Rounding up to the next standard overcurrent device size, and the three conditions Texas exam items are built on

What the exam asks about it

Nineteen of the five hundred questions in our bank turn on where a calculated figure is allowed to land, and they're spread across four areas: conductor sizing, conduit, motor circuits and load calculations. That spread is the point. Rounding is a subject nobody teaches on its own, so candidates meet the rules one at a time and never line them up beside each other.

The form those items take is almost always the same. A stem gives you a number that sits deliberately between two standard ratings, along with one fact about the circuit that decides which direction the number may move. The fact is easy to read past. It might be a receptacle, or a device rating above 800 amperes, or a conductor small enough to meet a separate cap.

The Texas papers scatter these across every calculation subject line, so no single heading on the PSI content outline counts them. They show up wherever a calculation finishes, which on the journeyman calculations portion is most of the paper.

Which numbers move and which stand

Line the rules up beside each other and the pattern is easier to hold than any single one of them.

Two of those rows move in opposite directions from the same starting figure, which is how an item can hand you one number and two answers that both look defensible. Deciding which rule owns the number is the whole of the question. The load figures that feed the last row come out of the arithmetic on our page about watts, volt-amperes and power factor.

The three conditions on 240.4(B)

240.4(B) is a permission with three conditions attached, and an item is usually testing one of the three.

1. The device rating sits at 800 amperes or below.
2. The conductor ampacity lands between two standard ratings in 240.6(A), leaving a gap to climb.
3. The circuit is a feeder, or a branch circuit of any kind except one supplying more than one receptacle for cord and plug connected portable loads.

A worked example sits in the bank. A feeder conductor's corrected and adjusted ampacity is 244 amperes, the device will stay at or below 800, and the feeder supplies no receptacles for portable loads. The standard ratings run through 225 and then 250, so the answer is a 250 ampere device. Take the permission away and the answer becomes 225, which is the largest standard rating at or below the ampacity.

Both figures are correct answers to different circuits. That is why the stem spends a whole sentence describing what the circuit supplies, and why reading the stem twice costs less than recalculating once. Where the permission is unavailable and the ampacity comes to 88 amperes, the ratings step 70, 80, 90 and the answer is the 80.

Above 800 amperes the direction reverses

Above 800 amperes the rule turns around. 240.4(C) requires the conductor ampacity to equal or exceed the rating of the device, so there's no climb into a device the conductors can't carry.

The bank has a set of conductors in parallel with a combined corrected and adjusted ampacity of 1,150 amperes. Standard ratings above 800 step 1,000 and then 1,200. Since 1,150 falls short of 1,200, the larger device is out and the answer is the 1,000 ampere device. Below 800 amperes those same two numbers would have gone the other way, which is why the threshold is a fact to check on every large device.

The threshold is checked against the device rating. An item can put the conductor ampacity close to the line and leave you to work out which side of it the answer sits on, and the options will carry one figure from each side.

The cap that outranks the permission

240.4(D) caps the overcurrent protection on the smallest conductors, it applies after correction and adjustment, and it only ever cuts. A derated result above the cap leaves the cap exactly where it was.

Order of operations settles every item in this family. Correct for ambient. Adjust for conductor count. Compare the result with the lowest rated termination in the path under 110.14(C), because a 75 degree lug pulls the usable ampacity down to the 75 degree column figure. Choose the device against whatever survived, applying 240.4(B) where its conditions hold. Apply the small conductor cap last.

The bank has a 14 AWG copper conductor corrected and adjusted to 19 amperes, with a candidate selecting a 20 ampere device on the strength of the rounding permission. The cap for that conductor sits below 20, so the cap governs. 240.4(B) is a permission to climb one step, and it can't climb past a limit another subdivision has already set. Look the cap figures up in your own book, because they're the sort of number that gets misremembered.

A separate floor sits underneath all of it. 210.19(A) puts the conductor's own ampacity at 125 percent of the continuous load, so a conductor at 46 amperes fails a 38 ampere continuous load even though a 50 ampere device would be permitted on it. The ampere figures that feed every one of these steps come out of the arithmetic on our page about Ohm's law and the exam.

Motor circuits round their own way

Motor branch circuits round in the opposite direction and with fewer strings attached, which is why an item can put a motor in the stem purely to change the answer.

240.4(G) is the hinge. 240.4 states the general requirement that conductors be protected at their ampacity, and then (G) carves out specific applications and sends them elsewhere. Motor circuit conductors go to Article 430. So a motor branch circuit with conductors sized at 100 amperes behind a 200 ampere short circuit and ground fault device is compliant, because that 200 ampere device is doing a different job: it handles faults and lets starting current through, while the overload device protects those conductors against sustained overcurrent.

Inside Article 430 two permissions run in order. 430.52(C)(1) lets you take the calculated figure up to the next standard rating. Where the motor still refuses to start, 430.52(C)(3) allows a further increase with a ceiling that depends on whether the device is a nontime delay fuse, a dual element fuse, an inverse time breaker or an instantaneous trip breaker. Read the ceiling for your own device in your own book.

Feeders turn around again. 430.62 builds a ceiling from the largest branch circuit device in the group plus the full load currents of the other motors, and a ceiling is a maximum, so you take the standard rating at or below it. Three motors at 44, 28 and 16 amperes with a 70 ampere device on the largest gives 114 amperes, and the largest motor is counted once through its device.

The conduit count and its 0.8 note

Conduit fill carries its own rounding note, and it behaves differently from everything above because a fraction of a conductor can't be pulled.

Divide the permitted conductor area by the area of one conductor and the ordinary result is truncated, so a division landing at 8.83 gives eight conductors. The note in Chapter 9 changes that where every conductor in the pull is the same size, meaning the same overall area with the insulation counted, and where the decimal reaches 0.8. Both conditions hold at 8.83, so the count goes up to nine. Lose either one, by mixing sizes or by landing on 0.79, and the answer stays at eight.

Same size is the condition candidates drop. A pull of one AWG size in two insulation types looks uniform and fails the test, because the overall areas differ.

The note also does nothing about a remainder. Six conductors of 0.0973 square inches in a raceway permitting 0.605 square inches leave 0.0212 square inches free, and a seventh conductor needs 0.0973 of its own. No rounding permission anywhere creates space, and the bank offers that leftover area as an option for anybody who read it as room.

What it costs, and what to drill

On the job the visible cost is a device pulled at inspection and a panel schedule that has to be redone. A 45 ampere breaker on a receptacle branch circuit is a finding an inspector can make from the schedule and the conductor size without opening anything, and the correction is a breaker, a relabel and a second visit.

The more expensive version runs the other way. A device chosen above what the terminations support passes a quick look and leaves the conductor protected at a value the lugs can't hold. That failure mode is where our page on interrupting rating and available fault current picks up, since the device that clears a fault has to be matched to the system it sits on as well as to the conductor behind it.

Three things before test day. Write the order of operations on a card and work every derating problem in that order until the order is automatic. Tab 240.4 and 240.6 in your own copy, because the conditions on 240.4(B) are the part people quote from memory and get wrong. And work a dozen of these under a clock, since the worked arithmetic for each one lives in the practice exam.

Texas runs the exam open book against the 2026 edition, adopted at 16 TAC 73.100 and enforced from 1 September 2026. Several articles moved for that edition, so read the section in front of you and treat a memorized number as a starting point.

Questions people ask

Can you round up to the next standard overcurrent device size?

Only where all three conditions in NEC 240.4(B) hold. The device rating has to sit at 800 amperes or below, the conductor ampacity has to land between two standard ratings so there's a gap to climb, and the circuit has to be something other than a branch circuit supplying more than one receptacle for cord and plug connected portable loads. Where any one of those fails, the device lands at the largest standard rating at or below the conductor ampacity. A feeder at 244 amperes takes a 250 ampere device, and the same figure on a portable load receptacle circuit takes 225.

What are the standard overcurrent device ratings?

They are listed at NEC 240.6(A), and you read them out of your own book, since the exam is open book and a memorized list is the thing that goes stale on you. What an exam item actually needs is the pair of ratings sitting either side of the figure you calculated, so the habit worth building is finding your own number on that list and reading one step in each direction before you choose. Our bank turns on those neighboring ratings repeatedly: a feeder at 244 amperes, a usable ampacity of 88, a parallel set at 1,150. Each one is decided by the two ratings around it.

Does a conductor ampacity ever get rounded?

No. There's no rule anywhere that rounds an ampacity. A corrected and adjusted figure of 78.4 amperes is the conductor as installed, and the device gets chosen against that figure. Doing it correctly reaches almost the same place, since the standard rating at or below 78.4 is 70 and 240.4(B) permits the next rating above where its conditions hold, which is 80. The difference matters because rounding the ampacity first would carry an 80 ampere device through even on a circuit that 240.4(B) shuts out.

Why do motor circuits round differently?

Because 240.4(G) sends motor circuit conductors to Article 430 for their protection. The general requirement in 240.4 is that conductors be protected at their ampacity, and (G) carves out specific applications. Inside Article 430 the branch circuit short circuit and ground fault device rounds up to the next standard rating under 430.52(C)(1) with no conditions on it, and may go higher again under 430.52(C)(3) within a ceiling that depends on the device type. That device handles faults and passes starting current, while a separate overload device protects the conductors.

When does the small conductor rule override the rounding permission?

Whenever the cap at 240.4(D) sits below the rating the permission would have given you. The cap applies after correction and adjustment and it only ever cuts, so the order of the steps decides the answer. A 14 AWG copper conductor corrected and adjusted to 19 amperes looks like a candidate for a 20 ampere device under 240.4(B), and the cap for that conductor sits below 20, so the cap governs. A permission to climb one standard rating can't climb past a limit another subdivision already set.

How do you round a conduit fill conductor count?

Truncate to the whole number, with one exception. Divide the permitted conductor area by the area of one conductor and drop the decimal, because a fraction of a conductor can't be pulled. The note in Chapter 9 permits the count to go up where every conductor in the pull is the same size, meaning the same overall area with insulation counted, and where the decimal reaches 0.8. A division at 8.83 on a uniform pull gives nine. Mixing two insulation types in one AWG size fails the uniformity test.