NEC Load Calculation Practice Questions

Each general-use outlet is figured at 180 VA. Divide the circuit's capacity by 180 and round down.

Volts × amps gives the circuit's VA; each receptacle is 180 VA; round down — you can't have a partial outlet. A 15 A circuit lands exactly on 10.

1. Rounding up.
2. Forgetting each outlet is 180 VA.
3. Using the wrong voltage.

Show windows are figured at 200 VA per linear foot of window.

Just multiply the window length by 200 VA. This is the load added for the show window, on top of the general lighting.

1. Using the 3 VA/ft² dwelling figure.
2. Measuring area instead of linear feet.

Each dryer is at least 5,000 VA. Table 220.54 gives a demand factor by the number of dryers — but 1 through 4 stay at 100% (no reduction).

Multiply the number of dryers by 5,000 VA, then apply the Table 220.54 factor. With 4 or fewer, the factor is 100%, so nothing changes.

1. Applying a reduction when there are 4 or fewer.
2. Using less than 5,000 VA per dryer.

Find the current (VA ÷ V), round the breaker up to the next standard size, then pick a wire that carries it.

Watts ÷ volts gives amps; the breaker rounds up to protect the circuit; 10 AWG copper is good for 30 A (and 240.4(D) caps 10 AWG at 30 A anyway).

1. Rounding the breaker down. Overcurrent rounds up.
2. Using 12 AWG — only good to 20 A.
3. Forgetting to convert kVA to VA first.

At least one receptacle within 18 in. of the top of each show window — and no point along the top can be more than 6 ft from a receptacle. One outlet reaches 6 ft each way, so it covers at most 12 ft of window.

Each end is 3.5 ft from an outlet ✓ — and the middle is 6 ft from either one, right at the limit ✓. No point is past 6 ft, so 2 outlets does it.

Divide the window length by 12 and round up — that is the fewest outlets that can keep every point within 6 ft. Then space them so no stretch between outlets is longer than 12 ft, and no end hangs more than 6 ft past an outlet.

1. Using the old "each 12 linear ft or major fraction" wording — the current rule is the 6-ft rule.
2. Rounding 1.6 down to 1 — 19 ft is more than one outlet can cover.
3. Mixing this up with show-window lighting load (200 VA per linear foot) — that is a different question.

Show-window lighting = 200 VA per linear foot of window.

Multiply the window length by 200 VA. This load is added on top of the general lighting for the space.

1. Using the 3 VA/ft² dwelling figure.
2. Measuring area instead of linear feet.

A multioutlet assembly whose receptacles are likely to be used at the same time = 180 VA per 1 ft (220.14(H)(2)).

When the loads run together, each foot of strip counts as 180 VA — round the length up to the next whole foot, then multiply. (If the loads are unlikely to run together, it's 180 VA per 5 ft instead.)

1. Using the 'per 5 ft' figure when the loads run together.
2. Not rounding the length up to the next foot.

Rule 220.53: where 4 or more fastened-in-place appliances are on the same feeder, apply a 75 % demand factor.

220.53 — appliance demand (75 %)

Four or more fastened-in-place appliances on a feeder get a 75 % demand factor (220.53). Multiple dryers use the Table 220.54 factor. Non-coincident loads (heat vs A/C) count only the larger of the two (220.60).

1. Applying the 75 % factor with fewer than 4 appliances.
2. Adding both heat and A/C — count only the larger.
3. Using the wrong dryer demand percentage.