Type any two of volts, amperes, ohms and watts and read the other two, each with the formula that produced it. Three-phase line values and a power factor field are in the same panel, and the code sections named further down are the 2026 edition.
The first row is the wheel proper, where watts and volt-amperes are the same number. The other two apply the power factor, and the three-phase row puts the square root of three in front of the product.
On three-phase, volts and amperes are line values and the ohms box steps aside, because a three-phase load is described by an impedance per phase rather than by one resistance across the line.
Series shares the current and divides the voltage. Parallel shares the voltage and divides the current. Name which one you have before you touch a formula, because that sentence is what the item is testing.
Leave it empty for the total resistance on its own. Fill it in and each element gets its current and its share of the voltage.
The power wheel on the calculators page takes two of the four quantities and stops there, which is the right shape for drilling. This page adds the three-phase form, the power factor field, the series and parallel helper, and the wheel drawn out so you can see which quadrant each formula came from.
Neither one is a code rule. Read the Ohm's law study page for what the exam does with this and where the volt-ampere and the watt stop being the same number.
Ohm's law is not in the code book. There is no section to cite for voltage equals current times resistance, and looking for one costs you a minute you do not have. The code assumes the arithmetic and then tells you which numbers to feed it.
Which voltage, first. NEC 120.5(A) lists the nominal system voltages to calculate branch-circuit and feeder loads with, and 120, 120/240, 208Y/120, 240, 480Y/277 and 480 are all on that list. Article 100 defines a nominal voltage as a value assigned to a circuit to name its voltage class, and its informational note says the voltage a circuit runs at can sit either side of the name.
Outside a load calculation, 110.4 is the one that governs: the voltage considered is the voltage the circuit operates at. So a stem that hands you 118 volts measured at a panel wants 118, and a stem that says 480Y/277 wants you to know which pair of conductors it means, which is what the Article 100 definition of voltage to ground is there for.
A note on the number itself, because a reader with an older book will go looking in the wrong place. The 2026 edition moved the load calculation article out of Chapter 2 and into Chapter 1 as Article 120, so those two rules are printed at 120.5(A) and 120.5(B). If your book still numbers that article in the 220s, look there for the same two, and see what changes with the 2026 code.
Watts against volt-amperes is the distinction the whole page turns on. On direct current, NEC 120.5(D) treats volt-amperes as watts and kilovolt-amperes as kilowatts for loads calculated under Article 120, so the two words describe one number. On alternating current they part company the moment the load stops being resistive, and the power factor field is where you say so.
The square root of three goes on the boundary between line values and phase values, and the code book uses it the same way you do. Informative Annex D, Example D3(a), turns 142,000 volt-amperes on a 480 volt three-phase feeder into 171 amperes by dividing by 480 times the square root of three, and takes a motor at 7.6 amperes on the same system up to 6,310 volt-amperes by multiplying by it. The single-phase example a few pages earlier divides 18,075 volt-amperes by 240 volts and gets 75 amperes, with no factor at all.
Annex D also states its own assumption about power factor, which is that all the loads in its examples share one. Take the same discipline into the exam. An item that hands you a power factor wants the conversion and is telling you so by handing it to you, and an item that does not is not inviting you to invent a value.
Order of operations, and it is short. Decide whether the load is resistive, pick the two quantities you were given, read the formula off the wheel, then check the answer by a second route. Squaring the current and multiplying by the resistance should agree with volts times amperes, and when the two disagree you have found a slipped decimal before you marked it.
What it is worth on the paper is modest and the indirect return is not. The testing vendor's content outline gives the journeyman knowledge portion 3 of 56 scored items under Definitions, Theory, and Plans, and the journeyman calculations portion 2 of 24 under Calculations and Theory. A master candidate sees 7 of 70 and 2 of 30. Every load calculation on the paper ends in an ampere figure, and this relationship is the last step of getting there.
Start on the first row of the picker, where the load is resistive. Put 120 volts across it and measure 10 amperes. Resistance is volts divided by amperes, so 120 over 10 is 12 ohms, and power is volts times amperes, so 120 times 10 is 1,200 watts.
Check it by the two forms people forget. Current squared times resistance is 100 times 12, which is 1,200. Voltage squared over resistance is 14,400 over 12, which is 1,200 again. Three routes, one answer, about ten seconds.
Now the three-phase row. A 480 volt three-phase load drawing 50 amperes per line at a power factor of 0.85 has an apparent power of the square root of three times 480 times 50, which is 41,569 volt-amperes, or 41.6 kVA. Multiply by the 0.85 and the true power is 35,334 watts, which is 35.3 kilowatts.
Leave the power factor at 1 and the same two numbers give 41.6 kW, which is the error to watch for. Use 3 in place of the square root of three and you are high by about 1.73. Leave the factor out altogether and you are low by the same amount.
For the second panel, put branches of 12, 6 and 4 ohms in parallel. The reciprocals are a twelfth, a sixth and a quarter, which come to a half, so the total is 2 ohms. Add 24 volts as the source and the branches carry 2, 4 and 6 amperes for a total of 12 amperes. Rewire the same three in series and the total is 22 ohms, which is eleven times the parallel figure and worth seeing once.
Four quantities, two relationships, twelve faces. The wheel is not twelve things to memorize. It is voltage equals current times resistance and power equals voltage times current, each one substituted into the other until every pair of knowns has a route to every unknown.
Read from the hub outward. The letter in the middle of a quadrant is what that quadrant solves for, and the three faces around it are the three ways to get there.
| Solving for | From two knowns | Formula |
|---|---|---|
| Volts | ||
| V | amperes and ohms | V = I × R |
| V | watts and amperes | V = P ÷ I |
| V | watts and ohms | V = √(P × R) |
| Amperes | ||
| I | volts and ohms | I = V ÷ R |
| I | watts and volts | I = P ÷ V |
| I | watts and ohms | I = √(P ÷ R) |
| Ohms | ||
| R | volts and amperes | R = V ÷ I |
| R | volts and watts | R = V² ÷ P |
| R | watts and amperes | R = P ÷ I² |
| Watts | ||
| P | volts and amperes | P = V × I |
| P | amperes and ohms | P = I² × R |
| P | volts and ohms | P = V² ÷ R |
| System | Apparent power, VA | True power, W | Amperes from power |
|---|---|---|---|
| Direct current | V × I | V × I | P ÷ V |
| Single-phase ac | V × I | V × I × pf | P ÷ (V × pf) |
| Three-phase ac | √3 × V × I | √3 × V × I × pf | P ÷ (√3 × V × pf) |
| Arrangement | Total resistance | Shared | Divided |
|---|---|---|---|
| Series, any number | R1 + R2 + R3 … | the current | the voltage |
| Parallel, any number | 1 ÷ (1/R1 + 1/R2 …) | the voltage | the current |
| Parallel, exactly two | (R1 × R2) ÷ (R1 + R2) | the voltage | the current |
| Parallel, all equal | R ÷ number of branches | the voltage | the current |
Voltage equals current times resistance, which is written V = I x R. Rearrange it and current is voltage divided by resistance, and resistance is voltage divided by current. Bring power in through P = V x I and those two relationships between them produce the twelve formulas drawn in the wheel further down this page.
Multiply them. On direct current, or on an alternating current load that is purely resistive, 120 volts at 10 amperes is 1,200 watts. Once the load is a motor or anything else with a winding in it, volts times amperes gives volt-amperes instead, and watts are that figure times the power factor.
Multiply the square root of three by the line voltage, the line current and the power factor. At 480 volts, 50 amperes and a power factor of 0.85 that comes to 35,334 watts, or 35.3 kilowatts. Leave the power factor out of the same product and you have apparent power instead, 41,569 volt-amperes, which is 41.6 kVA.
Volt-amperes are volts times amperes with nothing else applied to the product. Watts are the part of that the load turns into work, and the ratio between the two is the power factor. They are the same number on a heater and they are not on a motor, which is one reason the code book writes load calculations in volt-amperes.
No, and there is no section to cite for it. The code assumes the arithmetic and then says which numbers go into it: 120.5(A) lists the nominal voltages to calculate branch-circuit and feeder loads with, 120.5(B) permits the answer to be rounded to the nearest whole ampere, and 110.4 says the voltage considered is the one the circuit operates at. In the 2026 edition those live in Article 120, which is where the load calculation article went.
Add the reciprocals of the branches and invert the total. Branches of 12, 6 and 4 ohms give a twelfth plus a sixth plus a quarter, which is a half, so the total is 2 ohms. Two shortcuts save the fractions: exactly two branches are their product over their sum, and branches that are all equal are one branch divided by how many there are.
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