Every table & formula on one sheet

TRAP: 240.4(D) caps small conductors: 14 AWG → 15A, 12 AWG → 20A, 10 AWG → 30A OCPD max — regardless of table ampacity.
TIP: For the exam, memorize the 75°C column. Most terminations are rated 75°C.

Adjusted ampacity = Table 310.16 value × adjustment factor TRAP: Neutrals carrying ONLY unbalanced current and equipment grounds do NOT count as current-carrying conductors.

Final ampacity = Table 310.16 value × temp correction × conduit fill adjustment TIP: When BOTH derating and temperature correction apply, multiply them together. Use the 90°C column for derating, then cap at the 75°C termination value.

Box Fill Counting Rules:

• Each conductor entering the box = 1 volume (based on its AWG)
• All internal clamps combined = 1 volume (largest conductor size)
• All grounds combined = 1 volume (largest ground size)
• Each yoke/strap (device) = 2 volumes (largest conductor connected to it)
• Pigtails & conductors that begin AND end inside the box = 0 volumes

Total box fill = (conductors × vol) + (1 × clamp vol) + (1 × ground vol) + (devices × 2 × vol) TRAP: Equipment grounds count as ONE entry regardless of how many ground wires. Pigtails do NOT count.

Common Conductor Areas (THHN/THWN-2):

AWG	Area (in²)	AWG	Area (in²)
14	0.0097	4	0.0824
12	0.0133	3	0.0973
10	0.0211	2	0.1158
8	0.0366	1	0.1562
6	0.0507	1/0	0.1855

Common EMT Internal Areas:

Trade Size	Total Area (in²)	40% Fill (in²)
½"	0.304	0.122
¾"	0.533	0.213
1"	0.864	0.346
1¼"	1.496	0.598
1½"	2.036	0.814
2"	3.356	1.342

Required conduit area = total conductor area ÷ fill percentage (40% for 3+) TIP: Quick method — add up all conductor areas, divide by 0.40, then pick the conduit whose total area is ≥ that result.

NEC recommends max 3% drop for branch circuits, 5% total (feeder + branch).

Single-Phase: VD = (2 × K × I × D) ÷ CM Three-Phase: VD = (1.732 × K × I × D) ÷ CM Where:

• K = resistivity constant: 12.9 (copper) or 21.2 (aluminum)
• I = load current in amps
• D = one-way distance in feet
• CM = circular mils of conductor

Common Circular Mil Values:

AWG	CM	AWG	CM
14	4,110	4	41,740
12	6,530	3	52,620
10	10,380	2	66,360
8	16,510	1/0	105,600
6	26,240	4/0	211,600

TIP: To find minimum conductor size, rearrange: CM = (2 × K × I × D) ÷ max VD

Problem: 120V, 16A load, 150 ft one-way, #10 Cu. What is the voltage drop?
VD = (2 × 12.9 × 16 × 150) ÷ 10,380 = 61,920 ÷ 10,380 = 5.96V %VD = 5.96 ÷ 120 = 4.97% — exceeds 3% recommendation Solution: Upsize to #8 Cu (16,510 CM): VD = 61,920 ÷ 16,510 = 3.75V (3.1%) — still over 3%. Use #6 Cu for margin.

Step 1 — General Lighting (220.12) ____ sq ft × 3 VA/sq ft = ____ VA Step 2 — Small Appliance Circuits (210.11(C)(1)) 2 circuits minimum × 1,500 VA = 3,000 VA Step 3 — Laundry Circuit (210.11(C)(2)) 1 circuit × 1,500 VA = 1,500 VA Step 4 — Apply Table 220.42 Demand Factors Total of Steps 1-3 = ____ VA
First 3,000 VA at 100% = 3,000 VA
Remainder ____ VA at 35% = ____ VA
Subtotal A = ____ VA Step 5 — Fixed Appliances (220.53)
List all fixed appliances (dishwasher, disposal, water heater, etc.). If 4+ appliances, apply 75% demand.
Total fixed appliances: ____ VA × 75% = ____ VA (Subtotal B) Step 6 — Range/Oven (Table 220.55)
Single range ≤12 kW = 8,000 VA demand.
Range demand = ____ VA (Subtotal C) Step 7 — Dryer (Table 220.54)
Single dryer: use nameplate or 5,000 VA minimum, whichever is larger.
Dryer = ____ VA (Subtotal D) Step 8 — HVAC (220.60)
Use the LARGER of A/C or heat — do not add both.
Larger of A/C or heat = ____ VA (Subtotal E) Step 9 — Total Service Load A + B + C + D + E = ____ VA total demand
____ VA ÷ 240V = ____ amps
Select next standard service size: ____ A TRAP: NEVER add A/C and heating together. Use the LARGER load only (220.60). TRAP: Use Table 220.55 demand (8 kW) for range, NOT nameplate value. TIP: Standard residential service sizes: 100A, 150A, 200A, 400A.

Neutral load = total demand MINUS:

• 240V-only loads (A/C, water heater, baseboard heat) — 0% neutral
• Range/dryer neutral — 70% of demand
• Over 200A: additional derating at 70% for portion over 200A

Range neutral = Table 220.55 demand × 0.70

Single-Phase, 115V/230V:

HP	115V FLC	230V FLC
½	9.8A	4.9A
¾	13.8A	6.9A
1	16A	8A
1½	20A	10A
2	24A	12A
3	34A	17A
5	56A	28A

Three-Phase, 208V/480V (Table 430.250):

HP	208V FLC	480V FLC
1	4.6A	2.0A
3	11.0A	4.8A
5	16.7A	7.6A
10	30.8A	14.0A
25	74.8A	34.0A
50	143A	65.0A

TRAP: ALWAYS use FLC from NEC tables, NEVER use motor nameplate amps. This is the #1 motor question trap.

Conductor sizing (430.22): Motor conductor ampacity ≥ FLC × 1.25 Branch-circuit OCPD (Table 430.52):

OCPD Type	Max % of FLC
Dual-element (time-delay) fuse	175%
Inverse-time circuit breaker	250%
Instantaneous-trip breaker	800% (certain motors)
Non-time-delay fuse	300%

Motor breaker = FLC × 2.50 (round UP to next standard size if needed) Overload protection (430.32): Overload = motor nameplate FLA × 1.15 (SF ≥ 1.15) or × 1.25 (SF < 1.15) Multiple motors on one feeder (430.24): Feeder conductor = largest motor FLC × 1.25 + sum of all other motor FLCs TRAP: Overloads use NAMEPLATE amps. Everything else uses TABLE FLC. Know the difference.

Location	Min. Clearance
Above finished grade (pedestrian only)	10 ft
Above residential driveways	12 ft
Above commercial/public roads	15 ft
Over 0-600V: above roads (truck traffic)	18 ft

TRAP: Residential driveway = 12 ft, NOT 10 ft. 10 ft is for sidewalks/yards only.

• SE conductors sized per load calculation — minimum #8 Cu or #6 Al
• Max 6 disconnects per service (230.71) — "six-throw rule"
• Each disconnect must be suitable for conditions, accessible, nearest point of entry
• Main bonding jumper required at service — only point where neutral bonds to ground

TRAP: Max 6 disconnects, not 6 circuits. A single 6-breaker panel counts as 6 disconnects.

GEC is sized based on the largest service-entrance conductor:

Service Conductor (Cu)	GEC (Cu)	GEC (Al)
2 AWG or smaller	8 AWG	6 AWG
1 AWG or 1/0	6 AWG	4 AWG
2/0 or 3/0	4 AWG	2 AWG
Over 3/0 to 350 kcmil	2 AWG	1/0 AWG
Over 350 to 600 kcmil	1/0 AWG	3/0 AWG
Over 600 to 1100 kcmil	2/0 AWG	4/0 AWG

TRAP: Table 250.66 uses SERVICE CONDUCTOR size. Table 250.122 uses OCPD RATING. Different inputs!

EGC is sized based on the overcurrent device (OCPD) rating:

OCPD Rating (Amps)	EGC — Cu	EGC — Al
15	14 AWG	12 AWG
20	12 AWG	10 AWG
30	10 AWG	8 AWG
60	10 AWG	8 AWG
100	8 AWG	6 AWG
200	6 AWG	4 AWG
300	4 AWG	2 AWG
400	3 AWG	1 AWG

TIP: EMT, RMC, and IMC all share 10 ft max. MC = 6 ft. NM/AC = 4.5 ft. Secured within 3 ft of boxes. TRAP: Support spacing and securing distance from boxes are different requirements.

Power Law: P = V × I   |   P = I² × R   |   P = V² ÷ R Ohm's Law: V = I × R   |   I = V ÷ R   |   R = V ÷ I Three-Phase Power: P = V × I × 1.732 × PF   (balanced 3φ)
I = P ÷ (V × 1.732 × PF) Continuous Load Sizing: Conductor ≥ 1.25 × continuous + 1.00 × non-continuous
OCPD ≥ 1.25 × continuous + 1.00 × non-continuous Transformer Sizing: I = VA ÷ V   (single-phase)
I = VA ÷ (V × 1.732)   (three-phase) Efficiency: Output = Input × Efficiency
Input = Output ÷ Efficiency Voltage Drop (Single-Phase): VD = (2 × K × I × D) ÷ CM   |   K(Cu) = 12.9, K(Al) = 21.2 Voltage Drop (Three-Phase): VD = (1.732 × K × I × D) ÷ CM

The numbers that show up on every C-10 exam:

• 125% — continuous load sizing (conductors AND OCPD)
• 80% — inverse: standard OCPD continuous rating = 80% of trip
• 3 VA/sq ft — dwelling general lighting load
• 1,500 VA — per small appliance / laundry circuit
• 8,000 VA — single range demand (≤12 kW)
• 5,000 VA — minimum dryer demand
• 70% — neutral demand for range
• 75% — demand factor for 4+ fixed appliances
• 6 ft — max wall receptacle spacing (no point >6 ft from outlet)
• 24 in — max countertop receptacle spacing
• 25 ohms — ground rod resistance threshold (add 2nd rod)
• $25,000 — contractor bond amount
• 8,000 hours — C-10 experience requirement (4 years)
• 70% — CSLB exam passing score