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Power and voltage drop

20 original questions. The keyed answer, the article to open, and a short explanation are on the page. Confirm the section in your NEC 2026 book. These items are a sample. The rest of this topic stays in the practice app.

All free sections·Study notes·20-question practice test

1 · Power and voltage drop · NEC Knowledge
An informational note mentions a recommended voltage drop on a branch circuit. What is the status of that note, in plain language?
1. A. It is guidance in the book. It is not itself a mandatory rule.Keyed answer
2. B. It is a hard limit for every branch circuit, with no exceptions.
3. C. It applies only to services over 1000 V.
4. D. It replaces the conductor ampacity tables.
Informational notes explain or recommend. They are not the requirement. Voltage-drop notes for branch circuits sit with Article 210. Confirm the note in NEC 2026 before you treat a percentage as a pass-fail rule. The conductor still has to satisfy the ampacity rules.
Open NEC 2026 Article 210
2 · Power and voltage drop · NEC Knowledge
A transformer nameplate is marked in kVA, and the load you were given is in kW. Which statement matches those units?
1. A. kVA and kW are the same number on every load
2. B. kW is always larger than kVA
3. C. kVA already has the power factor divided out of it
4. D. kVA is apparent power. kW is real power. They match only when the power factor is 1.Keyed answer
Apparent power is the volts-times-amps figure a transformer is usually rated for. Real power is smaller when the power factor is below 1. Article 450 is the transformer article. Read the nameplate units before you divide. The power math itself is not a code table.
Open NEC 2026 Article 450
3 · Power and voltage drop · NEC Knowledge
The primary breaker was chosen from the secondary full-load current because that current was the larger number on the nameplate. The helper says the bigger current is the safe setting for every device on the transformer. Which side did he assign to the wrong device?
1. A. The primary current, which is prohibited from being used on the primary device.
2. B. Neither side. A transformer is protected only by the building's main breaker.
3. C. The secondary current does not set the primary device. Each side is protected from that side's own current. Bigger is not a reason to swap them.Keyed answer
4. D. The grounding electrode, which is sized from the secondary current.
Article 450 protects the primary and the secondary according to the side each device is on. Confirm the primary rules in NEC 2026. This question does not ask for a new current.
Open NEC 2026 Article 450
4 · Power and voltage drop · NEC Knowledge
Secondary conductors run from the transformer to a panel in another building with no overcurrent device at the transformer. The helper says the primary breaker is the secondary protection for that long run. The problem says the primary device does not satisfy the secondary permission. What is missing on the secondary run?
1. A. Secondary overcurrent protection for this run. The distant panel does not borrow the primary breaker when the problem says that permission fails.Keyed answer
2. B. Nothing. Primary protection is the required protection for every secondary conductor of any length.
3. C. A larger primary breaker, which is how secondary conductors are protected.
4. D. The voltage-drop note, which replaces secondary overcurrent protection.
Article 450 still requires the secondary protection this problem did not waive. Confirm where that protection has to be in NEC 2026. Do not treat the primary breaker as the secondary device.
Open NEC 2026 Article 450
5 · Power and voltage drop · NEC Knowledge
A remote disconnect for the transformer can be locked open, but nothing on the transformer says where that disconnect is. The helper says a lock is the whole remote permission and the installer will remember the room. What marking is missing?
1. A. A marking that prohibits locking, because remote disconnects must stay unlocked.
2. B. Nothing. Memory is the field marking Article 450 accepts.
3. C. A field marking on the transformer that identifies the remote disconnect. The lock alone does not tell the next person where the handle is.Keyed answer
4. D. The informational voltage-drop note, posted as the disconnect label.
Article 450 lets a transformer disconnect be remote only when that permission, including the field marking, is met. Confirm the remote rule in NEC 2026.
Open NEC 2026 Article 450
6 · Power and voltage drop · NEC Knowledge
An oil-insulated transformer was set in an ordinary electrical room and judged only by the dry-type rules. The helper says insulation is a catalog detail and every indoor transformer follows the dry-type article path. What insulation did he fail to account for?
1. A. The oil. An oil-insulated transformer is not located by the dry-type rules. Confirm the indoor rules for this insulation in NEC 2026.Keyed answer
2. B. Air only, because oil-insulated units are required to use the dry-type clearances and nothing else.
3. C. The secondary voltage, which removes the insulation from consideration.
4. D. No insulation. Oil-filled transformers are prohibited from any location rule.
Article 450 distinguishes an oil-insulated transformer from a dry-type. Confirm which indoor rule applies to this insulation in NEC 2026. Do not borrow the dry-type path.
Open NEC 2026 Article 450
7 · Power and voltage drop · NEC Knowledge
The load is stated in kilowatts and the helper orders a transformer with that same number of kilovolt-amperes. The power factor is not unity. He says kilowatts and kilovolt-amperes are two names for the transformer rating. What rating did he treat as real power?
1. A. The kilowatt figure, which is the required transformer nameplate unit in every case.
2. B. Nothing. Kilowatts are defined as kilovolt-amperes divided by the voltage.
3. C. The transformer rating. A transformer is selected in kilovolt-amperes. Kilowatts are real power and are not the transformer rating unless the power factor is unity.Keyed answer
4. D. The disconnect height, which is measured in kilowatts.
Article 450 uses the transformer kilovolt-ampere rating. Confirm the application in NEC 2026. This question does not ask you to convert the load. Do not treat kilowatts as the nameplate unit.
Open NEC 2026 Article 450
8 · Power and voltage drop · NEC Knowledge
The installation was rejected because the voltage drop exceeded the percentages in an informational note. Ampacity, overcurrent protection, and the transformer rules were otherwise satisfied. The helper says the note is a mandatory pass-or-fail limit. What is that note?
1. A. The mandatory maximum voltage drop, and any excess is an automatic violation.
2. B. Guidance, not a mandatory pass-or-fail rule. Missing the note does not by itself fail an installation that meets the mandatory rules.Keyed answer
3. C. A transformer impedance limit, which replaces the nameplate.
4. D. The secondary protection table, printed as a suggestion that is actually mandatory.
The voltage-drop informational note in Article 210 is not a mandatory pass-or-fail rule. Confirm the note's status in NEC 2026. Do not fail the job on the note alone.
Open NEC 2026 Article 450
9 · Power and voltage drop · NEC Knowledge
A three-phase transformer was sized with the single-phase relationship, voltage alone under the volt-amperes, leaving out the three-phase term. The helper says three-phase is three single-phase problems and the extra term is optional. Which relationship did he leave behind?
1. A. The single-phase relationship, which is required for every three-phase transformer.
2. B. A rule that three-phase current is the voltage multiplied by the power factor only.
3. C. The vault rule, which is the three-phase term.
4. D. The three-phase relationship. This transformer does not use the single-phase current relationship. Do not compute a new current to answer.Keyed answer
Use the three-phase relationship for this three-phase transformer. Article 450 does not turn it into a single-phase unit. Confirm the application in NEC 2026. The choice is the relationship, not a calculated ampere.
Open NEC 2026 Article 450
10 · Power and voltage drop · NEC Knowledge
The primary device was set at the transformer's inrush current and left there as the running protection. The helper says the device must hold the inrush, so the inrush number becomes the rating. What did he turn the inrush into?
1. A. The required continuous rating, because inrush is the full-load current.
2. B. The overcurrent setting. Inrush explains why a device must tolerate starting, and it is not the transformer's running overcurrent rating.Keyed answer
3. C. The secondary voltage, which is equal to the inrush.
4. D. A mandatory voltage-drop failure.
Article 450 sets transformer overcurrent protection from the transformer rules, not from the inrush used as the rating. Confirm the protection in NEC 2026. Do not calculate a new inrush to answer.
Open NEC 2026 Article 450
11 · Power and voltage drop · NEC Knowledge
Primary and secondary voltages are equal on this two-winding transformer, and the helper omitted secondary protection because equal voltages mean the primary device is already the right size for both windings. The problem says the primary device does not meet the secondary permission. What did equal numbers fail to cancel?
1. A. The primary device, which must be removed when the voltages match.
2. B. Nothing. Equal voltages prohibit secondary protection.
3. C. The disconnect, which is required only when the voltages differ.
4. D. Secondary protection. A one-to-one voltage ratio is not the permission to protect the secondary from the primary device.Keyed answer
Article 450 still judges secondary protection by the permission in the rules, not by matching voltages. Confirm that permission in NEC 2026. This problem says it is not met.
Open NEC 2026 Article 450
12 · Power and voltage drop · NEC Knowledge
A grounding autotransformer that is there to derive a neutral was used as the general supply for a new lighting feeder. The helper says any transformer with a neutral terminal can carry the building load. What duty did he add that the unit was not there to do?
1. A. The correct duty, because every grounding autotransformer is a general lighting transformer.
2. B. A general power supply. Deriving a neutral is not the same as using the grounding autotransformer as an ordinary step-down transformer for the feeder.Keyed answer
3. C. A voltage-drop calculation, which replaces the transformer.
4. D. Class I sealing, which a neutral terminal requires.
Article 450 and the grounding-autotransformer rules limit that use. Confirm them in NEC 2026. Do not treat the neutral-deriving unit as an ordinary power transformer.
Open NEC 2026 Article 450
13 · Power and voltage drop · NEC Knowledge
One percentage the helper memorized was used for the primary device of every transformer on the job. Some units have secondary protection and some do not, and the impedances are not the same. He says the percentage is universal. What did the single percentage ignore?
1. A. Nothing. Every transformer primary is protected at the same percentage.
2. B. The disconnect location only, which is what the percentage sets.
3. C. A requirement to divide the percentage by the power factor before it is used.
4. D. The conditions that pick the primary protection: the voltage class, the impedance, and whether the secondary is protected. One memory is not the table.Keyed answer
Article 450 selects the maximum primary protection from the conditions the table uses. Confirm the row in NEC 2026. Do not invent a percentage or reuse one memory for every unit.
Open NEC 2026 Article 450
14 · Power and voltage drop · NEC Knowledge
The current measured with the transformer unloaded was written down as the full-load current. The helper says a quiet secondary is the nameplate condition, so the no-load reading replaces the full-load current. What current did the clamp miss?
1. A. No-load current, which is defined as the full-load current.
2. B. Full-load current. No-load current is not the nameplate full-load current and does not size the protection.Keyed answer
3. C. The voltage drop, which is measured in amperes on an unloaded winding.
4. D. The vault volume, which is the no-load current.
Article 450 uses the rated full-load current, not a no-load measurement. Confirm the nameplate current in NEC 2026. Do not compute a new value from the clamp.
Open NEC 2026 Article 450
15 · Power and voltage drop · Calculations
A three-phase load draws 30 A at 208 V with a power factor of 0.80. Using P = V × I × 1.732 × PF, what is the power?
1. A. 6,240 W
2. B. 8,646 WKeyed answer
3. C. 10,808 W
4. D. 12,470 W
208 × 30 = 6,240. Times 1.732 is about 10,808. Times 0.80 is 8,646 W. The 1.732 factor is the square root of 3 for three-phase power. Stopping before the power factor, or forgetting 1.732, produces the other choices.
Open NEC 2026 Article 120
16 · Power and voltage drop · Calculations
A three-phase transformer is rated 36 kVA. The problem tells you to use 360 as the product of the secondary line-to-line voltage and 1.732. What is the secondary full-load current?
1. A. 100 AKeyed answer
2. B. 83 A
3. C. 150 A
4. D. 36 A
36 kVA is 36,000 VA. 36,000 / 360 = 100 A. For three-phase you divide by voltage times 1.732, which the stem already combined. Article 450 is where transformer overcurrent protection starts once you have this current.
Open NEC 2026 Article 450
17 · Power and voltage drop · Calculations
A continuous single-phase load is 3,600 W at 120 V with a power factor of 0.9. Conductor ampacity is 125% of I = P / (V × PF). What ampacity do you need, to the nearest tenth of an ampere?
1. A. 33.3 A
2. B. 37.5 A
3. C. 41.7 AKeyed answer
4. D. 30 A
120 × 0.9 = 108. 3,600 / 108 = 33.3 A. × 1.25 = 41.7 A. Skipping the power factor and then applying 125% gives 37.5 A. Article 215 is the feeder factor. The formula is printed in the stem.
Open NEC 2026 Article 215
18 · Power and voltage drop · Calculations
A three-phase continuous load is 21,600 W. The problem says V × 1.732 × PF = 720. Conductors are 125% of I = P / 720. What conductor ampacity do you need?
1. A. 37.5 AKeyed answer
2. B. 30 A
3. C. 51.96 A
4. D. 90 A
21,600 / 720 = 30 A. × 1.25 = 37.5 A. The 720 figure already includes 1.732 and the power factor, so do not multiply by 1.732 again. Article 215 uses this current for the feeder.
Open NEC 2026 Article 215
19 · Power and voltage drop · Calculations
A single-phase transformer is rated 15 kVA. The secondary is 120 V. Full-load current is (kVA × 1,000) / V and does not use power factor. The secondary conductors in this drill are 125% of that current. What conductor ampacity do you need?
1. A. 125 A
2. B. 156.25 AKeyed answer
3. C. 250 A
4. D. 62.5 A
15 kVA is 15,000 VA. / 120 = 125 A. × 1.25 = 156.25 A. Stopping at full-load current leaves the conductor factor off. Doubling or halving that full-load current is not a step in the stem. Article 450 is where transformer protection starts. This item asks for the conductor ampacity the stem defined.
Open NEC 2026 Article 450
20 · Power and voltage drop · Calculations
A single-phase transformer primary is 48 kVA at 480 V. Full-load current is (kVA × 1,000) / V. This drill allows a primary breaker at 125% of that current, then the smallest standard rating that covers it. Ratings: 110, 125, 150, 175. What do you select?
1. A. 110 A
2. B. 100 A
3. C. 125 AKeyed answer
4. D. 156.25 A
48,000 / 480 = 100 A full-load. × 1.25 = 125 A. The smallest listed rating at or above that is 125 A. The full-load current itself may be under the percentage the stem allowed. Confirm the transformer protection table in NEC 2026 Article 450. The percentage here is printed.
Open NEC 2026 Article 450

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