Read and apply the National Electrical Code: navigation, definitions, adoption, conductor sizing, voltage drop, load calculations, box fill, grounding, GFCI/AFCI, motors and exam prep.

Electrical Code & Installations

## Why it matters

A conductor that is too small for its load overheats. Insulation ages, cracks and eventually fails, and that is how many electrical fires start. Sizing the conductor correctly is one of the most frequent calculations an electrician does, and one of the most tested on the journeyman exam. The steps are always the same: find the load, apply the continuous-load rule, look up the ampacity, then correct and adjust it for the real conditions.

> **Awareness level only.** This lesson is paper and calculator practice. Selecting and installing conductors on a real job is done under a licensed electrician's supervision, with the code edition your jurisdiction has adopted.

## Step 1: the load

Start from the load the circuit will carry. If it is **continuous** (maximum current for three hours or more), the general rule is to size the conductor at **125%** of the continuous load plus 100% of any non-continuous load. A 32 A continuous EV charger load needs a conductor good for at least 32 × 1.25 = 40 A before any corrections.

## Step 2: the base ampacity

Ampacity tables in Article 310 give a **base ampacity** for each conductor size (AWG or kcmil), material (copper or aluminium) and insulation temperature rating (60 °C, 75 °C, 90 °C), under standard conditions: a set ambient temperature and no more than three current-carrying conductors together.

Two limits apply before you even correct anything:

- **Terminal ratings (110.14(C)).** The final ampacity cannot exceed what the equipment terminals are rated for. Many terminals are rated 75 °C, and smaller equipment may be rated only 60 °C. You may use the 90 °C column as the **starting point for derating**, but the result must not exceed the ampacity at the terminal rating.
- **Small-conductor rule (240.4(D)).** For the smallest common sizes (14, 12 and 10 AWG copper), the overcurrent device is capped at specific values regardless of the table. Look these up in your code book.

## Step 3: correct for temperature

Hotter surroundings mean less room for the conductor's own heat. Article 310 gives **temperature correction factors**: in ambient temperatures above standard, multiply the base ampacity by a factor less than 1. Conductors on rooftops or in hot attics are the classic case.

## Step 4: adjust for bundling

When more than three **current-carrying conductors** share a raceway or cable for a significant length, they cannot shed heat as well. Article 310 gives **adjustment factors** that reduce the ampacity as the count rises. Neutrals that carry only unbalanced current and equipment grounding conductors usually do not count, but there are exceptions, so read the section.

## The formula

All of it combines into one line:

```text
adjusted ampacity = base ampacity × temperature correction factor × adjustment factor
```

The conductor is acceptable when the adjusted ampacity is at least the required ampacity from Step 1 (and the terminal and overcurrent limits are met).

## Worked example (illustrative values)

The numbers below are **practice values made up for this course**, chosen to look realistic. They are **not** the values in the NEC tables, which are copyrighted and change between editions. On the job and in the exam, use the tables in your code book.

A 30 A continuous load runs in a raceway with six current-carrying conductors through a space that runs hot.

- Required ampacity: 30 × 1.25 = **37.5 A**.
- Illustrative temperature factor: **0.91**. Illustrative adjustment factor for 4–6 conductors: **0.80**. Combined: 0.91 × 0.80 = 0.728.
- Try a conductor with an illustrative base ampacity of 38 A: 38 × 0.728 = 27.7 A. Too small.
- Try the next size up, illustrative base 52 A: 52 × 0.728 = 37.9 A. That is at least 37.5 A, so it passes this check.

Then confirm the terminal limit (the conductor's ampacity at the terminal temperature rating must also carry 37.5 A) and choose an overcurrent device that protects the conductor.

## Common mistakes

- Forgetting the 125% for continuous loads, or applying it twice.
- Applying correction and adjustment to a column the terminals cannot use, then not checking the terminal limit.
- Counting equipment grounding conductors as current-carrying.
- Using a factor table from another edition.
- Treating the derated value as the breaker size without checking the overcurrent rules in Article 240.

## Quick recap

Required ampacity comes from the load (125% if continuous). Adjusted ampacity = base × temperature factor × bundling factor. Pick the smallest conductor whose adjusted ampacity meets the requirement, then check terminal ratings and overcurrent limits. Always take the real values from your edition's tables ([NFPA 70](https://www.nfpa.org/70); checked on 2026-10-23).

## Try it

Open the **Do** tab. The conductor picker is written for you except for one line: the adjusted-ampacity formula.


Size a conductor for a continuous load with temperature correction and bundling adjustment.

Conductor sizing and ampacity

Derating of Conductors Explained

Ampacity Table, NEC 2020 - [310.15], (13min:26sec)

## Why it matters

Every conductor has resistance, so current flowing through it loses some voltage along the way. On a long run, a motor at the far end may get noticeably less than 120 V or 240 V: it runs hot, lights dim, and electronics misbehave. Voltage drop is why a conductor that passes the ampacity check can still be too small for a long run, and it is a common calculation on the job and in exams.

> **Awareness level only.** This is paper and calculator practice. Choosing conductors for real installations is done under a licensed electrician's supervision.

## Is it a code rule?

Mostly, no. For general branch circuits and feeders, the NEC's voltage-drop guidance appears in **informational notes** (for example in Article 210 for branch circuits and Article 215 for feeders). Those notes suggest a maximum of about **3%** on a branch circuit and **5%** total for feeder plus branch circuit. Because they are informational notes, they are not enforceable on their own. Some specific articles, such as those for certain sensitive equipment and fire pumps, and many engineering specifications and energy codes, do make limits mandatory. Good practice is to design within the 3% / 5% guidance anyway.

## The formula

The standard trade formula for a **single-phase** circuit is:

```text
VD = (2 × K × I × L) / CM
```

- **VD**: voltage drop in volts.
- **K**: the resistance of a one-mil-foot of conductor in ohms (about **12.9** for copper and **21.2** for aluminium at typical operating temperature; these are trade approximations).
- **I**: the load current in amps.
- **L**: the **one-way** length of the run in feet. The 2 accounts for the current going out and coming back.
- **CM**: the conductor's area in **circular mils**. Chapter 9 of the NEC lists conductor properties; for example 12 AWG is about 6,530 circular mils and 10 AWG is about 10,380.

For **three-phase** circuits, replace the 2 with √3 (about 1.732).

Percent drop is:

```text
% VD = VD / source voltage × 100
```

## Worked example

A 120 V, 16 A load sits 100 ft (one way) from the panel, on 12 AWG copper.

```text
VD = (2 × 12.9 × 16 × 100) / 6,530
   = 41,280 / 6,530
   ≈ 6.32 V
% VD = 6.32 / 120 × 100 ≈ 5.3 %
```

That is well over the 3% suggestion. Try 10 AWG:

```text
VD = 41,280 / 10,380 ≈ 3.98 V
% VD ≈ 3.3 %
```

Closer, but still above 3%. You might go up to 8 AWG, shorten the run or split the load. Note that the ampacity check alone would have accepted 12 AWG; voltage drop is what made the difference.

## Solving for the conductor size

Rearranged, the same formula gives the **minimum circular mils** for a target drop:

```text
CM = (2 × K × I × L) / allowed VD
```

For a 3% limit on 120 V, the allowed drop is 3.6 V. With the example above: CM = 41,280 / 3.6 ≈ 11,467 circular mils. 10 AWG (10,380) is just short, so the next size up meets it.

## Why it also matters for safety

Voltage drop is not just about performance. A motor starved of voltage draws more current to make the same power, which heats its windings. Long runs also raise the impedance of the fault path, which slows how quickly a breaker trips on a ground fault. That is why long runs often need a larger equipment grounding conductor too (Article 250 has a rule for increasing it in proportion).

## Common mistakes

- Using the round-trip length for L and then also multiplying by 2.
- Using 2 instead of 1.732 on three-phase.
- Mixing copper and aluminium values of K.
- Forgetting to convert the volt drop to a percentage of the **source** voltage.
- Treating the 3% / 5% guidance as optional good-to-have on long motor or pump circuits, where it really matters.

## Quick recap

Single-phase VD = 2KIL / CM; three-phase uses 1.732 instead of 2. Divide by the source voltage for percent. The NEC's 3% / 5% guidance lives in informational notes, but good design follows it, and long runs often need a bigger conductor than the ampacity check alone suggests. Use conductor areas from Chapter 9 of your edition ([NFPA 70](https://www.nfpa.org/70); checked on 2026-10-23).

## Try it

Open the **Do** tab. The calculator has every function but one: write the single-phase voltage-drop formula.


Compute the voltage drop on a branch-circuit run and choose a conductor that keeps it within 3%.

Voltage drop

Everything you need to know to solve Voltage Drop Calculations!!

## Why it matters

Before a house gets its electrical service, someone has to decide how big that service must be: 100 A, 150 A, 200 A or more. Too small, and the main breaker trips or the service overheats; too big, and the owner pays for copper and equipment they never use. The NEC gives a step-by-step **load calculation** for this. Permit offices often ask for it, and the journeyman exam almost always includes one.

> **Awareness level only.** This lesson teaches the method on paper. Sizing a real service is done by or under a licensed electrician or engineer, using the edition and amendments your jurisdiction enforces.

## Where it lives (and why the edition matters)

In the 2023 edition, branch-circuit, feeder and service load calculations are in **Article 220**. The 2026 edition reorganised Chapter 1 and moved these calculations into a new Article 120. That is a good example of why you must check which edition is adopted where you work. The method below follows the **standard method** for a single dwelling. There is also an **optional method** for dwellings that is often simpler and gives a smaller result; your exam may ask for either.

## The standard method, step by step

The calculation works in volt-amperes (VA), and the result is divided by the service voltage (240 V for a typical single-phase home) to get amps.

1. **General lighting and receptacles.** Multiply the dwelling's floor area (outside dimensions, not counting open porches, unfinished spaces not adaptable for future use, or garages) by the **unit load** in VA per square foot from the code's table.
2. **Small-appliance and laundry circuits.** Add a fixed VA amount for each required kitchen small-appliance branch circuit (at least two) and for the laundry circuit.
3. **Apply the lighting demand factors.** Not every light and receptacle is used at once, so the total from steps 1 and 2 is reduced in tiers: the first block of VA counts in full, the next block counts at a lower percentage, and anything above that at a lower percentage still.
4. **Fixed appliances.** Water heater, dishwasher, disposal and similar. When there are four or more, a demand factor may apply.
5. **Cooking equipment.** A household range is taken from its own table rather than its nameplate.
6. **Dryer.** At least a minimum VA or the nameplate, whichever is larger.
7. **Heating or air conditioning.** Use the **larger** of the two, because they do not run at the same time (non-coincident loads).
8. **Total** the VA and divide by 240 V for the minimum service in amps, then select the next standard service size.

## Worked example (illustrative values)

The unit load, appliance amounts and demand percentages below are **illustrative practice values** that follow the shape of the standard method. They are not a copy of any code table: use the tables in your edition for real work.

A 2,000 sq ft house, illustrative values:

| Item | VA |
|---|---|
| General lighting: 2,000 sq ft × 3 VA | 6,000 |
| Two small-appliance circuits × 1,500 VA | 3,000 |
| Laundry circuit | 1,500 |
| **Subtotal before demand** | **10,500** |

Illustrative demand tiers: first 3,000 VA at 100%, the next VA up to 120,000 at 35%, the rest at 25%.

```text
first 3,000 VA × 100%           = 3,000
remaining 7,500 VA × 35%        = 2,625
lighting after demand           = 5,625 VA
```

Add the other loads (illustrative):

| Item | VA |
|---|---|
| Lighting after demand | 5,625 |
| Range (illustrative table value) | 8,000 |
| Dryer | 5,000 |
| Larger of heat (0) or air conditioning (5,000) | 5,000 |
| Water heater | 4,500 |
| **Total** | **28,125** |

```text
28,125 VA / 240 V ≈ 117 A
```

The minimum is about 117 A, so a 125 A or (more commonly) 150 A or 200 A service would be chosen, leaving room for future loads such as an EV charger or heat pump.

## Why demand factors are tiered

The first few thousand VA of lighting and receptacles in any home are likely to be in use together (lights, fridge on a small-appliance circuit, TV). As the house grows, the chance that every outlet is loaded at once drops. Tiered demand factors capture that. The same idea, with different numbers, applies to cooking equipment and to groups of fixed appliances.

## Common mistakes

- Using inside dimensions or including the garage in the floor area.
- Applying the lighting demand factor to the range, dryer or air conditioner.
- Adding both heat and air conditioning instead of the larger one.
- Dividing by 120 V instead of 240 V for a 120/240 V service.
- Forgetting EV chargers, hot tubs and heat pumps, which now often dominate a home's load.
- Copying values or article numbers from a different edition (the 2026 edition moved the calculations to Article 120).

## Quick recap

A dwelling load calculation adds general lighting (area × unit load), small-appliance and laundry circuits, applies tiered demand factors to that group, adds fixed appliances, range, dryer and the larger of heat or cooling, and divides by 240 V. Check the article and table values in your adopted edition ([NFPA 70](https://www.nfpa.org/70); checked on 2026-10-23).

## Try it

Open the **Do** tab. The load calculator is complete except for the tiered lighting demand factor.


Run a basic standard-method service load calculation for a single-family dwelling.

Dwelling load calculations

NEC Dwelling Service Calculation Example

## Why it matters

Stuff too many wires into a box and you get crushed insulation, loose splices and heat with nowhere to go. Article 314 limits how many conductors, devices and fittings a box may hold, and Chapter 9 limits how full a conduit may be. Inspectors check both at rough-in, and both are favourite exam questions because they are pure arithmetic once you know what counts.

> **Awareness level only.** This lesson is paper and calculator practice. Installing boxes and pulling wire is done under a licensed electrician's supervision with your employer's training.

## Box fill: the idea

Every box has a **volume** in cubic inches, either marked on the box or listed in a table of standard boxes in Article 314. Each item inside the box "uses up" a set volume, based on the size of the conductors. The total used must not exceed the box's volume.

```text
required volume = allowance units × volume per conductor (for the size used)
box OK if: required volume ≤ box volume
```

The code's table gives the **volume allowance per conductor** for each AWG size. Look it up in Table 314.16(B) of your edition.

## What counts (the counting rules)

Section 314.16(B) tells you how many "allowance units" each item uses. In plain English:

- **Each conductor** that starts, ends or is spliced in the box: **1** unit. A conductor that passes through without a splice: 1 unit.
- **Pigtails and conductors that never leave the box**: 0.
- **All equipment grounding conductors together**: **1** unit (based on the largest). Some editions add a quarter unit for each additional EGC beyond the first four; check your edition.
- **Each device yoke** (a receptacle or switch strap): **2** units, based on the largest conductor connected to it.
- **All internal cable clamps together**: **1** unit.
- **Each support fitting** (fixture stud or hickey): **1** unit.
- **Small fixture wires** (from a luminaire canopy) may not count; there are conditions.

When conductors of different sizes are in the box, each item uses the allowance for its own size (devices and grounds use the largest connected or present).

## Worked example (illustrative values)

The volumes below are **illustrative practice values**; take the real ones from your code book.

A switch box marked **18.0 in³** holds:

- two 12 AWG cables, each with a black, a white and a bare ground,
- one duplex receptacle,
- internal cable clamps.

Count:

| Item | Units |
|---|---|
| 4 insulated 12 AWG conductors (2 black + 2 white) | 4 |
| All grounds together | 1 |
| Receptacle yoke | 2 |
| Internal clamps | 1 |
| **Total** | **8** |

With an illustrative allowance of 2.25 in³ per 12 AWG conductor: 8 × 2.25 = **18.0 in³**. It fits exactly. Add one more 12 AWG conductor and the requirement becomes 9 × 2.25 = 20.25 in³: too much, so you would need a deeper box or a box extension.

## Conduit fill: the idea

Conduit fill compares the **cross-sectional area** of the conductors with the **internal area** of the raceway:

```text
percent fill = total conductor area / conduit internal area × 100
```

Chapter 9 sets the maximum fill by the number of conductors. The best-known limit is **40%** for more than two conductors; one conductor and two conductors have their own limits (two conductors is the lowest, because two round wires jam easily). Chapter 9 lists the area of each conductor size and insulation type, and of each trade size and type of raceway. Informative Annex C does the arithmetic for you when all conductors are the same size and type, but you must be able to do it by hand when they are mixed.

## Worked example (illustrative values)

Three conductors, each with an **illustrative** area of 0.0133 in², in a conduit with an **illustrative** internal area of 0.304 in²:

```text
total area = 3 × 0.0133 = 0.0399 in²
fill       = 0.0399 / 0.304 × 100 ≈ 13.1 %
```

Well under 40%, so it passes. Remember that equipment grounding conductors in the conduit **do** count for conduit fill, even though they do not count as current-carrying for ampacity adjustment.

## Common mistakes

- Counting each ground as one unit (all of them together count as one).
- Forgetting the device yoke counts as two.
- Counting pigtails.
- Using the 40% limit for two conductors.
- Leaving equipment grounding conductors out of conduit fill.
- Using the wrong raceway type's area (EMT, PVC and rigid conduit of the same trade size have different internal areas).

## Quick recap

Box fill: count allowance units (conductors 1 each, all grounds 1, each device yoke 2, all clamps 1, each support fitting 1), multiply by the per-conductor volume for the size, and compare with the box volume. Conduit fill: total conductor area divided by raceway area, compared with the Chapter 9 limit (40% for more than two conductors). Use your edition's tables ([NFPA 70](https://www.nfpa.org/70); checked on 2026-10-23).

## Try it

Open the **Do** tab. The counting rules and the conduit-fill check are written; you add the box-fill volume formula.


Count box-fill allowances, check a box's volume and work out conduit fill percentage.

Box and conduit fill

Box Fill Calculations - Electrical Box Fill (THE EASY WAY)

Conduit Fill / Different Insulation types

The everyday code calculations: conductor ampacity with derating, voltage drop, dwelling service load calculations, and box and conduit fill.