I just recently tore out an installation where someone chicken-choked two 12/3 NM into a 3/4" EMT conduit. I just tore the conduit out wires and all; I cannot get the NM out of the conduit, it's in there so tight. I can't imagine how it was ever accomplished! It carries no more current than 4x THHN wires, which go effortlessly through 3/4" EMT.
For two 12/2 NM cable, you need at least 1" conduit (by calculations below) but it'll still be a difficult pull. For two 12/2 UF, you'll need at least 1-1/4" conduit.
Get pi out of the picture: Think in circular inches
When you get into large wires, the copper dimension starts to be expressed in a unit called K C Mil, or M C M.
- K or M = 1000 of ...
- C means circular area, instead of square area.
- Mil or M means "Mil", or 1/1000 (of an inch).
So a circular mil means the area of a circle 1/1000" across. KCMil or MCM mean 1000 of those circular mil units. A circular inch is 1 million CMil or 1000 KCMil.
So a "circular inch" is not a weird unit, and is in fact used in electrical already in the fashion of the KCMil.
Figuring out cross sections the easy way
Suppose your wire is W inches width in the widest dimension. The area of the wire is
W * W
and our unit is circular inches (or circular mils if we started with mils). Wow, that was easy. OK, so say your conduit is I width in interior cross section. The conduit is *I * I* obviously.
With 1 wire in a conduit, the one wire can't fill more than 53% of the conduit. That means the max wire area can be
I * I * .53
That means to be legal, this must be true:
I * I * .53 >= W * W
Algebra algebra algebra, and we get
I >= W * 1.38
Assuming 1 wire in conduit, the conduit ID must be 138% of the wire width.
Now, this extends for 2 wires in conduit, where only 31% conduit fill is allowed (because of the strong tendency for 2 wires to bind, especially when they are wide cables). So if the other cable is Y width, it's literally
I * I * .53 >= W * W + X * X
And if both wires are the same size W, then minimum I evaluates to 2.54 * W for 2 wires of W width in conduit.
Now since 12-2NM is 0.40" wide (400 mil), that means 2 in conduit requires 1.016" conduit ID. Fortunately most 1" trade size conduit is somewhat larger than 1" actual. That's your answer.
For 3 same-sized cables in conduit (40% fill allowed), it evals to 274% of cable max width.
For 4 same-sized cables in conduit (ditto), it evals to 317% of cable max width.