current. Which setup is correct?
A 50 kW three-phase load operates at 480 volts with 0.80 power factor. What is the approximate line current?
Which statement is the best master-level warning about electrical theory formulas?