Chapter XVII

General Trade Knowledge & Theory (Washington)

Journeyman Practice study guide with diagrams.

General Trade Knowledge & Theory (Washington Journeyman)

Learning Objectives

Apply Ohm's law and Watt's law to solve for voltage, current, resistance, and power in dc and single-phase ac circuits.
Solve series, parallel, and series-parallel networks using the correct rules for current, voltage, and resistance.
Calculate impedance, power factor, true power, and apparent power, and explain current-voltage phase relationships.
Compute synchronous speed, percent slip, and rotor speed, and interpret motor nameplate data.
Use the turns ratio to find transformer primary and secondary voltages and currents, and state the rules of Article 450.
Select conductors and overcurrent protection using Table 310.16, 240.4(D), 310.15(B)(1), and 310.15(C)(1).
Select and verify test instruments, and describe safe work practices and an electrically safe work condition.

1.1 Ohm's Law, Watt's Law, and the Power Formulas

Every calculation in this chapter grows out of two relationships. Ohm's law ties voltage, current, and resistance together: E = I × R, I = E / R, and R = E / I. Watt's law ties power to voltage and current: P = E × I. Combining the two gives the forms that save the most time on an open-book exam: P = I² × R and P = E² / R.

Worked examples:

A 120-volt branch circuit drawing 12 amperes has R = 120 / 12 = 10 ohms.
A 1,500-watt, 120-volt heater draws I = P / E = 1,500 / 120 = 12.5 amperes.
A 3,000-watt, 240-volt heating element draws 12.5 amperes, so its resistance is R = E / I = 240 / 12.5 = _19.2 ohms_.
A 120-volt circuit with 8 ohms of resistance draws I = 120 / 8 = 15 amperes and dissipates P = E² / R = 14,400 / 8 = 1,800 watts.

Keep the units straight, because the exam tests the distinction constantly. Watts are real power, volt-amperes are apparent power, and volt-amperes reactive are reactive power. For a purely resistive load the power factor is 1.0 and the two are numerically equal; for any other load they are not, so a question that gives voltage and amperes but no power factor can only be answered in volt-amperes. NEC load calculations rest on the same relationships, which is why a load in volt-amperes is converted with I = VA / E.

1.2 Series Circuits

A series circuit has one path for current, and four rules control every series problem: current is the same through every component; total resistance is the sum of the individual resistances; source voltage divides in proportion to resistance; and individual powers add to the total.

Three resistors of 10, 20, and 30 ohms on a 120-volt source: R total = 60 ohms, so I = 2 amperes.
Two resistors of 40 and 80 ohms on a 240-volt source: R total = 120 ohms and I = 2 amperes, so the drop across the 80-ohm resistor is 160 volts and the drop across the 40-ohm resistor is 80 volts. The drops add back to the source voltage.

Two behaviors are asked as theory rather than arithmetic. If one component of a series string opens, current becomes zero and the entire source voltage appears across the open point, because no current is left to produce drops in the remaining components. Series dividers are also the principle behind dual-voltage motors, where two windings per phase sit in series so each sees half of the line voltage.

1.3 Parallel Circuits

A parallel circuit gives each load its own path, and the rules invert: voltage is the same across every branch; total current is the sum of the branch currents; equivalent resistance comes from 1/R total = 1/R1 + 1/R2 + …, or from the product-over-sum shortcut for two branches; and total power is the sum of the branch powers.

Three 12-ohm resistors in parallel: R = 12 / 3 = 4 ohms. With 12, 6, and 4 ohms, R = 2 ohms.
10 and 15 ohms in parallel: R = 150 / 25 = 6 ohms, so a total current of 5 amperes means 30 volts across the circuit.
Power in a parallel bank is the sum of the branch powers: 10 and 15 ohms across 60 volts dissipate 360 and 240 watts, for 600 watts total.

If one branch of a parallel circuit opens, the voltage across the remaining branches is unchanged, so their currents are unchanged and only the total current drops. Two unequal loads on the same 240-volt feeder add their currents; the feeder never carries their average.

1.4 AC Theory: Reactance, Impedance, Phase, and Power Factor

In ac circuits the opposition to current is impedance (Z), built from resistance (R) and reactance (X). Inductive reactance comes from coils and magnetic fields; capacitive reactance comes from capacitors and electric fields. Resistance and reactance are 90 degrees apart, so they combine as Z = √(R² + X²), never arithmetically.

The phase relationships are core theory. In a purely resistive circuit, current and voltage are in phase and the power factor is 1.0. In a purely inductive circuit, current lags the voltage by 90 degrees. In a purely capacitive circuit, current leads the voltage by 90 degrees.

R = 8 ohms and XL = 6 ohms in series: Z = √(64 + 36) = √100 = 10 ohms.
R = 30 ohms and XL = 40 ohms: Z = √(900 + 1,600) = 50 ohms, and PF = R / Z = 30 / 50 = 0.80.
A three-phase load of 50 kW at a power factor of 0.80 draws kVA = kW / PF = 62.5 kVA.

Power factor is the cosine of the angle between voltage and current, so it can never exceed 1.0, and a load at unity power factor draws the least current for a given wattage. The power triangle pairs real power, reactive power, and apparent power. In a series RLC circuit, resonance occurs when XL equals XC: the reactances cancel, impedance falls to its minimum, current reaches its maximum, and the power factor is unity. Distractors often offer sine or tangent values (X/Z or X/R) in place of the power factor, and mixing up apparent power and watts is the other standard error: at 120 volts, any current other than 16 amperes, for example _13.33 A_, yields a different volt-ampere load than the one stated.

1.5 Three-Phase Relationships and Load Calculations

Three-phase systems add one factor, √3, which is 1.732. Line-to-neutral voltage equals line-to-line voltage divided by 1.732, so a 480-volt wye system has 277 volts to neutral and a 208-volt, 4-wire system has 120 volts to neutral. Apparent power is VA = 1.732 × E(line-to-line) × I(line).

480 volts and 50 amperes: VA = 1.732 × 480 × 50 = 41,570, which is _41.6 kilovolt-amperes_. Using E × I alone gives 24.0 kVA, using a factor of 3 instead of 1.732 gives _72.0 kilovolt-amperes_, and dividing by 3 gives 13.9 kilovolt-amperes.
480 volts and 60 amperes: kVA = 1.732 × 480 × 60 ÷ 1,000 = 49.9 kilovolt-amperes. Omitting the 1.732 factor gives 28.8 kVA, a wrong multiplier gives _86.4 kilovolt-amperes_, and dividing by an extra factor gives 14.4 kVA.
True power is P = 1.732 × E × I × power factor. At 460 volts, 24 amperes, and 0.85 power factor, P = 16,255 watts, or 16.3 kilowatts. Omitting the power factor gives 18.8 kilowatts, and applying 1.732 twice gives 28.2 kilowatts.

Apparent power is the figure used for transformer and feeder sizing, because NEC load calculations work in volt-amperes; to find line current from kVA, rearrange the formula as I = VA ÷ (1.732 × E).

1.6 Induction Motors: Synchronous Speed, Slip, and Torque

The rotating magnetic field in a polyphase induction motor turns at synchronous speed, Ns = 120 × f / P, where f is frequency in Hz and P is the number of poles per phase. At 60 Hz a 4-pole motor runs at 1,800 rpm, a 6-pole motor at 1,200 rpm, and an 8-pole motor at 900 rpm. Frequency drives the answer: the same 4-pole motor on a 50 Hz supply drops to 1,500 rpm, and a 6-pole motor on 50 Hz runs at 1,000 rpm.

The rotor always turns slightly slower than the field. That difference, as a percentage of synchronous speed, is slip: slip = (Ns − Nr) / Ns × 100, and Nr = Ns × (1 − slip).

At 1,800 rpm synchronous speed with 5 percent slip: Nr = 1,800 × 0.95 = 1,710 rpm.
At 1,200 rpm with 4 percent slip: Nr = 1,200 × 0.96 = 1,152 rpm; at 900 rpm with the same slip, Nr = 864 rpm.
A nameplate full-load speed of 1,750 rpm on a 4-pole, 60 Hz motor: slip = (1,800 − 1,750) / 1,800 × 100 = 2.78, a 50 rpm difference on a 1,800 rpm field. The distractors come from the wrong speed difference: 5.00 percent matches 1,710 rpm, 2.50 percent matches 1,755 rpm, and 1.39 belongs to 1,775 rpm.

Theory questions cluster around three ideas. Slip exists only because the rotor is slower than the field: at synchronous speed there is no relative motion, no induced rotor voltage, and no torque, while a rotor turning faster than the field describes a generator. In the normal operating range torque increases as slip increases, up to the breakdown (maximum) torque, after which the motor stalls. At the instant of starting the rotor is at standstill, so slip is 100 percent and the motor draws locked-rotor current. Starting methods manage that inrush: across-the-line starting applies full voltage with a starting torque higher than full-load torque, while wye-delta starting reduces the voltage across each winding, lowering both starting current and torque. Neither method changes synchronous speed or rotation direction, which is set by swapping two supply leads.

Nameplate data completes the picture. Service factor, for example 1.15, is a multiplier on rated horsepower that tells you how much overload a motor may carry at nameplate voltage and temperature without exceeding its insulation temperature limit. A dual-voltage motor marked 230/460 volts places its per-phase windings in parallel for 230 volts and in series for 460 volts, so each winding sees half of the line voltage.

1.7 Motor Full-Load Current, Code Letters, and Branch-Circuit Conductors

Motor calculations do not start with the nameplate current; they start with the tables. Table 430.248 provides full-load currents for single-phase ac motors and Table 430.250 for three-phase ac motors, with voltage ranges of 110–120, 220–240, 440–480, and 550–600 volts. Table values to know: a single-phase, 230-volt, 5 hp motor draws 28 amperes, and a three-phase, 460-volt, 25 hp motor draws 34 amperes.

Table 430.7(B) assigns locked-rotor code letters using kilovolt-amperes per horsepower. Code letter G covers 5.6 to 6.29 kilovolt-amperes per horsepower, F covers 5.0 to 5.59, and H covers 6.3 to 7.09, so knowing the neighborhood of the given range is enough to pick the letter.

Section 430.22 sets the conductor rule for a single continuous-duty motor: branch-circuit conductors may not have an ampacity less than 125 percent of the motor full-load current rating. A 5 hp, 230-volt single-phase motor with a full-load current of 28 amperes requires 28 × 1.25 = 35 amperes, and a 25 hp, 460-volt three-phase motor with a table current of 34 amperes requires 34 × 1.25 = 43 amperes. Protection builds on the same table value: an inverse-time breaker at 300 percent of 34 amperes gives 102 amperes, moving up to the next standard size. Keep the two sources distinct: the tables are used for conductor ampacity, disconnecting means, and short-circuit and ground-fault protection, while the higher nameplate full-load current is used for overload protection.

1.8 Transformers: Turns Ratio, Current, and Article 450 Requirements

For an ideal transformer the turns ratio equals the voltage ratio and the inverse of the current ratio: Vp / Vs = Np / Ns, Ip / Is = Ns / Np, and Vp × Ip = Vs × Is, because power is conserved.

A 10:1 step-down transformer on 600 volts: 600 / 10 = 60 volts, and a 300-turn primary at 240 volts with a 24-volt secondary needs Ns = 300 × (24 / 240) = 30 turns.
A 10 kVA transformer at 240 volts on the primary draws _41.7 A_; on a 120-volt secondary the current is _83.3 A_. Answering 20.8 A confuses the primary current with a value computed at the wrong voltage, and _166.7 A_ comes from using a 4:1 ratio instead of 2:1.
A 25 kVA transformer at 480/120 volts: the primary current is 25,000 / 480 = 52.1 A and the secondary current is _208.3 A_. The distractor _104.2 A_ doubles the primary current.
A 15 kVA transformer at 480 volts on the primary draws 31.25 amperes, and a 5 kVA transformer with a 240-volt secondary draws 20.8 A. Because current varies inversely with the turns ratio, a step-down in voltage always means a step-up in current.

Article 450 governs the installation. Section 450.13(A) requires each transformer to have a nameplate giving the name of manufacturer, the rated kilovolt-amperes, the frequency, and the primary and secondary voltage. Section 450.21(A) covers dry-type transformers installed indoors and rated 112 1/2 kVA or less: they need a separation of at least 300 mm (12 in.) from combustible material. Section 450.21(B) covers individual dry-type transformers rated over 112 1/2 kVA: they must be installed in a transformer room of fire-resistant construction with a minimum fire rating of 1 hour.

Transformer vaults follow 450.45. Ventilation openings under 450.45(B) may be arranged with roughly half of the required area near the floor and the remainder in the roof or in sidewalls near the roof, or with all of the area in or near the roof; all of the area near the floor is not permitted. The same section requires a door sill or curb of approved height that will confine the oil from the largest transformer within the vault, never less than 100 mm (4 in.). Section 450.45(C) gives the numerical rule for a naturally ventilated vault: the combined net area of all openings, after deducting the area occupied by screens, gratings, and louvers, must be not less than 1900 mm2 (3 in.2) per kVA of transformer capacity in service, and never less than 0.1 m2 (1 ft2) for any capacity under 50 kVA. Halving or doubling the per-kVA figure is the standard distractor.

1.9 Ampacity, Ambient Correction, Adjustment, and Small-Conductor Limits

Table 310.16 is the ampacity table for insulated conductors rated up to 2,000 V with not more than three current-carrying conductors in a raceway, cable, or direct burial, organized by size (AWG or kcmil) and by the 60°C, 75°C, and 90°C temperature columns. Values to carry into the exam: copper at 60°C is rated 15 amperes at 14 AWG, 20 amperes at 12 AWG, and 30 amperes at 10 AWG. The 75°C column is higher for the same size, but a 60°C termination rating limits you to the 60°C ampacity. Section 240.4(D) then caps the overcurrent protection for small conductors, after any ambient correction and adjustment factors have been applied: 18 AWG copper = 7 amperes, 16 AWG copper = 10 amperes, 14 AWG copper = 15 amperes, 12 AWG copper = 20 amperes, and 10 AWG copper = 30 amperes.

Ampacity taken from the table is valid only at the table's assumed conditions. Section 310.15(B)(1) requires ambient temperature correction using Table 310.15(B)(1)(1), which is based on an ambient of 30°C; hotter locations reduce ampacity, and the correction factor multiplies the table value for the conductor's temperature rating. Section 310.15(C)(1) requires adjustment when more than three current-carrying conductors are installed in a raceway or cable without maintaining spacing, with factors that drop as the count rises: 80 percent for 4 through 6 conductors, 70 percent for 7 through 9, and 50 percent for 10 through 20.

A worked example ties this together. A 2,000-watt, 240-volt water heater draws 8.33 A. Because it is a continuous-type fixed appliance, the branch circuit is sized at 125 percent of the load: 8.33 × 1.25 = 10.4 amperes. A 12 AWG copper conductor with 60°C insulation carries 20 amperes per Table 310.16 and may be protected at 20 amperes under 240.4(D), so it covers 10.4 amperes with margin; 14 AWG is limited to 15 amperes by 240.4(D), 10 AWG is larger than required, and 8 AWG is oversized. Section 110.14(D) also requires terminals to be tightened to the torque value marked on the equipment using a calibrated torque tool.

1.10 Test Instruments and Safe Work Practices

Instrument questions test both selection and verification. A digital multimeter is the general-purpose meter; start on the highest voltage range when it is not autoranging and step down, so an unexpected voltage cannot damage the meter or cause an arc. A clamp-on ammeter measures current by sensing the magnetic field around a single conductor, so the circuit is never opened, and the jaws must encircle only the ungrounded conductor — encircling both the ungrounded and grounded conductors cancels the fields and reads near zero. A megohmmeter applies a high dc test voltage for insulation resistance, testing each phase conductor against the grounded frame with the motor disconnected and locked out; for a 600-volt winding, 500 volts DC is the usual test voltage, while 1,000 V AC is wrong on both counts and 2,500 volts DC stresses the winding. A non-contact voltage tester is a supplemental check only: it can indicate that a conductor is energized, but absence of voltage must be confirmed with a contact-type tester that has been verified on a known live source.

The safe work rules that this chapter's questions cite as 110.6(A)(1) and 110.6(A)(2) come from the electrical safety standard and are tested heavily. The tester must be verified on a known live source or proving unit both before and after the absence-of-voltage test, because a failed tester that reads nothing produces a false "dead" reading. Test instruments must be rated for the circuit voltage and the available fault current, not merely for the voltage: a 480-volt motor circuit calls for a CAT III or CAT IV rating, not CAT I or CAT II. Lockout/tagout belongs to the same family: each authorized worker applies a personal lock and tag and verifies the absence of voltage, and a tag alone never substitutes for a lock.

Note the numbering when you look up an article. In the NEC 2026, 110.1 is the scope of Article 110 and 110.6 addresses conductor sizes, expressed in AWG or kcmil; the test-instrument requirements above come from the electrical safety standard, while the safe work questions cite 120.5 for an electrically safe work condition, including verification of de-energization. Within the NEC, Article 120 covers load calculations, and 120.11 limits the total load on a branch circuit, with 120.11(B) stating the maximum load in volt-amperes for lighting units operating at less than 100 percent power factor — the product of voltage and current (120 × 16 = 1,920), not the true power in watts.

Code Navigation

TopicNEC SectionWhat it covers
Scope of general requirements110.1What Article 110 covers
Conductor sizes110.6Sizes expressed in AWG or kcmil
Test instruments — verification110.6(A)(1)Verified on a known live source before and after use
Test instruments — ratings110.6(A)(2)Rated for circuit voltage and available fault current
Terminal torque110.14(D)Terminals tightened to the marked torque value
Electrically safe work condition120.5De-energization, isolation, and voltage verification
Branch-circuit maximum load120.11Total load may not exceed the branch-circuit rating
Lighting units below 100 percent PF120.11(B)Maximum load stated in volt-amperes
Small-conductor overcurrent limits240.4Protection limits for 18 AWG through 10 AWG
Specific small-conductor limits240.4(D)14 AWG = 15 amperes, 12 AWG = 20 amperes, 10 AWG = 30 amperes
Ampacity table310.16Ampacities by size and 60°C / 75°C / 90°C rating
Ambient temperature correction310.15(B)(1)Correction factors, based on a 30°C ambient
Conductor adjustment310.15(C)(1)More than three current-carrying conductors
Motor locked-rotor code letters430.7(B)kVA per horsepower ranges for code letters
Motor branch-circuit conductors430.22Not less than 125 percent of motor full-load current
Single-phase motor full-load current430.248Full-load currents for single-phase ac motors
Three-phase motor full-load current430.250Full-load currents for three-phase ac motors
Transformer nameplate450.13(A)Manufacturer, rated kVA, frequency, and voltages
Dry-type, 112 1/2 kVA or less450.21(A)300 mm (12 in.) separation from combustible material
Dry-type, over 112 1/2 kVA450.21(B)Transformer room with a 1-hour fire rating
Vault ventilation openings450.45(B)Opening locations and the 100 mm (4 in.) sill or curb
Vault ventilation area450.45(C)Not less than 1900 mm2 (3 in.2) per kVA of capacity

Common Exam Traps

Adding resistance and reactance directly. Z = √(R² + X²), not R + X: 8 and 6 ohms give 10 ohms, not 14, and R = 30 with XL = 40 ohms gives Z = 50 ohms and PF = 0.80.
Treating watts and volt-amperes as interchangeable. A 240-volt, 15-ampere load at 0.80 power factor consumes 2,880 watts but 3,600 of apparent power. Multiply by the power factor for watts; divide for volt-amperes.
Dropping or doubling √3 in three-phase calculations. 1.732 × 480 × 50 is _41.6 kilovolt-amperes_, not 24.0 kVA (which omits the factor), not _72.0 kilovolt-amperes_ (which uses 3), and not 13.9 kilovolt-amperes (which divides by 3).
Mixing table full-load current with nameplate current. Sizing conductors under 430.22 uses Table 430.248 or Table 430.250 times 125 percent — 28 amperes becomes 35 and 34 becomes 43 — while overload protection uses the nameplate value.
Using the wrong speed difference for slip. Slip is (Ns − Nr) / Ns × 100. On a 1,800 rpm field a 50 rpm difference is 2.78, while 5.00 corresponds to 90 rpm and 2.50 to 45 rpm.
Reading Table 430.7(B) by the wrong range. Letter G covers 5.6 to 6.29 kilovolt-amperes per horsepower; F stops at 5.59 and H starts at 6.3, so a one-letter error is the usual mistake.
Confusing the two dry-type transformer rules. 450.21(A) covers transformers rated 112 1/2 kVA or less, while 450.21(B) covers those rated over 112 1/2 kVA and requires a 1-hour fire-rated room. The same pattern appears in 450.45 between (B) and (C).

Practical Field Points

Compute both the running current and the expected table value before opening a motor circuit; the quick check for an overload trip is a clamp-on ammeter on each phase compared with the nameplate full-load amperes.
Verify every voltage tester on a known live source or proving unit both before and after the test. A tester that fails silently is the most dangerous instrument on the truck, and a non-contact pen alone is never proof of an electrically safe condition.
When clamping a circuit, close the jaws around one ungrounded conductor only; around a cable carrying both conductors the fields cancel and the meter reads near zero, which leads to a false "no load" conclusion.
Keep the kVA rating and the winding voltage together when converting transformer currents: divide kVA × 1,000 by the voltage of the winding in question, so a 10 kVA transformer gives _41.7 A_ at 240 volts and _83.3 A_ at 120 volts.
Treat Table 310.16 ampacity as a starting point only: hot ambient temperatures bring 310.15(B)(1) correction factors into play, bundled conductors bring 310.15(C)(1) adjustment factors, and 240.4(D) caps the overcurrent device.

Summary

This chapter is the toolbox the rest of the code depends on. Ohm's law and Watt's law, together with P = I² × R and P = E² / R, convert between voltage, current, resistance, and power. Series circuits add resistance and divide voltage; parallel circuits hold voltage constant, add current, and use the reciprocal formula or R / n for equal resistors. In ac circuits, reactance and resistance combine as Z = √(R² + X²), current leads voltage by 90 degrees in a capacitive circuit, lags it by 90 degrees in an inductive circuit, and is in phase with voltage in a resistive circuit; power factor is true power divided by apparent power. Three-phase work adds the 1.732 factor and divides line-to-line voltage by 1.732 to reach line-to-neutral voltage. Induction motors turn at synchronous speed Ns = 120 × f / P, run below it by the percentage slip, and develop torque as slip rises to breakdown torque, with 125 percent of the table full-load current setting branch-circuit conductor ampacity under 430.22. Transformers obey the turns ratio for voltage and the inverse ratio for current, and Article 450 governs nameplates, indoor clearances, vault ventilation, and vault openings. Conductors come from Table 310.16 as modified by 310.15(B)(1) and 310.15(C)(1) and capped by 240.4(D). Instruments and safe work practices — verified testers, category ratings matched to available fault current, and an electrically safe work condition — turn the arithmetic into work that can be done safely.

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