Ohm's law is five scored items on the Texas journeyman exam and the arithmetic inside a hundred more

What the exam asks about it

Thirty-nine of the five hundred questions in our bank sit in the theory area, and Ohm's law is the machinery inside most of them. The scored count on the Texas papers is smaller than that, and it's worth knowing exactly. PSI publishes a content outline for each portion in its candidate information bulletin for the TDLR electricians program, updated 9 July 2026. On the journeyman knowledge portion, the subject line called Definitions, Theory, and Plans carries 3 of the 56 scored items. On the journeyman calculations portion, Calculations and Theory carries 2 of the 24. That is five scored items out of the eighty a journeyman sits across both papers. A master candidate gets 7 of 70 and 2 of 30.

Reading those five items as the size of the topic is the mistake that costs candidates the paper. Every load calculation on the calculations portion finishes in an ampere figure, and the last step of getting there is this relationship. So the five items measure how often theory is the subject of a question, and they say nothing at all about how often it's the work inside one.

One relationship, four rearrangements

Voltage equals current times resistance. Rearrange that one sentence and you have the current, or you have the resistance, and that covers most of what a theory item wants from you.

• Current from voltage and resistance: divide the voltage by the resistance.
• Current from power and voltage: divide the power by the voltage.
• Power from current and resistance: square the current, then multiply by the resistance.
• Power from voltage and resistance: square the voltage, then divide by the resistance.

Work it once with numbers you can check. Put 240 volts across 12 ohms and the current is 20 amperes. Power is 240 times 20, which is 4,800 watts. Square the current and multiply by the resistance: 400 times 12 is 4,800. Square the voltage and divide by the resistance: 57,600 over 12 is 4,800. Three forms agreeing is your check, and it costs about ten seconds.

The worked arithmetic for every one of those rearrangements lives in the ohms law question set, which is the page to read when you want the derivations set out step by step. This page is about what the Texas exam does with them.

The wrong answers people actually pick

Our bank carries a line on every wrong option saying what a person who picked it did. Read across the theory questions and a handful of errors produce almost all of them.

The third row repays attention, because it's two quantities with two different units coming out of the same two numbers. Power lost in a conductor is the current squared times the resistance, so a feeder with a quarter of an ohm of conductor resistance carrying 45 amperes turns 506 watts into heat. Multiply the same two figures without squaring and you get 11.25, which is the voltage dropped along that run. Both are real quantities and only one of them answers the question.

Only the current is squared. That is why doubling the load on a run quadruples the heat in it while doubling the voltage drop, and why long runs are the ones that force a size increase.

Where watts and volt-amperes split

Volts times amperes gives volt-amperes. It gives watts where the load is resistive, meaning a heater or an incandescent lamp, anything that holds the current and the voltage in step with each other. Keep that qualifier attached and the power wheel stays useful.

One question in the bank asks under what condition the two figures are the same number, and the answer is a power factor of one. Every wrong option on it comes from treating the two words as interchangeable. A second question hands you 4,200 volt-amperes at a power factor of 0.80 and offers 840 as an option, which is what you get by multiplying by one minus the power factor. That figure is a power quantity of no kind at all, because apparent power, true power and reactive power sit on a right triangle, and the leftover side comes from subtracting squares.

An item that hands you a power factor is telling you it wants the conversion. An item that leaves one out is asking you to work at unity. Our page on watts, volt-amperes and power factor takes that split apart properly, including both three-phase forms.

Where the square root of three lands

The factor lives on the boundary between line quantities and phase quantities. Cross that boundary and it appears. Stay on one side of it and it stays out of your arithmetic.

• Wye: the line to line voltage is the square root of three times the line to neutral voltage, and the line current equals the winding current.
• Delta: the line current is the square root of three times the winding current, and the line voltage equals the winding voltage.
• Apparent power on either connection: the square root of three times line voltage times line current.

A three-phase load drawing 30 amperes at 480 volts is 24.9 kilovolt-amperes, because 480 times 30 is 14,400 and 14,400 times 1.732 is about 24,900. The bank marks the option that omits the factor as the single most common theory error, and it sits right beside an option that multiplied by three where the square root belonged. Those two wrong answers are a factor of three apart from each other, which is how you spot a stem written to catch this.

Keep the square root of two clear of it. That one relates the peak of a sine wave to its RMS value, so a 480 volt circuit puts about 679 volts on the insulation twice every cycle. A question offering you 339 volts is offering the same factor used as a division.

What getting it wrong costs on the job

Two consequences show up in the bank as job conditions, with the arithmetic sitting underneath.

A resistance heater rated 4,500 watts at 240 volts, connected to a 208 volt supply, produces about 3,380 watts. The element resistance holds steady, so the power follows the square of the applied voltage, and 208 over 240 squared is 0.751. Scale the power by the voltage ratio alone and you get 3,900 watts, which overstates the output by about 520 watts. A customer who was promised a heat output gets three quarters of it, and the repair is a different element.

The same squaring runs through motor starting. A motor at the end of a run with 8 percent voltage drop starts on 92 percent of nominal, and torque follows the square of the voltage, so it develops about 85 percent of its rated starting torque. A motor that fails to come up to speed keeps drawing starting current, which deepens the drop that caused the trouble, so the condition feeds itself until the overload device opens. Our page on series and parallel circuits covers why the conductor resistance sits in series with the load in the first place.

The clock, and what a retake costs

The journeyman calculations portion allows 110 minutes for 26 items, which is 24 scored plus two unscored pretest items you have no way of picking out. That is a little over four minutes each, and the arithmetic is the part you should have stopped thinking about.

TDLR's fiscal 2025 figures put the pressure in context. The journeyman calculations portion tested 6,328 candidates and passed 1,301, a rate of 20.6 percent. The journeyman NEC portion tested 5,731 and passed 1,402, or 24.5 percent. Both portions have to be passed, and a candidate under the two-part format sits two exams.

A retake costs a seat and a wait. The examination fee goes to the department's testing vendor at the amount current when you book, which is why 16 TAC 73.80 carries no examination fee at all, and your thirty dollar journeyman application fee is non-refundable under 16 TAC 73.80(e). The expensive part is the calendar. You may sit at 7,000 hours under 16 TAC 73.21(b) and you hold the license at 8,000 under Tex. Occ. Code 1305.155, so a failed portion eats a window you were probably counting on.

What to do about it before test day

Four drills, and all of them are about speed.

1. Work the four rearrangements until you can name which one you need from the units in the stem alone. Time yourself at thirty seconds an item.
2. Build the habit of checking one answer two ways. Power from volts and amperes, then power from amperes squared times ohms, and the agreement is your proof.
3. Write the denominator out in full on every three-phase current problem before you divide. Two multipliers sit in it, the square root of three and the power factor, and dropping either one inflates the answer.
4. Learn the sanity direction for each operation. A parallel combination lands below the smallest branch. A series total lands above the largest element. A multiplication of two known values lands above both of them.

Texas runs an open book exam, so bring your own current copy of the code. The 2026 edition is what the state enforces from 1 September 2026 under 16 TAC 73.100, and none of the arithmetic on this page moved with it. What moved is where the load calculation rules live, which is worth a look before you sit down.

Questions people ask

How many Ohm's law questions are on the Texas journeyman exam?

Five scored items across both papers carry a theory label. The PSI content outline for the TDLR electricians program gives Definitions, Theory, and Plans 3 of the 56 scored items on the journeyman knowledge portion, and Calculations and Theory 2 of the 24 on the calculations portion. A master candidate gets 7 of 70 and 2 of 30. Those counts describe items whose subject is theory. The relationship itself does the work inside a much larger share of the paper, because every load calculation ends in an ampere figure and this is the step that produces it.

Is the Texas electrician exam open book?

Yes, and you bring your own copy. The edition Texas enforces is the one adopted at 16 TAC 73.100, which is the 2026 National Electrical Code from 1 September 2026. Ohm's law itself sits in no section of the book, so there's nothing to look up and nothing to tab for it. That is why speed matters here more than on a code item, since you have no index to buy the time back with. Candidates who tab well and calculate slowly still run out of clock on the calculations portion, which allows 110 minutes for 26 items.

What is the most common wrong answer on theory questions?

An inverted division, with a missing square root of three close behind it. Our bank carries a line on each wrong option naming the error behind it, and reading across the theory area those two signatures produce most of them. The inverted division shows up whenever a stem hands you two values and asks for a third, since any answer smaller than both of the numbers you were given has a division where a multiplication belonged. The missing square root of three shows up on three-phase items and leaves the answer low by about 1.73.

Do I have to calculate motor current from horsepower?

Go to the full load current table. NEC 430.6 settles which motor current applies to which calculation, and motor conductor sizing comes off the tables in Article 430. Horsepower describes what leaves the shaft, about 746 watts of it per horsepower, and what enters at the terminals is larger by the efficiency while the line current is larger again by the power factor. Our bank walks a 15 horsepower motor at 88 percent efficiency to 12,716 watts at its terminals, and three of the four options come from doing one step of that chain backward.

How fast do I have to be on the calculations portion?

A little over four minutes an item. The journeyman calculations portion allows 110 minutes for 26 items, 24 of which are scored and two of which are unscored pretest items you have no way of identifying. TDLR recorded 6,328 attempts on that portion in fiscal 2025 with 1,301 passes, which is 20.6 percent. Most of those items are code lookups with arithmetic on the end, so the arithmetic has to be automatic if the lookup is going to get the time it needs.

Does the square root of three ever appear in single-phase work?

It stays out of single-phase arithmetic entirely. The factor comes from the phase relationship in a three-phase system, so it appears when you cross between line quantities and phase quantities and it appears nowhere else. Two other numbers get confused with it. The square root of two relates the peak of a sine wave to its RMS value, which is an insulation question. The figure two belongs to a voltage drop round trip, where the tape measure reads one way and the formula wants both conductors, so a one-way run of 180 feet goes in as 360.