Electrical Calculations and Theory
Core relationships
Nearly every calculation question reduces to Ohm's law (E = I × R) and the power equation (P = E × I). For single-phase loads, I = VA ÷ E; for balanced three-phase loads, I = VA ÷ (1.732 × E_LL). Apparent power in kVA equals kW ÷ power factor, so a load's current rises as power factor falls. Conductors in series add resistance; in parallel add the reciprocals. Know that conductor resistance increases with length and temperature, which is why voltage drop grows with distance: VD (single-phase) = 2 × K × I × L ÷ CM, and VD (three-phase) = 1.732 × K × I × L ÷ CM.
Code-driven math
Many questions are arithmetic wrapped in a Code rule. Branch-circuit and feeder conductors for continuous loads are sized at 125% of the load; motor branch conductors use 125% of Table 430.250 full-load current; storage-type water heaters and fixed space heaters use 125%; the standard method (Article 220) applies demand factors to ranges, dryers, and receptacle outlets, while the optional method (220.82/220.84) uses a single demand factor on total connected load. Box fill (314.16) counts each conductor by Table 314.16(B) volume, all equipment grounding conductors as one, each yoke as two, and internal clamps as one.
Traps
Mixing 1.732 into single-phase problems, using nameplate current where the Code demands table values, forgetting the 125% multiplier, ignoring power factor, and using 120 V when the circuit is 240 V. Also watch the difference between volt-amperes and watts.
Worked example
A 25 kW, 480 V, three-phase heater at unity power factor: I = 25,000 ÷ (1.732 × 480) = 30.1 A. At 125% for continuous duty, 37.6 A, so 8 AWG copper (50 A at 75 °C) with a 40 A breaker is adequate.