Exam Cheat Sheet · Quick Reference
Oklahoma Limited Electrical Contractor
Oklahoma · PSI Services Contractor
Oklahoma State Portion 100 questions
Key Distinctions
In a series circuit total resistance is the sum of the individual resistances and total current equals the current in any other part of the circuit (IT = I1 = I2 = I3), while in a parallel circuit the voltage is the same across each branch, total current is the sum of the branch currents, and total resistance is always less than the resistance of any branch.
Ugly's Electrical References 2020, Series Circuits, Book Page 3Both use the noncontinuous load plus 125 percent of the continuous load, but 210.20(A) sets the minimum rating of the branch-circuit overcurrent device (with no noncontinuous load, 16 A × 1.25 = 20 A), while 215.2(A)(1) sets the minimum feeder conductor ampacity (80 A noncontinuous + 60 A × 1.25 = 155 A).
NEC 2023 Section 210.20(A)Branch-circuit conductors to a single motor are sized at 125 percent of that motor's full-load current, while feeder conductors are sized at 125 percent of the largest motor's FLC plus the sum of the FLCs of the other motors.
NEC 2023 Section 430.22, Table 430.248Dwelling units use a minimum unit load of 3 volt-amperes per square foot, while retail occupancies use 1.9 volt-amperes per square foot (offices 1.3, automotive/dormitory/library 1.5, manufacturing/religious 2.2).
NEC 2023 Section 220.41A disconnect serving as a one-family dwelling unit's supply must be rated at least 100 amperes, but a feeder disconnect at a separate structure such as a detached garage (more than two circuits) requires a rating of not less than 60 amperes (one-circuit = 15 A minimum, two-circuit = 30 A minimum).
NEC 2023 Section 225.39(D)Per NEC 305.5, shielded or lead-covered conductors shall not be bent to a radius less than 12 times the overall diameter, while nonshielded conductors have a minimum of 8 times the overall diameter.
NEC 2023 Section 305.5Direct burial under general conditions requires 750 mm (30 in.) of cover, while circuits over 22 kV through 40 kV installed in RMC/IMC (Column 3) require only 150 mm (6 in.).
NEC 2023 Table 305.15(A)Per NEC 315.40, Type MV cable terminated in equipment, pull boxes, or vaults must be secured within 1.5 m (5 ft) of terminations, and supports elsewhere may not exceed 1.8 m (6 ft) apart.
NEC 2023 Section 315.40Per NEC 250.28(B), when a main or system bonding jumper is a screw only, the screw shall be identified with a green finish (not red) visible with the screw installed.
NEC 2023 Section 250.28(B)NEC 358.26 (renumbered 358.24(B) in NEC 2023 and 2026) limits EMT to 360 degrees of bends between pull points; because a conduit body is a pull point, adding one lets each section on either side have up to 360 degrees of bends.
NEC 2023 Section 358.24(B)Crane conductors use the nameplate FLA of the largest motor (or group for a single crane motion) plus 50 percent of the next largest, while general motor feeders use 125 percent of the largest motor FLC plus 100 percent of the others.
NEC 2023 Section 610.14(E)(2)NEC 220.5(A) lists 120, 120/240, 208Y/120, 240, 347, 480Y/277, 480, 600Y/347, and 600 volts as nominal voltages for load calculations; '277/480' is not a listed designation.
NEC 2023 Section 220.5(A)Key Terms
Formulas to Know
RT = R1 + R2 + R3 (e.g., 20 + 25 + 15 = 60 ohms)ET = E1 = E2 = E3; each branch I = ET ÷ R; IT = I1 + I2 + I3 (120/60 + 120/80 + 120/120 = 2 + 1.5 + 1 = 4.5 A); total resistance is always less than any branchP = E × I (120 V × 10 A = 1200 W); I = P ÷ E; R = E ÷ I, or R = E² ÷ P (14,400 ÷ 1200 = 12 ohms)KW = (E × I × PF × 1.73) ÷ 1000 (460 × 52 × 0.80 × 1.73 ÷ 1000 ≈ 33 kW); 1.73 ≈ √3I = (KW × 1000) ÷ (E × PF × 1.73); I = (HP × 746) ÷ (E × %EFF × PF × 1.73)Cost = Watts × Hours Used × Rate per kWh ÷ 1000 (5000 W × 360 h × $0.15 ÷ 1000 = $270 per month)Branch OCPD rating ≥ noncontinuous load + 125% × continuous load (continuous only: 16 × 1.25 = 20 A); feeder ampacity ≥ noncontinuous + 125% × continuous (80 + 60 × 1.25 = 155 A; 30 + 100 × 1.25 = 155 A)Branch-circuit conductors = 125% × motor FLC from the FLC tables (28 A × 1.25 = 35 A); feeder = (125% × largest motor FLC) + sum of other motor FLCs (65 × 1.25 + 40 = 121.25 A; 15.2 × 1.25 + 15.2 + 15.2 = 49.4 A)Largest motor (or group for a single crane motion) nameplate FLA + 50% of next largest (40 + 0.5 × 30 = 55 A; 40 + 0.5 × 20 = 50 A)Receptacle demand = first 10 kVA at 100% + remainder at 50% (10,000 + 8,000 × 0.5 = 14,000 VA); lighting = area × unit load (5,000 sq ft × 1.9 VA = 9,500 VA)