Watts, volt-amperes and power factor, and which one a Texas exam item is actually asking for

What the exam asks about it

Thirty-nine of the five hundred questions in our bank sit in the theory area, and power is the subject of about a third of them. Five hand you a power factor. Two work the right triangle between apparent, true and reactive power. Three ask for a three-phase current or a three-phase power from line quantities. Two more test whether the watt figure and the volt-ampere figure are the same number.

PSI gives theory five scored items across the two journeyman papers, in the candidate information bulletin for the TDLR electricians program updated 9 July 2026: 3 of the 56 on the knowledge portion under Definitions, Theory, and Plans, and 2 of the 24 on the calculations portion under Calculations and Theory. Master candidates get 7 of 70 and 2 of 30. Every load calculation on the paper is a power calculation underneath, which is the part those counts hide from you.

Why the code counts in volt-amperes

The code counts in volt-amperes because it declines to assume anything about the load. Volt-amperes is what the conductor carries, whatever the load does with it. Watts is what gets converted into useful work. On a heater those are one figure. On a motor they separate, because the current and the voltage fall out of step with each other.

A conductor heats according to current, and current follows volt-amperes, which is why the load calculation rules are written in that unit. In the 2026 edition those rules moved. The dwelling unit minimum unit load for branch circuit calculations sits at NEC 120.13, and the feeder and service counterpart sits at NEC 120.41. That material was in Article 220 through the 2023 edition, so a study guide printed two years ago sends you to the wrong chapter of the book you're allowed to carry into the room.

The two sections carry different figures for different purposes, and using one where the other belongs produces a plausible wrong answer. Our page on rounding to the next standard size picks the arithmetic up from there, at the point where a calculated figure has to land on a device rating.

Power factor, and the current it adds

Power factor is the ratio of watts to volt-amperes. At one they're equal. Below one the circuit carries more current than the useful power suggests, and how much more is worth working once with numbers.

Take 8,000 watts on a 240 volt single-phase circuit. At unity the current is 8,000 over 240, or 33.3 amperes. At 0.8 the apparent power is 8,000 over 0.8, which is 10,000 volt-amperes, and the current is 10,000 over 240, or 41.7 amperes. Same useful work, a quarter more current, and the conductor gets sized on the current.

So a load with poor power factor wants a bigger conductor for the same amount of work, and that's the practical reason the concept is on this exam at all. It's also why the bank keeps the arithmetic pointed one way at a time. A stem going from volt-amperes to watts multiplies by the factor. A stem going from watts to current divides by it. Each direction has its own signature wrong answer sitting in the options.

The single-phase and three-phase forms

Single phase: volt-amperes is the voltage across the load times the current through it.

Three phase: volt-amperes is the line to line voltage times the line current times the square root of three. The word line is load bearing. Hand that formula a phase voltage and the answer is wrong by a factor of 1.73, which the bank marks as the single most common theory error.

Written from the other side, the same quantity is three times the phase voltage times the phase current, with no square root in it anywhere. Two forms, one result. Where yours disagree by roughly 1.73, you've taken one quantity from each side.

Going the other way, from power to current, both multipliers move into the denominator. A balanced three-phase load of 45 kilowatts at 208 volts with a power factor of 0.9 draws 138.8 amperes in each line conductor, because 208 times 1.732 is 360.3, times 0.9 is 324.2, and 45,000 divided by that is 138.8. Drop the square root of three and you get 240.4. Drop the power factor and you get 124.9. Both figures are sitting in the options, which is why writing the denominator out in full before dividing is worth the five seconds.

The triangle that decides the leftover

Apparent power, true power and reactive power sit on a right triangle with apparent power as the hypotenuse. They combine as sides of that triangle, so arithmetic that adds or subtracts them along a line is treating a triangle as a straight line.

A feeder carrying 30 kilovolt-amperes into a load at 0.8 power factor is carrying 24 kilowatts of true power and 18 kilovars of reactive power. The 18 comes from the square root of 900 minus 576. Subtract 24 from 30 and you get 6, which is the option the bank sees picked, and what makes it wrong has nothing to do with arithmetic accuracy.

The same shape decides an impedance question. Eight ohms of resistance in series with 6 ohms of inductive reactance gives 10 ohms of impedance, from the square root of 64 plus 36. The power factor is the resistance over the impedance, 8 divided by 10, which is 0.8. Taking the reactance over the impedance gives 0.6, a real ratio with a different name and a place in the options. Our page on series and parallel circuits covers why resistance and reactance in series behave differently from two resistances in series.

A horsepower rating makes it a motor item

An item that hands you a horsepower rating is a motor item, and motors carry their own rules that supersede this arithmetic. NEC 430.6 settles which current applies to which calculation, and conductor sizing comes off the full load current tables in Article 430.

The reason is a chain of steps sitting between a horsepower figure and an ampere figure. Horsepower describes what leaves the shaft, about 746 watts of it per horsepower. What enters at the terminals is larger by the efficiency. The line current is larger again by the power factor. Our bank walks that chain: a 15 horsepower motor at 88 percent efficiency puts 11,190 watts on the shaft and draws 12,716 watts at its terminals, because the input is the output divided by the efficiency. Multiply by 0.88 and you get 9,847 watts, which would have the motor delivering more than it takes in.

Carry that same motor onto a 480 volt three-phase supply at 0.85 power factor and the line current is 18 amperes. The options include 15.3 from a missing power factor and 31.2 from a missing square root of three. Both come from the denominator, which is where this whole family of items is decided.

Reading which unit the stem wants

Five signals, and each one settles the unit before you calculate anything.

That last row carries the most weight and it's the one candidates argue with. Reading a horsepower figure as an invitation to calculate is how a four minute item becomes a ten minute item that ends on a wrong answer. The worked arithmetic for all of these forms lives in the power calculations question set, which is the page to read when you want each one derived.

What it costs, and what to drill

On the job the cost is a conductor running warmer than the design expected. A load worked in watts on equipment with a power factor below one understates the current the conductor carries, and the conductor is sized on current. That gap is the exact error the code removed by writing the calculation rules in volt-amperes in the first place.

On the paper the cost is time. The journeyman calculations portion allows 110 minutes for 26 items, 24 of them scored and two unscored pretest items you have no way of picking out. TDLR recorded 6,328 attempts on that portion in fiscal 2025 and 1,301 passes, a rate of 20.6 percent. The knowledge portion ran at 24.5 percent on 5,731 attempts, and both have to be passed.

Four things before test day. Write the three-phase denominator out in full on every current problem. Say the direction of the power factor step out loud before you apply it. Work one power triangle a day until subtracting the two known sides stops feeling reasonable. And open your own copy of the 2026 edition at NEC 120.13 and 120.41, the edition Texas enforces from 1 September 2026 under 16 TAC 73.100, so that the move out of Article 220 is something you've already seen with your own eyes.

Questions people ask

What is the difference between watts and volt-amperes?

Volt-amperes is what the conductor carries and watts is what gets converted into useful work. On a resistive load such as a heater the two are one figure, because the current and the voltage stay in step. On a motor they separate, and the ratio between them is the power factor. The code writes its load calculation rules in volt-amperes so that the answer holds whatever the connected equipment turns out to be, since a conductor heats according to current and current follows volt-amperes. An item that hands you a power factor is asking you to move between the two.

Where does the square root of three go in a three-phase calculation?

It lives on the boundary between line quantities and phase quantities, so it appears whenever you cross that boundary. Apparent power on a three-phase system is the line to line voltage times the line current times the square root of three. Written from the phase side the same quantity is three times the phase voltage times the phase current, with no square root in it. Going from power back to current, both the square root of three and the power factor sit in the denominator, and dropping either one inflates the answer.

Why does the NEC calculate in volt-amperes?

Because it refuses to assume anything about the load. A conductor heats according to the current it carries, and that current follows volt-amperes whatever the phase relationship inside the equipment happens to be. Writing the rules in watts would make every answer depend on a power factor nobody knows at the design stage. In the 2026 edition the dwelling unit minimum unit load for branch circuit calculations sits at NEC 120.13 and the feeder and service counterpart at NEC 120.41, material that lived in Article 220 through the 2023 edition.

How do I find the current of a three-phase motor?

Go to the full load current tables in Article 430. NEC 430.6 settles which motor current applies to which calculation, and conductor sizing comes off those tables. Working it from horsepower puts three assumptions in a row between you and the answer, since horsepower describes shaft output at about 746 watts each, terminal input is larger by the efficiency, and line current is larger again by the power factor. Exam items built on that chain offer an option for each step done backward, and the table exists so you skip the chain entirely.

What is reactive power and how do you calculate it?

It's the third side of a right triangle whose hypotenuse is the apparent power and whose base is the true power. A feeder carrying 30 kilovolt-amperes at 0.8 power factor carries 24 kilowatts of true power, and the reactive power is the square root of 30 squared minus 24 squared, which is 18 kilovars. The three quantities combine as sides of a triangle, so subtracting 24 from 30 to get 6 treats the triangle as a straight line. That 6 is a standing option in the bank.

Did the load calculation sections move in the 2026 code?

Yes. The dwelling unit minimum unit load for branch circuit calculations is at NEC 120.13 in the 2026 edition, and the feeder and service counterpart is at NEC 120.41. Both were in Article 220 through the 2023 edition. Texas adopted the 2026 edition at 16 TAC 73.100 with effect from 1 September 2026, so that's the book the exam is written against and the book you carry in. A memorized Article 220 section number will send you to the wrong part of the index under time pressure.