3.3 Electrical-Electronic Fundamentals & Theory

Why This Matters on the Exam

The NEC tells you what to install; theory tells you why the rules exist and lets you compute values the tables do not list. The Virginia Journeyman blueprint assigns 2 questions to Electrical-Electronic Fundamentals, but the formulas appear in every load calculation, voltage-drop, and transformer question on the rest of the exam. Skip this section and you will not be able to verify any answer arithmetically.

Ohm's Law

Ohm's law: E = I × R, where E is voltage in volts, I is current in amperes, and R is resistance in ohms. Rearrangements:

• I = E / R
• R = E / I

If a 120V circuit draws 12 A, the total resistance is R = 120 / 12 = 10 Ω. If the same circuit's resistance rises to 12 Ω (a loose connection), current falls to 10 A — the drop in current is a troubleshooting signal.

Power Formulas

The three power equations:

• P = I × E (power = current × voltage)
• P = I² × R (power lost as heat in resistance)
• P = E² / R (power from voltage across resistance)

A 120V heater drawing 12 A dissipates P = 12 × 120 = 1440 W. A 10 Ω resistor at 120V dissipates P = 120² / 10 = 1440 W — same answer, different entry point.

Series Circuits

In a series circuit:

• Resistance adds: R_total = R1 + R2 + R3 + ...
• Current is the same through every element.
• Voltage divides across each resistor proportional to its resistance.

Three resistors (4 Ω, 6 Ω, 10 Ω) in series across 120 V: R_total = 20 Ω, I = 120/20 = 6 A. Voltage across the 10 Ω resistor = 6 × 10 = 60 V. Christmas-tree mini-lights are a classic series string — one open element kills the whole string.

Parallel Circuits

In a parallel circuit:

• Voltage is the same across every branch.
• Current divides among branches (inversely to resistance).
• Conductance adds: 1/R_total = 1/R1 + 1/R2 + 1/R3 + ...

Two 10 Ω resistors in parallel: 1/R_total = 1/10 + 1/10 = 0.2, so R_total = 5 Ω — half of either resistor. A practical shortcut for two resistors: R_total = (R1 × R2) / (R1 + R2) = (10 × 10) / (10 + 10) = 100/20 = 5 Ω.

Series vs. Parallel at a Glance

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The series loop carries 6 A through every resistor (20 Ω total); each parallel branch carries 12 A (120/10), for 24 A total at the source — the parallel circuit draws four times the series current.

Voltage Drop Calculation

The NEC recommends (Informational Notes, not enforceable) ≤3% voltage drop on branch circuits and ≤5% combined feeder + branch. The formula:

• 1-phase: Vd = (2 × L × I × R) / 1000
• 3-phase: Vd = (√3 × L × I × R) / 1000

Where L = one-way length in feet, I = load current in amperes, R = conductor resistance in Ω per 1000 ft (Chapter 9, Table 8 for copper; use AC values at 75°C for typical ampacity).

Worked Example: Voltage Drop and Parallel Resistance

Part A — Voltage drop. A 208V, 3-phase feeder carries 100 A for 150 ft through 1/0 AWG aluminum (≈0.201 Ω/kft at 75°C). Voltage drop:

Vd = (√3 × 150 × 100 × 0.201) / 1000 = (1.732 × 150 × 100 × 0.201) / 1000 ≈ 5.22 V

Percentage drop = 5.22 / 208 ≈ 2.5% — within the 3% feeder recommendation.

Part B — Parallel resistance. Two loads, 6 Ω and 12 Ω, share a 120V source. R_total = (6 × 12) / (6 + 12) = 72/18 = 4 Ω. Total current = 120/4 = 30 A; individual currents are 20 A (120/6) and 10 A (120/12), which sum to 30 A — Kirchhoff's current law verified.

AC vs. DC

Direct current (DC) flows in one direction at constant magnitude (batteries, solar PV). Alternating current (AC) reverses direction periodically; US power operates at 60 Hz (60 cycles/second). AC permits transformers to step voltage up for transmission and down for use; DC cannot be transformed directly. The NEC addresses AC and DC throughout — pay attention to which article applies (e.g., Article 690 for solar PV, Article 692 for fuel cells).

Single-Phase vs. Three-Phase

• Single-phase, 2-wire, 120V — one ungrounded conductor and one grounded (neutral) conductor; serves small loads.
• Single-phase, 3-wire, 120/240V — two ungrounded conductors 180° out of phase plus a neutral; common in dwellings (ranges, dryers use 240 V, lighting uses 120 V).
• Three-phase, 4-wire, 208Y/120V — three ungrounded conductors 120° apart plus a neutral; common in commercial buildings.
• Three-phase, 4-wire, 480Y/277V — common in industrial and large commercial; 277 V line-to-neutral serves lighting.

Three-phase delivers power more efficiently (three conductors carry the power of six) and produces a constant total power (no 120 Hz pulsation as in single-phase).

Power Factor and Efficiency

Power factor (PF) = W / VA, ranging 0 to 1.0. Resistive loads (heaters) have PF = 1.0; inductive loads (motors, ballasts) have PF < 1.0. A 10 kW motor at PF 0.8 draws 10,000 / 0.8 = 12,500 VA — the conductors and overcurrent device must carry the 12,500 VA equivalent current, not the 10 kW.

Efficiency (η) = P_out / P_in. A 5 HP motor (3730 W output) drawing 4500 W input has η = 3730/4500 ≈ 0.83 (83%). The 770 W difference is lost as heat.

Transformers: Turns Ratio

For an ideal transformer (no losses):

• V_primary / V_secondary = N_primary / N_secondary (turns ratio)
• I_primary × V_primary = I_secondary × V_secondary (power in = power out)

A 480V:120V transformer has a 4:1 turns ratio. If the secondary delivers 40 A at 120V (4800 VA), the primary draws 10 A at 480V (also 4800 VA). The VA is constant; only voltage and current trade. Real transformers have losses (core + copper) and efficiency typically 95–99%.