Journeyman Electrician

Direct current (DC) flows in one direction with constant polarity. Batteries, photovoltaic modules, fuel cells, and rectifiers produce DC. Alternating current (AC) reverses direction periodically, following a sine wave. Nearly all utility power is AC because AC voltage can be raised and lowered efficiently with transformers, allowing high-voltage, low-current transmission over long distances.

Term	Meaning
Cycle	Onecompletesinewave:apositivealternationfollowedbyanegativealternation(360electrical
	degrees) .
Frequency	Cyclespersecond measuredinhertz(Hz) StandardUS frequencyis60Hz socurrentchanges , . . . ,
	direction120timespersecondandonecyclelasts1/60second(about167milliseconds) . .
Peakvalue	Themaximuminstantaneousvoltageorcurrentreachedinonealternation .
Peak -to-peak	Fromthepositivepeaktothenegativepeak:2×peak .

Term	Meaning
RMS(effective)value	TheACvaluethatproducesthesameheatinginaresistorasthesameDCvalue Voltmeters . ,
	ammeters nameplates andNECvoltagesareRMSvaluesunlessstatedotherwise , , .
Phaseangle	Thedisplacement inelectricaldegrees betweentwowaveforms suchasvoltageandcurrent or , , , ,
	between the phases of a three-phase system .
PEAK AND RMS Peak = RMS × 1 414 . Peak -to-peak = Peak ×	RMS = Peak × 0 707 . 2

1.414 is √2 and 0.707 is 1 ÷ √2. Average value (0.637 × peak) is rarely tested.

120 V RMS: Peak = 120 × 1.414 = 169.7 V (about 170 V)

Peak-to-peak = 169.7 × 2 = 339.4 V

240 V RMS: Peak = 240 × 1.414 = 339.4 V

Oscilloscope reads 340 V peak: RMS = 340 × 0.707 = 240.4 V

Insulation must withstand the peak voltage, which is why a "120 V" circuit actually stresses insulation at about 170 V on every cycle.

Single-Phase Systems

The most common dwelling service is 120/240 V, single-phase, 3-wire. It comes from a transformer secondary with a center tap that is grounded and serves as the neutral. Each ungrounded (hot) conductor is 120 V to the neutral, and the two hot conductors are 240 V to each other because their voltages are 180° apart. Lighting and receptacle loads connect line-to-neutral at 120 V; ranges, dryers, water heaters, and air-conditioning equipment connect line-to-line at 240 V.

On a 3-wire, 120/240 V system, the neutral carries only the unbalanced current. If one leg carries 30 A and the other 20 A of line-to-neutral load, the neutral carries 10 A. If the loads are perfectly balanced, the neutral current is zero. An open neutral on this system puts the two 120 V legs in series across 240 V, and the side with the higher resistance (the lighter load) receives the higher voltage, which can destroy electronics.

Three-Phase Systems

Three-phase power uses three ungrounded conductors whose voltages are 120 electrical degrees apart. Because the three waveforms overlap, the total power delivered is nearly constant rather than pulsing, which makes three-phase motors smoother, smaller, and more efficient than single-phase motors of the same rating. Three-phase also delivers more power per pound of conductor, so it is standard for commercial and industrial services. The two transformer connections are wye (star) and delta.

One end of each winding connects to a common point, the

The windings connect end-to-end in a closed triangle. Line

neutral. Line voltage = phase voltage × 1.732; line current =

voltage = phase voltage; line current = phase current × 1.732.

phase current. Common systems: 208Y/120 V (120 × 1.732 =

A plain delta (240 V or 480 V, 3-wire) has no neutral. Center-

208) and 480Y/277 V (277 × 1.732 = 480). The neutral carries

tapping one winding creates the 4-wire high-leg delta.

only unbalanced current on balanced linear loads.

Wye: E(line) = E(phase) × 1.732 I(line) = I(phase)

Delta: E(line) = E(phase) I(line) = I(phase) × 1.732

√3 = 1.732 1 ÷ √3 = 0.577

To find line-to-neutral voltage in a wye from line-to-line voltage, divide by 1.732 (208 ÷ 1.732 = 120; 480 ÷ 1.732 = 277).

The High-Leg (Red-Leg or Wild-Leg) Delta

In a 120/240 V, 3-phase, 4-wire delta, one of the three 240 V windings (A to C) is center-tapped and grounded. Phases A and C each measure 120 V to the neutral. Phase B, the vertex opposite the tapped winding, measures about 208 V to ground (240 × 0.866 = 207.8 V). This phase is the high leg. It must never feed a 120 V line-to-neutral load, and single- pole breakers must never be installed on it.

Marking (110.15): the conductor or busbar with the higher voltage to ground must be durably marked by an outer finish that is orange in color or by other effective means, wherever a connection is made and the grounded conductor is also present.

Phase position (408.3(E)): on switchboards, switchgear, and panelboards, the high leg must be the B phase (the center position). An exception allows a different arrangement in the same section as metering equipment, which is why a utility meter socket may place the high leg in the C position.

Service conductors (230.56): a service conductor with the higher voltage to ground must also be marked orange or by other effective means at each termination or junction point.

Know the voltages to ground of each common system: 120/240 V single-phase (120 V to ground); 208Y/120 V (120 V to ground on every phase); 480Y/277 V (277 V to ground); 120/240 V high-leg delta (120 V on A and C, about 208 V on B). Questions often ask which phase is the high leg (B), what color identifies it (orange), and what it measures to ground (about 208 V).

System	Phase/Wires	AvailableVoltages	TypicalUse
120/240V	1φ 3-wire ,	120 V line-to-neutral; 240 V line-to-line	Dwellings smallcommercial ,
208Y/120V	3φ 4-wire wye ,	120 V line-to-neutral; 208 V line-to-line (3φ	Offices retail apartment , ,
		or1φ)	buildings
480Y/277V	3φ 4-wire wye ,	277 V line-to-neutral (lighting); 480 V line-	Largecommercial industrial ,
		to-line (motors HVAC) ,
120/240 V high -leg delta	3φ 4-wire delta ,	120VfromAandCtoneutral;about208V	Oldercommercialandlight
		from B to neutral; 240 V line-to-line	industrial
240Vor480Vdelta	3φ 3-wire ,	Line-to-line only; no neutral	Motorloads industrial ,

A single-phase load connected to two phases of a 208Y/120 V system sees 208 V, not 240 V. On a single-phase, 3-wire circuit taken from a 208Y/120 V system (two phases and the neutral), the neutral carries current even when the two line- to-neutral loads are equal, because the phases are 120° apart rather than 180° apart.

Three-Phase Power and Current

For balanced three-phase loads, power and current calculations use line-to-line voltage and the 1.732 multiplier. Most exam problems give a load in kVA or kW and ask for the line current, which then leads to conductor or overcurrent sizing.

VA = E × I × 1.732 W = E × I × 1.732 × PF

I = VA ÷ (E × 1.732) I = W ÷ (E × 1.732 × PF)

Single-phase: I = VA ÷ E I = W ÷ (E × PF)

Convert kVA or kW to VA or W by multiplying by 1,000. Common products: 208 × 1.732 = 360; 240 × 1.732 = 416; 480 × 1.732 = 831.

A balanced 50 kVA load is supplied at 208 V, three-phase.

I = VA ÷ (E × 1.732)

I = 50,000 ÷ (208 × 1.732) = 50,000 ÷ 360.3 = 138.8 A

Same 50 kVA at 480 V: I = 50,000 ÷ 831.4 = 60.1 A

Raising the voltage from 208 V to 480 V cuts the current by more than half for the same load, allowing smaller conductors.

A 36 kW, 480 V, three-phase load has a power factor of 0.90.

I = W ÷ (E × 1.732 × PF)

I = 36,000 ÷ (480 × 1.732 × 0.90)

I = 36,000 ÷ 748.2 = 48.1 A

At unity PF the current would be 36,000 ÷ 831.4 = 43.3 A

When power is given in watts, the power factor must be included. When it is given in VA or kVA, the power factor is already accounted for and is not used.

Power Factor

In a purely resistive AC circuit, voltage and current rise and fall together (in phase), and every volt-ampere does useful work. Inductive loads, such as motors, transformers, and magnetic ballasts, cause current to lag voltage. Capacitive loads cause current to lead voltage. When current and voltage are out of phase, part of the current only builds and collapses magnetic or electric fields, returning energy to the source each cycle without doing work. The source and conductors must still carry that current.

Three kinds of power result, and they form the power triangle. Power factor (PF) is the ratio of true power to apparent power, equal to the cosine of the phase angle θ. It ranges from 0 to 1.0 (or 0% to 100%).

Power	Unit	Description
Truepower(P)	W kW ,	Powerthatperformswork(heat light motion) Measuredbyawattmeter , , . .
Reactivepower(Q)	VAR kVAR ,	Powerthatestablishesmagneticfields(inductors)orelectricfields(capacitors)and
		isreturnedtothesourceeachcycle .
Apparentpower(S)	VA kVA ,	Volts×amperes:whatthesource transformer andconductorsmustactuallysupply , ,
		andcarry .

A 240 V single-phase motor draws 20 A. A wattmeter reads 4,080 W.

VA = E × I = 240 × 20 = 4,800 VA

PF = W ÷ VA = 4,080 ÷ 4,800 = 0.85 (85%)

VAR = √(VA² − W²) = √(4,800² − 4,080²)

VAR = √(23,040,000 − 16,646,400) = √6,393,600 = 2,529 VAR

The ratio W ÷ VA is always 1.0 or less. If a calculated PF exceeds 1, the watts and volt-amperes were reversed.

Effects of low power factor: more current for the same useful power, larger conductors and transformers, higher I²R losses and voltage drop, and possible utility penalty charges.

Correction: capacitors connected in parallel with inductive loads supply leading reactive power that offsets the lagging reactive power, raising the power factor toward 1.0 and reducing line current upstream of the capacitors.

Memory aid: "ELI the ICE man." In an inductive (L) circuit, voltage (E) leads current (I); in a capacitive (C) circuit, current (I) leads voltage (E).