Journeyman Electrician

Electrical theory is the foundation for every calculation on the journeyman exam. This section covers the behavior of voltage, current, and resistance; power, energy, and efficiency; AC and three-phase systems; power factor; inductance, capacitance, and transformers; series, parallel, and combination circuits; and conductor properties and voltage drop. Roughly one-third of a typical exam is calculation-based, and most of those questions reduce to a handful of formulas: Ohm's Law, the power formulas, the three-phase current formula, the transformer relationships, and the voltage drop formula. Expect to identify the correct formula for a stated situation, solve it quickly with a calculator, and choose the next standard conductor size or the nearest answer.

Electricity is the controlled movement of electric charge, carried in metal conductors by electrons. Every electrical system, from a doorbell transformer to a utility substation, exists to move charge through a load so that the load can do useful work: produce heat, light, motion, or a signal. Understanding what makes charge move, and what opposes that movement, explains nearly every rule in the NEC that deals with conductor sizing, overcurrent protection, and grounding.

Atomic Structure

All matter is made of atoms. The nucleus holds protons (positive charge) and neutrons (no charge). Electrons (negative charge) occupy shells around the nucleus. The electrons in the outermost shell, the valence electrons, determine how well a material conducts. When valence electrons are loosely held, an applied voltage can push them from atom to atom, and current flows. When they are tightly held, very little current flows.

One to three valence electrons that

Four valence electrons. Silicon and

Five to eight tightly bound valence

move freely. Silver, copper, gold, and

germanium conduct under certain

electrons. Rubber, thermoplastic, glass,

aluminum are the best conductors.

conditions and are used in diodes,

porcelain, mica, and dry wood resist

Copper has one valence electron and is

transistors, rectifiers, and solid-state

current flow. Conductor insulation

the standard building-wire material

controls such as dimmers and variable-

keeps current in its intended path and

because of its conductivity, strength,

frequency drives.

protects people.

and cost.

Voltage, Current, and Resistance

Three quantities describe every circuit. The exam uses E for voltage (from electromotive force), although some references use V. A useful comparison is a water system: voltage is the pressure, current is the flow rate, and resistance is the restriction of the pipe.

Quantity	Symbol	Unit	Defi nitionandKeyFacts
Voltage	E(orV)	Volt(V)	Electricalpressureorpotentialdifferencebetweentwopoints Voltagepushes .
			currentbutdoesnotflow;itexistsacrosstwopointsevenwhennocurrent
			flows asatanopenswitch , .
Current	I	Ampere(A)	Therateofchargeflow Oneampereisonecoulomb(about624×1018 . .
			electrons)passingapointpersecond Currentflowsonlythroughacomplete .
			(closed)circuit .
Resistance	R	Ohm(Ω)	Oppositiontocurrentflow Resistanceconvertselectricalenergyintoheat . .
			Everyconductorhassomeresistance .
Power	P	Watt(W)	Therateatwhichelectricalenergyisconvertedtoworkorheat .

Electron flow describes electrons moving from the negative terminal of a source to the positive terminal. Conventional current flow, used in most engineering drawings and in the arrow on a diode symbol, describes current moving from positive to negative. Both describe the same circuit and give identical calculation results; the exam rarely depends on which convention is used.

Factors That Determine Resistance

The resistance of a conductor depends on four factors. These factors explain why long runs need larger conductors, why aluminum is sized larger than copper, and why ampacity is reduced in hot locations. They combine in the relationship R = (K × L) ÷ CM, which is the basis of the voltage drop formula in 1.6.

Factor	EffectonResistance
Material	Eachmaterialhasaspecificresistance(resistivity) Aluminumhasroughly16timestheresistanceof . .
	copperofthesamesize .
Length	Resistanceisdirectlyproportionaltolength Doublingthelengthdoublestheresistance . .
Cross-sectional area	Resistance is inversely proportional to area Doubling the circular-mil area cuts the resistance in half . .
Temperature	Forcopperandaluminum resistancerisesastemperaturerises TheconductorresistancesinNEC , .
	Chapter9 Table8arelistedat75°C , .

Ohm's Law

Ohm's Law states that current in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance. If voltage increases while resistance stays the same, current increases in proportion. If resistance increases while voltage stays the same, current decreases. Any one of the three values can be found when the other two are known.

E = I × R I = E ÷ R R = E ÷ I

Cover the unknown in the E-over-I-R triangle: E on top, I and R on the bottom. If E is covered, multiply I × R; if I or R is covered, divide E by the other.

A 120 V heating element has a resistance of 12 Ω. Find the current.

I = E ÷ R = 120 V ÷ 12 Ω = 10 A

A 240 V load draws 20 A. Find its resistance.

R = E ÷ I = 240 V ÷ 20 A = 12 Ω

A 6 Ω load carries 20 A. Find the voltage across it.

E = I × R = 20 A × 6 Ω = 120 V

Always convert values to base units first: 2.5 kΩ = 2,500 Ω, 500 mA = 0.5 A, 4.16 kV = 4,160 V.

Ohm's Law applies to resistive loads such as heaters, incandescent lamps, and resistors. In AC circuits containing motors, transformers, or capacitors, replace R with impedance (Z): E = I × Z. Questions that give a resistance and a voltage and ask for current are almost always straight Ohm's Law problems.

Power, Energy, and Efficiency

Power is the rate of doing work, measured in watts (W). In a resistive circuit, all of the power is converted into heat or light. Combining the basic power formula (P = E × I) with Ohm's Law produces the twelve formulas of the power wheel, which let you find any one of P, I, E, or R from any two of the others.

ToFind	Known:EandI	Known:IandR	Known:EandR	Known:Pandoneother
Power(P) watts ,	P=E×I	P=I²×R	P=E²÷R	—
Current(I) amperes ,	—	—	I=E÷R	I=P÷E;I=√(P÷R)
Voltage(E) volts ,	—	E=I×R	—	E=P÷I;E=√(P×R)
Resistance(R) ohms ,	R=E÷I	—	—	R=E²÷P;R=P÷I²

A 1,500 W portable heater is rated 120 V.

Current: I = P ÷ E = 1,500 W ÷ 120 V = 12.5 A

Resistance: R = E² ÷ P = (120 × 120) ÷ 1,500

R = 14,400 ÷ 1,500 = 9.6 Ω

Check: P = I² × R = 12.5² × 9.6 = 156.25 × 9.6 = 1,500 W

Checking the answer with a second formula from the wheel catches most calculator errors.

Power at Other Than Rated Voltage

A resistive load's resistance stays essentially constant, so its power changes with the square of the voltage (P = E² ÷ R). Doubling the voltage on a fixed resistance quadruples the power; cutting the voltage in half reduces the power to one- quarter. This explains a common field problem: a 240 V water heater or range element installed on a 208 V system produces only about three-quarters of its rated output.

A 4,800 W, 240 V water heater element is connected to 208 V.

Step 1 R = E² ÷ P = 240² ÷ 4,800 = 57,600 ÷ 4,800 = 12 Ω

Step 2 P = E² ÷ R = 208² ÷ 12 = 43,264 ÷ 12 = 3,605 W

Step 3 3,605 ÷ 4,800 = 0.75 (about 75% of rated output)

Current at 208 V: I = E ÷ R = 208 ÷ 12 = 17.3 A (not 20 A)

Always find the resistance at the rated values first, then apply the new voltage. Do not assume the current stays the same.

The same square relationship applies to current: line loss in a conductor equals I² × R. Halving the current reduces the conductor loss to one-quarter. This is why utilities transmit power at high voltage and low current, and why higher-voltage branch circuits and feeders (240 V instead of 120 V, 480 V instead of 208 V) deliver the same power with smaller conductors and less loss.

Energy and Energy Cost

Energy is power used over time. Utilities bill energy in kilowatt-hours (kWh): one kilowatt used for one hour. Because watts are a rate, a 100 W lamp uses the same energy in 10 hours as a 1,000 W heater uses in 1 hour: 1 kWh. Energy in kWh = (watts × hours) ÷ 1,000, and cost = kWh × the rate per kWh. For heating and cooling problems, 1 W equals approximately 3.412 BTU per hour.

A 1,500 W heater runs 8 hours per day for 30 days.

The utility rate is $0.14 per kWh.

Energy = 1,500 W × 8 h × 30 days ÷ 1,000 = 360 kWh

Cost = 360 kWh × $0.14 = $50.40

If the problem gives current and voltage instead of watts, find watts first: P = E × I.

Horsepower and Motor Efficiency

Motor output is rated in horsepower (hp), which is mechanical power delivered at the shaft. One horsepower equals 746 watts. A motor's nameplate horsepower is its output; the electrical input is always larger because some energy is lost as heat in the windings, core, and bearings, and as friction and windage. Efficiency is the ratio of output to input and is always less than 100%.

Efficiency = Output ÷ Input

Input = Output ÷ Efficiency

Output = Input × Efficiency

Losses = Input − Output

Use efficiency as a decimal (85% = 0.85). Dividing by a decimal less than 1 always makes the input larger than the output; if your input is smaller, the formula was inverted.

A 10 hp motor operates at full load with 85% efficiency.

Output = 10 hp × 746 W = 7,460 W

Input = Output ÷ Efficiency = 7,460 ÷ 0.85 = 8,776 W

Losses = 8,776 − 7,460 = 1,316 W (given off as heat)

At 230 V single-phase, unity PF: I = 8,776 ÷ 230 = 38.2 A

Theory questions use this method. For conductor and overcurrent sizing, the NEC requires motor full-load current from the tables in Article 430 rather than a calculated or nameplate value; motor circuits are covered in their own section.

A question that asks for the watts a motor consumes wants input (output ÷ efficiency). A question that asks what a motor delivers wants output (hp × 746). Many wrong answers are simply hp × 746 with the efficiency step left out.

AC Systems, Three-Phase Power, and Power Factor