How Calculations Are Tested
Roughly a third of a journeyman exam is arithmetic with a code section behind it. The California General Electrician outline puts calculations inside its Installation section (66 of 100 questions) rather than in a separate block, so they arrive mixed in with the rest: a conductor to size, a box to fill, a motor breaker to pick, a dwelling service to total. The code is on screen and an online calculator is provided, but with under three minutes a question there is no time to learn a method at the terminal — only to confirm a value.
The twelve problems below are the ones that recur, each solved the way the 2023 NEC solves it, with the section to confirm on exam day and the wrong answer the distractors are built from. The reference tables themselves — ampacities, GEC and EGC sizes, box volumes — are collected in our California exam guide; how to find any of them in the on-screen code fast is in the NEC navigation guide. Practise on our free electrician questions, where every calculation shows its working.
The Constants
| Constant | Value | Where |
|---|---|---|
| Continuous load factor | 125% of the continuous load (3 hours or more) for conductors and OCPD | 210.19(A)(1), 210.20(A), 215.2(A)(1) |
| Small-conductor limits | 14 AWG 15 A · 12 AWG 20 A · 10 AWG 30 A (copper) | 240.4(D) |
| Standard OCPD sizes | 15, 20, 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150, 175, 200, 225, 250, 300, 350, 400 … | 240.6(A) |
| Bundling adjustment | 4–6 current-carrying conductors 80% · 7–9 70% · 10–20 50% | Table 310.15(C)(1) |
| Temperature correction, 75 °C conductors | 31–35 °C 0.94 · 36–40 °C 0.88 · 41–45 °C 0.82 · 46–50 °C 0.75 | Table 310.15(B)(1)(1) |
| Conduit fill | 1 conductor 53% · 2 conductors 31% · more than 2 40% | Chapter 9, Table 1 |
| Voltage-drop constant K | 12.9 copper · 21.2 aluminium (Ω·cmil/ft) | Industry convention; the NEC gives resistance in Chapter 9 Table 8 |
| Recommended voltage drop | 3% branch circuit, 5% feeder plus branch combined | 210.19(A) and 215.2(A) Informational Notes — a recommendation, not a requirement |
| Motor conductors | 125% of table full-load current — never nameplate | 430.22, Tables 430.248/430.250 |
| Motor branch OCPD (typical squirrel-cage) | Inverse-time breaker 250% · dual-element fuse 175% · non-time-delay fuse 300% | Table 430.52(C)(1) |
| Motor overloads | 125% of nameplate FLA (service factor ≥ 1.15 or rise ≤ 40 °C), else 115% | 430.32(A)(1) |
| Dwelling general lighting | 3 VA per square foot | 220.41 (2023) |
| Dwelling demand | First 3,000 VA at 100%, 3,001–120,000 VA at 35%, remainder 25% | Table 220.45 |
Problems 1–3: Sizing Conductors
1. A continuous load
A 208 V single-phase branch circuit supplies 36 A of continuous lighting. Copper THHN, 75 °C terminations. What conductor and breaker?
Minimum ampacity before adjustment: 36 × 1.25 = 45 A. Table 310.16, 75 °C copper: 8 AWG is 50 A — 8 AWG. Breaker: 210.20(A) also requires 125%, so at least 45 A — not a standard rating in 240.6(A) — giving 50 A, which the 50 A conductor supports. (When a conductor's ampacity itself falls between standard sizes, 240.4(B) allows the next size up for circuits up to 800 A.) Trap: 6 AWG, from remembering 8 AWG as "40 A" — its 60 °C rating.
2. Temperature correction and bundling together
Nine current-carrying 12 AWG THHN copper conductors share a raceway in a 42 °C space. What is each conductor's ampacity?
Correction and adjustment start from the conductor's own insulation rating — 90 °C for THHN — not the 75 °C termination column: 12 AWG at 90 °C is 30 A. Temperature (Table 310.15(B)(1)(1), 90 °C column, 41–45 °C): 0.87. Bundling (Table 310.15(C)(1), 7–9 conductors): 0.70. Ampacity: 30 × 0.87 × 0.70 = 18.3 A. The circuit may still be protected at 20 A only if the load does not exceed 18.3 A; otherwise the conductor must be upsized. Trap: starting from 25 A (75 °C) or 20 A (240.4(D)) and getting 15.2 or 12.2 A — the derating starts from the 90 °C value, and the 240.4(D) cap applies to the overcurrent device, not the derating base.
3. When the derated value governs
Same raceway, but the load is 22 A non-continuous. 12 AWG at 18.3 A is too small. Try 10 AWG: 40 A at 90 °C × 0.87 × 0.70 = 24.4 A — enough for 22 A, and 240.4(D) then limits the breaker to 30 A. 10 AWG on a 25 A or 30 A breaker. Always check the derated ampacity against the load and the termination column (10 AWG at 75 °C is 35 A, so the terminations are fine).
Problems 4–5: Box Fill and Conduit Fill
4. Box fill
A device box contains two 12/2 NM cables with ground (four 12 AWG insulated conductors, two 12 AWG bare EGCs), one duplex receptacle, and two internal cable clamps. Minimum box volume?
Count per 314.16(B): conductors 4 × 1 = 4; all EGCs together = 1; internal clamps (any number) = 1; the receptacle yoke = 2, based on the largest conductor connected to it (12 AWG). Total 8 × 2.25 in³ (Table 314.16(B)(1) for 12 AWG) = 18.0 in³. A 4 × 1½ in square box (21.0 in³) works; a 3 × 2 × 2½ in device box (12.5 in³) does not. Traps: counting each EGC separately, or counting the device as one.
5. Conduit fill from Chapter 9
Six 6 AWG THHN and two 10 AWG THHN copper conductors in EMT. Minimum trade size?
Table 5 areas: 6 AWG THHN 0.0507 in² × 6 = 0.3042; 10 AWG THHN 0.0211 × 2 = 0.0422; total 0.3464 in². More than two conductors, so 40% fill (Table 1). Table 4, EMT, 40% column: 1 in is 0.346 in² — 0.0004 short — so 1¼ in EMT (0.598 in²). Traps: reading the 100% column, or using Annex C, which only works when every conductor is the same size.
Problems 6–7: Voltage Drop
6. Find the drop
A 120 V, 20 A branch circuit runs 150 ft one way to an outdoor receptacle on 10 AWG copper. Percentage drop at full load?
VD = 2 × K × I × L ÷ CM = 2 × 12.9 × 20 × 150 ÷ 10,380 = 7.46 V, which is 7.46 ÷ 120 = 6.2% — twice the 3% recommended. Trap: forgetting the factor 2 (single-phase, out and back) and getting 3.1%.
7. Find the conductor
Same circuit; what copper size holds the drop to 3%?
Allowed drop: 120 × 0.03 = 3.6 V. CM = 2 × K × I × L ÷ VD = 2 × 12.9 × 20 × 150 ÷ 3.6 = 21,500 cmil. Chapter 9 Table 8: 8 AWG is 16,510 cmil (too small); 6 AWG is 26,240 — 6 AWG. And because the ungrounded conductors were increased for voltage drop, 250.122(B) requires the equipment grounding conductor to be increased in proportion: 12 AWG (6,530 cmil) × (26,240 ÷ 10,380) = 16,509 cmil → 8 AWG EGC. That second step is a question on its own.
Problems 8–9: A Complete Motor Circuit
8. Conductors, breaker and overloads
A 10 hp, 460 V, three-phase squirrel-cage motor, nameplate FLA 12.6 A, service factor 1.15, inverse-time circuit breaker, copper THHN at 75 °C.
- Conductors (430.22): 125% of the table current. Table 430.250, 10 hp at 460 V: 14 A. 14 × 1.25 = 17.5 A → Table 310.16 at 75 °C: 14 AWG (20 A) — and here 240.4(D) does not apply, because 240.4(G) sends motor circuits to Article 430.
- Branch-circuit short-circuit and ground-fault protection (430.52): inverse-time breaker, 250% of table current = 14 × 2.5 = 35 A → 35 A breaker. If 35 A did not hold the starting current, 430.52(C)(1) Exception 2 allows up to 400% for breakers where the calculated value is 100 A or less.
- Overloads (430.32(A)(1)): service factor 1.15, so 125% of the nameplate current: 12.6 × 1.25 = 15.75 A.
The pattern to hold onto: conductors and short-circuit protection use the table; overloads use the nameplate.
9. Several motors on one feeder
Feeder to three motors, table currents 34 A, 14 A and 7.6 A. Conductor ampacity (430.24): 125% of the largest plus the sum of the rest: 34 × 1.25 + 14 + 7.6 = 64.1 A → 4 AWG copper at 75 °C (85 A) is generous; 6 AWG (65 A) is the minimum. Feeder OCPD (430.62): the largest branch-circuit device (34 × 2.5 = 85 A → 90 A) plus the other motors' currents: 90 + 14 + 7.6 = 111.6 → not more than 110 A (round down here — 430.62 says "not greater than").
Problems 10–11: Dwelling Load Calculations
10. Standard method (220.40 – 220.55)
A 2,000 ft² house, 120/240 V single-phase, with a 12 kW range, a 5 kW dryer, a 4.5 kW water heater, 10 kW of electric space heating and a 5 kW air conditioner.
| Step | Working | VA |
|---|---|---|
| General lighting (220.41) | 2,000 × 3 | 6,000 |
| Small-appliance circuits (220.52(A)), two minimum | 2 × 1,500 | 3,000 |
| Laundry circuit (220.52(B)) | 1 × 1,500 | 1,500 |
| Subtotal, then Table 220.45 demand | 10,500 → 3,000 + (7,500 × 0.35) | 5,625 |
| Range (Table 220.55, Column C, one range ≤ 12 kW) | — | 8,000 |
| Dryer (220.54), 5,000 VA or nameplate, larger | — | 5,000 |
| Water heater (220.53 does not apply — fewer than four appliances) | 100% | 4,500 |
| Heating vs. air conditioning (220.60), larger of non-coincident loads | 10,000 vs 5,000 | 10,000 |
| Total | 33,125 |
33,125 ÷ 240 = 138 A → 150 A service (230.79(C) sets the 100 A minimum for a one-family dwelling; the calculated load sets the actual size). Traps: applying the 35% demand to the range and dryer (they have their own tables), adding both heating and cooling, or forgetting the laundry circuit.
11. Optional method (220.82)
Same house. General loads: 6,000 + 3,000 + 1,500 + 12,000 (range nameplate) + 5,000 + 4,500 = 32,000 VA. 220.82(B): first 10,000 at 100% plus the remainder at 40% = 10,000 + 8,800 = 18,800. Add the largest of the heating/cooling options per 220.82(C): with four or more separately controlled heating units 40% applies, but with fewer, central electric heating is at 65% — 10,000 × 0.65 = 6,500 — versus the air conditioner at 100%, 5,000; take 6,500. Total 25,300 ÷ 240 = 105 A → 110 A, so a 125 A or 150 A service. The optional method almost always comes out lower — which is why the exam asks for it by name; read which method the question wants.
Problem 12: A Range over 12 kW, and the Grounding Conductors
A single 15.5 kW range. Table 220.55 Note 1: for ranges over 12 kW up to 27 kW, increase the Column C value by 5% for each kW (or major fraction) above 12. 15.5 − 12 = 3.5 → 4 kW (a major fraction rounds up) → 20%. 8,000 × 1.20 = 9,600 VA; the branch circuit is 9,600 ÷ 240 = 40 A, so 8 AWG copper at 75 °C on a 40 A breaker, with 210.19(A)(3) allowing the neutral at 70% of the ungrounded conductors' ampacity.
For the 150 A service above with 1/0 AWG copper service conductors: the grounding electrode conductor to a water pipe or building steel is 6 AWG copper (Table 250.66, 1 or 1/0 AWG row), and to a rod it may stop at 6 AWG in any case (250.66(A)). The equipment grounding conductor for a 150 A feeder is 6 AWG copper (Table 250.122, 200 A row covers 110–200 A). Trap: reading Table 250.66 by the overcurrent device — it is keyed to the service conductor size — and Table 250.122 by the conductor size, when it is keyed to the overcurrent device.
Working Method on Exam Day
- Name the section first. "Motor conductors — 430.22, table current, 125%." If you cannot name it, open the code's contents to the article, not the search box.
- Identify the base current — table or nameplate, continuous or not, current-carrying conductor count, ambient — before any arithmetic.
- Round in the code's direction: conductors and branch OCPD up to the next standard size (240.4(B)); feeder OCPD for motors down (430.62); range kW major fractions up.
- Check the second constraint. A derated ampacity still has to satisfy the termination column; an upsized conductor drags the EGC with it; a 30 A cap on 10 AWG applies to the device, not the derating.
- Distrust an answer that matches a number in the question. The distractors are the nameplate current, the 60 °C column and the pre-demand total.
Frequently Asked Questions
Do I derate from the 75 °C or 90 °C column?
From the conductor's insulation rating — 90 °C for THHN/THWN-2 — then check that the result does not exceed the 75 °C (or 60 °C) termination ampacity, per 110.14(C) and 310.15(A). Derating from the 75 °C column gives an answer that is too small.
Is 3% voltage drop an NEC requirement?
No. The 3% branch and 5% total figures are Informational Notes to 210.19(A) and 215.2(A), which are recommendations. Some sections do require it — 647.4(D) for sensitive electronic equipment, and 695.7 for fire pumps — and energy codes such as California's Title 24 impose limits separately.
Why are motor conductors sized from the table and not the nameplate?
430.6(A)(1) requires Tables 430.247–430.250 for conductor, switch and short-circuit protection sizing so that a replacement motor of the same horsepower fits the circuit. Overload protection under 430.32 uses the nameplate, because it protects the specific motor installed.
When do I use the optional dwelling calculation?
220.82 may be used for a dwelling with a 120/240 V or 208Y/120 V service of at least 100 A, and the exam tells you which method it wants. The optional method usually gives a smaller load, so a question that names it expects the smaller answer.
How many calculation questions are on the California electrician exam?
DLSE does not publish a count; calculations are spread through the Installation section, which is 66 of the 100 questions. Expect roughly 25–35 questions that need arithmetic, most of them the twelve types worked above.
Sources
- NFPA 70, National Electrical Code, 2023 edition — Articles 210, 215, 220, 230, 240, 250, 310, 314 and 430, and Chapter 9 Tables 1, 4, 5 and 8. Section numbers cited are the 2023 numbering; Table 310.16 was 310.15(B)(16) before 2020, and dwelling lighting moved to 220.41 in 2023.
- California DLSE Electrician Certification Unit — the General Electrician examination outline and the open-book reference list.
- Every problem is QuizCram's own; no exam item or NEC text is reproduced, and the on-screen 2023 code governs on exam day.
