# Ampacity In Ungrounded Conductors, (Sub-Panels, and Loads)

I am 2nd year 1st semester student at Independent Electrical Contractors of Rocky Mountain. This question is from our final Review:

This is a fairly straightforward Ohm's law question. Each ungrounded conductor is carrying how many amps? 8200 kW / 240 V = 34.16 A; we round up to 34.2 A and we're good.

This question is similar: how many amps in each conduit on A phase are being carried to the subpanel? A phase is carrying 210 A; we divide by two, and come up with 105 A. Looking at the questions individually they make sense, but when they are side by side I get confused.

Looking at the first question we know that AC current is flowing back or forth on the conductors headed to breakers 6 and 8. The amps is 34.2 A.

On the second question, we have four conductors headed to a 2 pole breaker. I don't know a lot about panels and breakers, but I think each breaker corresponds to a phase in this case. Will call the breaker the red wire goes to A phase, so I guess if that subpanel is pulling 210 A, 105 A will flow through each conductor.

I think I'm overthinking this. I also think I might be missing something because the connections are in parallel on the bottom and in series on the top.

• what is the question? Dec 6, 2021 at 2:40
• @MathKeepsMeBusy I think it's whether or not it is (a), (b), (c), (d), or (e) in #9.
– jonk
Dec 6, 2021 at 4:35

I'm assuming we are talking about North American "split-phase" AC panels, where you have two conduits using equal wire diameters to a split-phase subpanel. In this case, you can assume that each conduit will equally share currents to the subpanel.

I'm also assuming that $$\A\phi\$$ is what I normally think of as L1/H1 (or "hot 1") and that $$\B\phi\$$ is what I normally think of as L2/H2 (or "hot 2.")

See the following:

simulate this circuit – Schematic created using CircuitLab

Above, $$\I_{_\text{P}}\$$ (I used P for power, implying things like a stove, for example) represents all active 240 VAC loads. A stove, for example, will likely have the oven coil directly across 240 VAC, but may have a clock that runs off of 120 VAC, using either $$\A\phi\$$ or $$\B\phi\$$ for that purpose. $$\I_{_\text{AB}}\$$ represents the portion of the active 120 VAC $$\A\phi\$$ loads that are matched up with active $$\B\phi\$$ loads. $$\I_{_\text{A}}\$$ represents the residual difference between the active $$\A\phi\$$ loads and the active $$\B\phi\$$ loads, which we know to be $$\40\:\text{A}\$$.

So $$\I_{_\text{A}}=40\:\text{A}\$$ and $$\I_{_\text{P}}+I_{_\text{AB}}=170\:\text{A}\$$. We cannot know the values of $$\I_{_\text{P}}\$$ or $$\I_{_\text{AB}}\$$, separately. That's not given. But we do know that the $$\A\phi\$$ total is $$\I_{_\text{P}}+I_{_\text{AB}}+I_{_\text{A}}=210\:\text{A}\$$ and that the $$\B\phi\$$ total is $$\I_{_\text{P}}+I_{_\text{AB}}=170\:\text{A}\$$ and so the difference, $$\I_{_\text{A}}=40\:\text{A}\$$, must be returned over the neutral wire.

Since the conduit wiring is all the same (1/0), and likely the same length, we can assume that the currents will split roughly equally over the two conduits. The above diagram shows this.

The question is then about what current is carried by the $$\A\phi\$$ conductor in each conduit. Which is $$\105\:\text{A}\$$, as shown.