You bring the factors. These bring the method.

Four calculators. Every value that lives in a code table is typed in by you, off the page you read it from, and the label tells you which table that is. What runs here is the arithmetic and the order it happens in, which is the half that actually gets failed.

Why it asks instead of knowing

A calculator with the tables built into it is a copy of the tables with a form on the front. That is the one behavior worth avoiding, and avoiding it costs you nothing here: you have the book open in the exam room anyway, and looking a factor up is the skill being tested.

So the fields marked you read this one want a number out of your own book, and each one names the table it is on. Everything after that is arithmetic, plus the rules about which figure caps which, which is where the points really go.

ampacityConductor deratingCorrection for ambient temperature, adjustment for conductor count, and the termination limit that outranks both.

Two factors multiply against the table ampacity, in either order, because multiplication does not care. Then a third rule caps the answer, and that one is not a multiplication at all. Candidates lose this question by stopping at the multiplication.

Base ampacity in amperes, read in the column for the insulation rating of the conductor itself, from Table 310.16 in your code bookyou read this one A

Start in the ninety degree column if the conductor is a ninety degree conductor. Correction and adjustment are applied from there, not from the terminal rating.

Ambient temperature correction factor, from Table 310.15(B)(1)(1) in your code bookyou read this one ×

Table 310.16 is a thirty degree table, so it pairs with the thirty degree correction table named above. The forty degree correction table pairs with the free-air ampacity tables instead. Reading the right factor off the wrong table is a quiet way to be wrong, because the arithmetic still works.

Adjustment factor for the number of current-carrying conductors, from Table 310.15(C)(1) in your code bookyou read this one ×

Count current-carrying conductors, not conductors. A grounded conductor that carries only the unbalanced current of the other conductors is not counted, and an equipment grounding conductor is not counted at all. Enter one if adjustment does not apply.

Ampacity of the same conductor read in the termination temperature column, from Table 310.16 in your code bookyou read this oneoptional A

Optional, and it is the step people skip. Read the same size again in the column that matches the lowest rated terminal in the circuit. NEC 110.14(C) caps the final answer at that figure.

The load this conductor has to carry, in amperesoptional A

Optional. Supply it and the working shows whether the corrected ampacity covers the load.

The method

Start at the ampacity for the insulation rating of the conductor itself, not at the terminal rating.
Multiply by the ambient correction factor.
Multiply by the conductor-count adjustment factor.
That product is the corrected and adjusted ampacity.
Now read the same conductor again in the termination temperature column. Under NEC 110.14(C) the lower of the two figures is the one you are allowed to use.

What still limits this answerA derated ampacity is not the final answer on its own. NEC 110.14(C) holds the circuit to the lowest temperature rating in the path, and that is usually a lug or a breaker terminal rather than the wire. Overcurrent protection is a separate step in Article 240, with its own rounding rule, and this does not do it.

article 314Box fillConductors, yokes, clamps, support fittings, grounds and terminal blocks, added the way NEC 314.16(B) adds them.

Six categories get added together. Two of them are counted in a way that surprises people every single time: a yoke is a double allowance, and the first four equipment grounding conductors are one allowance between them.

Free volume of the box in cubic inches, marked on the box itself or from Table 314.16(A) in your code bookyou read this one in³

Add any plaster ring, extension ring or domed cover that is marked with its own volume. A securely installed barrier takes volume away rather than adding it.

Conductors of the first size that count toward fill count

A conductor that comes in from outside and terminates or splices in the box counts once. One that passes through unbroken counts once. A loop long enough to be twice the free-conductor length counts twice. A pigtail made up entirely inside the box counts not at all.

Volume allowance in cubic inches for that conductor size, from the volume allowance table printed with 314.16(B) in your code bookyou read this one in³

Conductors of a second size that count toward filloptional count

Volume allowance in cubic inches for the second size, from the volume allowance table printed with 314.16(B) in your code bookyou read this oneoptional in³

Yokes or straps carrying a device or a piece of equipmentoptional count

Each yoke is a DOUBLE allowance, sized by the largest conductor connected to a device on that yoke. A device wider than a single device box gets the double allowance for every gang it occupies, so count the gangs here rather than the devices.

Volume allowance for the largest conductor connected to a device on a yoke, from the volume allowance table printed with 314.16(B) in your code bookyou read this oneoptional in³

One or more internal cable clamps are present in the box

One allowance in total no matter how many clamps, and it is sized by the largest conductor in the box rather than by the clamp. A connector whose clamping mechanism sits outside the box gets nothing.

Types of support fitting in the box, a luminaire stud and a hickey being two typesoptional types

Per type, not per fitting. Two studs are one allowance. A stud and a hickey are two.

Volume allowance for the largest conductor in the box, from the volume allowance table printed with 314.16(B) in your code bookyou read this oneoptional in³

This is the figure the clamp allowance and each support-fitting allowance are both sized from.

Equipment grounding conductors entering the boxoptional count

Up to four of them share a single allowance. Every one after that adds a quarter of an allowance. Counting each ground as a full allowance overstates the fill badly and is the second most common error here.

Volume allowance for the largest equipment grounding conductor, from the volume allowance table printed with 314.16(B) in your code bookyou read this oneoptional in³

Terminal block assemblies in the boxoptional count

Volume allowance for the largest conductor landed on a terminal block, from the volume allowance table printed with 314.16(B) in your code bookyou read this oneoptional in³

The method

Conductor fill: each counted conductor at the allowance for its own size.
Clamp fill: one allowance in total where any internal clamp is present, sized by the largest conductor in the box.
Support fitting fill: one allowance for each TYPE of fitting, sized by the largest conductor in the box.
Device fill: two allowances for every yoke or strap, sized by the largest conductor connected to a device on that yoke.
Equipment grounding conductor fill: one allowance for the first four, then a quarter allowance for each one after that.
Terminal block fill: one allowance for each assembly, sized by the largest conductor landed on it.
Add the six together and compare against the free volume of the box.

What still limits this answerThe result means nothing without the marked volume of the box itself, which is why that is the first field rather than a lookup. Locknuts, wire connectors and bushings get no allowance. This handles NEC 314.16 only: minimum box depth in NEC 314.24 and the pull and junction box sizing in NEC 314.28 are separate rules with separate arithmetic.

chapter 9Voltage dropOne-way length, load current, conductor resistance, and the multiplier that changes with the number of phases.

The arithmetic is small. What people get wrong is the multiplier and the length. Use the one-way length and let the multiplier account for the return path, or use the round-trip length and no multiplier, but never both.

Single-phase or three-phase

Single-phase carries a multiplier of two, because the current goes out and comes back. Three-phase carries the square root of three, about 1.732, because the return is shared between the other two conductors, phase shifted, rather than piling up in one of them.

Load current in amperes A

One-way circuit length in feet ft

Conductor resistance in ohms per thousand feet, from the conductor properties table in Chapter 9 of your code bookyou read this one Ω/kft

Read the row for your conductor size, in the right metal. Copper and aluminum are different columns and swapping them is worth a lot of volts.

Conductor reactance in ohms per thousand feet, from the alternating-current resistance and reactance table in Chapter 9 of your code bookyou read this oneoptional Ω/kft

Optional. Leave it blank for a resistance-only answer, which is what most exam questions want. Fill it in with a power factor and the working switches to effective impedance instead.

Load power factor as a decimal, from the nameplate or the job specificationoptional pf

Only used when a reactance is supplied. At unity power factor the reactance contributes nothing and the two methods agree.

Sets of conductors in paralleloptional sets

System voltage, so the drop can be shown as a percentageoptional V

The method

Pick the multiplier: two for single-phase, the square root of three for three-phase.
Turn ohms per thousand feet into ohms for this run by multiplying by the one-way length and dividing by a thousand.
Where a reactance and a power factor are both given, use effective impedance instead of plain resistance: resistance times power factor, plus reactance times the sine of the angle.
Divide by the number of parallel sets, because parallel conductors share the current.
Multiply the multiplier by the current by that impedance. The answer is in volts.
Divide by the system voltage for a percentage.

What still limits this answerThe familiar three percent and five percent figures come out of informational notes, and Article 90 is clear that explanatory material of that kind is not a requirement of the code. Job specifications often are binding, and they are frequently tighter. This calculates a drop. It does not tell you whether the drop is acceptable, and it does not size a conductor.

theoryOhm's law and the power wheelGive it any two of volts, amperes, ohms and watts. It returns the other two and names the two formulas it used.

Twelve formulas in a wheel are four relationships wearing different hats. Fill in any two boxes and leave the other two empty.

Voltsoptional V

Amperesoptional A

Ohmsoptional Ω

Wattsoptional W

The method

Volts equal amperes times ohms.
Watts equal volts times amperes.
Watts equal amperes squared times ohms.
Watts equal volts squared divided by ohms.
Every other face of the wheel is one of those four rearranged.

What still limits this answerThis is the direct-current form, and it holds on alternating current only for a purely resistive load. Put a motor on it and watts are no longer volts times amperes: the power factor comes in, and on three-phase the square root of three comes in as well. A heater obeys the wheel. A motor does not.

Full practice exam Timed article drill