Electrical Theory Practice Test

AC fundamentals, three-phase relationships, power factor and conductor properties — the theory that every calculation on the exam quietly depends on.

Practise Theory

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What it is, and why electricians need it

Electrical theory accounts for only a handful of standalone exam questions, but it is load-bearing: the calculation questions assume it, and candidates who skip it end up memorising formulas they cannot check. Three areas earn the most return.

Three-phase relationships. In a wye system, line-to-line voltage is the square root of 3 times line-to-neutral voltage, while line current equals winding current. In a delta system it is exactly the reverse: line voltage equals winding voltage, but line current is the square root of 3 times winding current. That single pair of facts explains 208Y/120, 480Y/277, and why the square root of 3 appears in every three-phase power calculation. Omitting it is the most costly arithmetic error on the exam, because the resulting wrong answer is usually one of the offered options.

Real, apparent and reactive power. Watts do work. Volt-amperes are what the conductors and transformer actually have to carry. The ratio between them is the power factor. Equipment is rated in kVA rather than kW precisely because heating depends on current, and current follows apparent power regardless of how much of it is doing useful work.

Conductor behaviour. Resistance is directly proportional to length and inversely proportional to cross-sectional area, which is why voltage drop governs long runs even when ampacity is ample. Metals heat up and become more resistive as they do, which is why ampacity tables carry ambient correction factors. And aluminium conducts about 61 percent as well as copper, which is why the ampacity table lists it separately and two sizes lower.

Formulas and principles

  • Wye: Vline = 1.732 × Vphase, Iline = Iphase

    Vphase is line-to-neutral. 480 V line-to-line gives 277 V to neutral.

    The origin of the 208Y/120 and 480Y/277 designations.

  • Delta: Vline = Vphase, Iline = 1.732 × Iphase

    The mirror image of wye. Each line conductor feeds two windings.

  • Three-phase VA = 1.732 × Vline × Iline

    Apparent power from line values. Multiply by power factor for watts.

    Leaving out 1.732 understates the answer by about 42 percent.

  • PF = W ÷ VA

    Power factor, between 0 and 1. Often expressed as a percentage.

    kVA = kW ÷ PF, so apparent power is always at least as large as real power.

  • Vpeak = Vrms × 1.414

    For a sine wave. 120 V RMS peaks at about 170 V.

    Why insulation in a nominal 120-volt circuit must withstand roughly 170 volts.

  • 1 hp = 746 W

    Mechanical output. Electrical input is higher, by the efficiency.

    Never size motor conductors from this; use the full-load current tables.

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Worked examples

Scenario: A balanced three-phase load draws 50 amperes per line at 480 volts line-to-line. Find the apparent power, and the real power if the power factor is 0.9.

  1. Apparent power: VA = 1.732 × 480 × 50 = 41,569 VA, or about 41.6 kVA.
  2. Real power: W = VA × PF = 41,569 × 0.9 = 37,412 W, or about 37.4 kW.
  3. Note the difference: 41.6 kVA is what the conductors must carry, 37.4 kW is what does work.
  4. Had you omitted the 1.732, you would have got 24 kVA — a 42 percent error.

Answer: About 41.6 kVA apparent, about 37.4 kW real

Scenario: A 100 kVA transformer supplies a load at 0.8 power factor. How many kilowatts can it deliver, and why is it not 100 kW?

  1. Real power = kVA × PF = 100 × 0.8 = 80 kW.
  2. The winding current is set by the 100 kVA rating, not by the 80 kW.
  3. So the transformer is fully loaded thermally while delivering only 80 kW of useful power.
  4. Correcting the power factor to unity would let the same transformer deliver 100 kW.

Answer: 80 kW, because current and therefore heating follow apparent power

Common mistakes

  • Confusing 1.732 with 1.414. The square root of 3 is 1.732 and belongs in three-phase calculations. The square root of 2 is 1.414 and converts between RMS and peak on a sine wave. They are not interchangeable.
  • Swapping the wye and delta relationships. Wye multiplies voltage by 1.732 and leaves current alone; delta multiplies current and leaves voltage alone. Remember that the neutral exists in wye, which is what creates the line-to-neutral voltage.
  • Multiplying by power factor to get kVA from kW. Divide. kVA = kW ÷ PF. Apparent power is always greater than or equal to real power, so an answer smaller than the kW figure is impossible.
  • Sizing motor conductors from a horsepower-to-watt conversion. 746 watts per horsepower is mechanical output. The electrical input is larger and the Code requires conductors to be sized from the full-load current tables, not from a power conversion.
  • Assuming a current reading on one conductor represents a circuit. Clamp around a cable containing both the ungrounded and grounded conductors and the fields cancel, giving nearly zero. You must clamp a single conductor — the same cancellation is how a GFCI detects a fault.

Related practice topics

Topics that come up alongside this one on the exam.

Questions about this topic

How much electrical theory is on a licensing exam?

Less than you might expect as standalone questions, but it underpins everything else. The Texas journeyman NEC Knowledge outline allocates 3 of 56 scored items to Definitions, Theory and Plans, rising to 7 of 70 at master level. The real weight is indirect: you cannot do three-phase calculations without understanding wye and delta relationships, and you cannot size a transformer without understanding apparent power.

Why is three-phase power preferred for larger loads?

Three overlapping waveforms mean instantaneous power is nearly constant rather than pulsing to zero twice a cycle. That gives smoother motor torque and lets a three-phase motor start without the auxiliary winding or capacitor a single-phase motor needs. It also carries a given power with less conductor material, which is why services above modest sizes are almost always three-phase.

What is the difference between kW and kVA, and why does it matter?

Kilowatts are real power, the part doing useful work. Kilovolt-amperes are apparent power, the product of volts and amperes. For a reactive load they differ by the power factor. It matters because conductors and transformers are limited by current, which follows apparent power: a 100 kVA transformer delivers 100 kW into a unity power factor load but only 80 kW at 0.8 power factor, while the winding current and heating are identical.

Where does 277 volts come from?

It is the line-to-neutral voltage of a 480-volt three-phase wye system: 480 ÷ 1.732 = 277. That is why commercial lighting so often runs at 277 volts — it is simply one phase of the building's 480-volt service to neutral, and the higher voltage lets one circuit carry far more luminaires than it could at 120 volts.

Why does conductor resistance rise as it heats?

Metals have a positive temperature coefficient of resistance. The practical consequence is a feedback loop: a loaded conductor heats itself, which raises its resistance, which raises its voltage drop and its heating further. It is also why ampacity tables are published for a stated ambient temperature and must be corrected when the surroundings are hotter.

Does a transformer work on direct current?

No. Inducing a voltage in the secondary requires changing magnetic flux, and steady DC produces constant flux. Worse, with no counter-electromotive force to oppose it, the primary current is limited only by the winding's small resistance, so connecting a transformer to DC typically burns out the primary.

How do capacitors improve power factor?

They supply leading reactive power that offsets a motor's lagging reactive demand, so less reactive current has to travel through the service conductors and transformer. The motor still does the same work and consumes the same watts; what falls is the total line current, which frees up capacity and reduces losses.

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