---
id:
aliases: []
tags:
  - destiny/fleeting
  - status/incomplete
  - topic/construction/electrical
  - type/encyclopedia
  - authorship/original
title: Conductor Sizing
---
# Conductor Sizing

It is tempting to size circuit conductors
based on overcurrent protection,
but they are sized independently.

Conductors are sized to be suitable for the load,
overcurrent protection is sized to protect the conductors.

Allowable conductor ampacity restrictions
are not based on a wire's ability to carry current,
they are intended to protect its insulation
from damage due to excessive heating.

## "The 80% Rule"

> [!danger]
> This description is provided _for reference only_.
> 
> Like every NEC "rule"
> that isn't preceded by a section reference,
> it is _not_ code.
> 
> It is my opinion that this one should never be repeated,
> even as shorthand.

"The 80% Rule" is a rule of thumb
referring to a common convention of several articles
including:

* [[nfpa-70_210_branch-circuits#210.19(A)(1) General.]]

* [[nfpa-70_215_feeders#215.2(A)(1) General.]]
* [[nfpa-70_215_feeders#215.3 Overcurrent Protection.]]

* [[nfpa-70_430_motors#430.22 Single Motor.]]
* [[nfpa-70_430_motors#430.51 General.]]

which paraphrased states:

> ... the minimum conductor size shall have an ampacity
> not less than the noncontinuous load
> plus 125 percent of the continuous load

When the rule is repeated,
the noncontinuous load is ignored
and it is stated that conductors are suitable
for 80% their listed rating,
since 80% is the reciprocal of 125%.

The rule neglects important context and common exceptions,
namely transformers, whose feeder conductors are sized at 100%[^1].

[^1]: [[nfpa-70_215_feeders#215.2(B)(1) Feeders Supplying Transformers.]]

## Branch Circuits

### Receptacle Branch

There is no maximum number of receptacles per circuit _in any occupancy_.

It is a common misconception that the limit
can be calculated with a formula like

$$
\frac{1.25(180VA)}{120V} = 1.875A, \quad \frac{20A}{1.875A} = 10.\bar{6}
$$

but the 180VA per yoke load specified in
[[nfpa-70_220_load-calculations#220.14(I) Receptacle Outlets.|220.14(I)]]
is specifically for calculating service and feeder sizing.

Per [[nfpa-70_210_branch-circuits#210.19(A)(1) General.|210.19(A)(1)]]
a receptacle branch circuit's load
is the load of the equipment intended to be served by it.

Where general-use receptacles are provided
without specific equipment in mind,
circuits will be engineered at the minimum load.
If a receptacle circuit's load is a whole multiple of 180VA
there's a good chance that's the number of devices,
or at least was at some point in the design.

## Feeders

![[nfpa-70_250_grounding#250.122(A) General.]]

Apparently in the 2026 NEC First Draft Meetings,
Code Making Panel 5 clarified that the equipment grounding conductor (EGC)
never needs to be larger than the largest ungrounded conductor in any raceway
when installed in parallel.
I can not find a source to verify this.
Statements from other reputable sources including Mike Holt
are in contradiction to this idea.

***

Given a minimum ampacity, find all valid configurations.

[[nfpa-70_310_conductors_for_general_wiring]]

> [!cite] NEC Article 310 (emphasis added)
> #### 310.10(H) Conductors in Parallel.
> ##### (1) General.
> Aluminum, copper-clad aluminum, or copper conductors,
> for each phase, polarity, neutral, or grounded circuit
> shall be permitted to be connected in parallel
> (electrically joined at both ends)
> _only in sizes 1/0 AWG and larger_
> where installed in accordance with 310.10(H)(2) through (H)(6).

Rank by total cost of install.

### Complexity to Ignore

#### Conductor Material

Tinned copper and copper-clad aluminum conductors
can be assumed out of scope.

### Complexity to Respect

#### Equipment Grounding Conductor Material

Wire and EGC conductors are usually assumed to match,
but it is sometimes necessary to use a copper EGC with aluminum wires,
either for spec requirements or conduit fill considerations.

## Voltage Drop

[[voltage-drop]]

## Transformers

$$
I = \frac{S}{ \sqrt{3} \times V \times E }
$$

* $I$ = nameplate current rating
* $S$ = nameplate kVA rating
* $V$ = feeder voltage
* $E$ = efficiency

## Motors

1 electric horsepower = 746 watts

full-load current (FLC) / full-load amperes (FLA)

minimum circuit ampacity (MCA)

$$
\text{MCA} = 1.25 \times \text{FLC}
$$
