Electrical Theory and Fundamentals
The Electrical Theory and Fundamentals domain tests calculation fluency: Ohm's law, power, series and parallel circuits, AC impedance and power factor, three-phase line/phase relationships, and voltage drop. Questions often hide the real load behind a nameplate or a mixed circuit, so track units and phase.
Core relationships
- Ohm's law: V = I × R; power: P = V × I = I²R = V²/R.
- Series: one current path; resistances add; voltages add. Parallel: same voltage; currents add; 1/R_total = sum of reciprocals. In series the largest resistance dissipates the most power; in parallel the lowest resistance carries the most current.
- AC: Z = √(R² + X²), power factor = R/Z, true power P = V × I × PF, reactive power Q = V × I × sinθ, apparent power S = V × I.
- Balanced three-phase: wye V_line = √3 × V_phase, I_line = I_phase; delta I_line = √3 × I_phase, V_line = V_phase; P = √3 × V_line × I_line × PF.
- Voltage drop: single-phase VD = 2KID/CM; three-phase VD = √3KID/CM. K is about 12.9 for copper and 21.2 for aluminum.
Worked example
A 240 V single-phase 30 A load is 100 ft from the panel on 8 AWG Cu with 16,510 circular mils. VD = 2 × 12.9 × 30 × 100 / 16,510 = 4.69 V. Percent drop = 4.69/240 = 1.95%, below the common 3% branch-circuit target. If the same load were three-phase 240 V, VD = √3 × 12.9 × 30 × 100 / 16,510 = 4.06 V. The current per phase is 30 A only if the load is specified as line current; verify the nameplate before substituting.
Traps
- Using 120 V when the circuit is 240 V, or using line voltage where phase voltage belongs.
- Adding parallel resistances directly instead of using reciprocals.
- Treating kVA and kW as interchangeable when PF is not 1.0.
- Forgetting the return-path factor of 2 in single-phase voltage drop.
- Using nameplate current for a motor when the question asks for conductor sizing from Table 430.250.