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

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

## "The 80% Rule"

"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_430_motors#430.22 Single Motor.]]

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%.

## Branch Circuits

### Receptacle Branch

> [!important]
> 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

> [!cite] NEC Article 250 (emphasis added)
> ### 250.122 Size of Equipment Grounding Conductors
> #### (A) General.
> Copper, aluminum, or copper-clad aluminum
> equipment grounding conductors of the wire type
> shall not be smaller than shown in Table 250.122,
> but in no case shall they be required to be larger
> than the circuit conductors supplying the equipment...

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.

> [!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

> [!info] Ohm's Law
>
> $$
> V = I \times R, \quad R = \frac{V}{I}, \quad I = \frac{V}{R}
> $$

### Step 1: Effective Impedance $Z$

$$
Z = R \cos(\theta) + X \sin(\theta)
$$

where
* $Z$ = Effective impedance
* $R$ = AC resistance
* $X$ = Reactance
* $\theta$ = Power factor angle = $\arccos(PF)$

#### Parallel Runs

The equivalent resistance of parallel resistances is given by

$$
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}
$$

For $n$ parallel resistances of value $R$

$$
\begin{align*}
\frac{1}{R_{\text{eq}}} &= n \times \left(\frac{1}{R}\right) \\
&= \frac{n}{R} \\
R_{\text{eq}} &= \frac{R}{n}
\end{align*}
$$

### Step 2: Voltage Drop

> [!important]
> This section assumes a 3-phase
> 208Y/120V or 480Y/277V voltage system

> [!info] 3-Phase Voltage
>
> $$
> V_{LL} = \sqrt{3} \times V_{LN}, \quad V_{LN} = \frac{V_{LL}}{\sqrt{3}}
> $$

3% allowable voltage drop for a 120V line-to-neutral load:

$$
\text{Max} \Delta V = 0.03 \times 120 \text{V}_{LN} = 3.6 \text{V}_{LN}
$$

3% allowable voltage drop for a 208V line-to-line load:

$$
\text{Max} \Delta V = 0.03 \times 208 \text{V}_{LL} = 6.24 \text{V}_{LL}
$$

#### Line to Neutral Loads

$$
\Delta V_{LN} = I \times Z \times 2L
$$

#### Line to Line Loads

$$
\Delta V_{LL} = \sqrt{3} \times \left( I \times Z \times 2L \right)
$$

#### 3-Phase Loads

$$
\Delta V_{3\phi} = \sqrt{3} \times \left( I \times Z \times L \right)
$$

where
* $\Delta V$ = Voltage drop in volts ($V$)
* $I$ = Current in amperes ($A$)
* $L$ = Length of wire one way in feet ($\text{ft}$)

> [!important]
> "Current" is not the OCPD rating,
> but the actual load.

***

It is often more useful to know the maximum length
a certain wiring configuration is suitable for.

$$
L = \frac{ \Delta V }{ I \times M } \times \frac{1}{Z}
$$

## 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)

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