Electrical Theory and Calculations
This domain tests the arithmetic and circuit theory behind nearly every calculation question: Ohm's law, power, series and parallel circuits, AC impedance, three-phase relationships, voltage drop, and the load and fill calculations that later domains build on. Expect numeric answers with units, and expect just enough data to make a wrong choice tempting.
Core relationships
- Ohm's law: V = I × R; power: P = V × I = I²R = V²/R.
- Single-phase power: P = V × I × pf. Three-phase: P = √3 × V_L × I_L × pf.
- Series: one current path, current is common, voltages add, R_T = R1 + R2 + ...
- Parallel: voltage is common, currents add, 1/R_T = 1/R1 + 1/R2 + ...
- AC impedance: Z = √(R² + X²); X_L = 2πfL; X_C = 1/(2πfC).
- Voltage drop: single-phase VD = 2KIL/cmil; three-phase VD = √3KIL/cmil. K = 12.9 for copper and 21.2 for aluminum at typical conditions.
Worked example
A 20 A, 240 V single-phase load is 150 ft from the panel. Use 8 AWG copper (16,510 cmil). VD = 2 × 12.9 × 20 × 150 / 16,510 = 77,400 / 16,510 = 4.69 V. Percent drop = 4.69 / 240 × 100 = 1.95%, acceptable for a branch circuit under the 3% recommendation. With 10 AWG (10,380 cmil), VD = 2 × 12.9 × 20 × 150 / 10,380 = 7.46 V = 3.1%, which exceeds the recommendation. The reasoning: longer or smaller conductors raise drop; size voltage drop from actual load current, not breaker rating.
Traps that cost marks
- Forgetting the 2 in single-phase voltage drop or the √3 in three-phase voltage drop.
- Using 120 V when the load is line-to-line, or mixing line and phase values in a wye system.
- Applying a power factor to a purely resistive load.
- Using motor nameplate current where Article 430 requires table values for conductors and short-circuit protection.
- Derating for temperature and then forgetting bundle adjustment, or applying adjustments in the wrong order.
- Rounding too early; carry decimals and round only at the end.