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DE Electrician study materials
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DE Electrician: Electrical Theory and Principles

DE Electrician · Work through your exam, one topic at a time.

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Electrical Theory and Principles

This domain tests the physical math behind code rules: Ohm's law, power, series and parallel circuits, AC reactance and impedance, power factor, three-phase relationships, transformers, and voltage drop. Expect calculation items and concept items that ask why a conductor, overcurrent device, or neutral behaves as it does.

Core relationships

  • Ohm's law: E = I × R; I = E/R; R = E/I.
  • DC power: P = E × I; P = I²R; P = E²/R. Keep units consistent: volts, amperes, ohms, watts.
  • Series: one path; current is the same; voltages add; R_total = R1 + R2 + ...
  • Parallel: voltage is the same; branch currents add; 1/R_total = 1/R1 + 1/R2 + ...
  • AC: 60 Hz sine wave; RMS values are used for power and code calculations; peak = RMS × √2.
  • Inductive reactance X_L = 2πfL; capacitive reactance X_C = 1/(2πfC); impedance Z = √(R² + (X_L − X_C)²).
  • Power factor = P/S = cos θ. True power is in watts; apparent power is in volt-amperes; reactive power is in VAR.
  • Voltage drop: single-phase VD = 2KID/CM; three-phase VD = √3KID/CM, where K is the conductor constant, I is amperes, D is one-way distance, and CM is circular mils.

Three-phase and transformers

For a balanced three-phase load, P = √3 × V_L × I_L × PF. In a wye, line voltage = √3 × phase voltage and line current = phase current. In a delta, line voltage = phase voltage and line current = √3 × phase current. Transformer turns ratio: V_p/V_s = N_p/N_s; current ratio is inverse; the kVA rating does not change with voltage ratio.

Worked example

A 240-volt single-phase load draws 30 amperes at 0.80 power factor. True power = 240 × 30 × 0.80 = 5,760 W. Apparent power = 240 × 30 = 7,200 VA. Reactive power = √(7,200² − 5,760²) = 4,320 VAR. If the same 5,760 W load were balanced three-phase at 240 V and 0.80 PF, line current would be I = P/(√3 × V × PF) = 5,760/(1.732 × 240 × 0.80) ≈ 17.3 A.

Traps

  • Mixing RMS and peak values.
  • Using the single-phase formula on a three-phase load.
  • Forgetting power factor when converting watts to amperes.
  • Adding resistances directly in parallel.
  • Treating transformer kVA as fixed current at every voltage.
  • Ignoring the return path in voltage-drop calculations.