Wiring Methods & Materials — Texas Journeyman Practice Questions

4 questions · 8% of the Calculations exam

Worked questions

1. A box-fill calculation has six 12 AWG conductor equivalents. The supplied allowance is 2.25 cubic inches per equivalent. What minimum volume is required?

A27 cubic inches
B8.25 cubic inches
13.5 cubic inches
D2.67 cubic inches

Why C is correct

The problem supplies both the counted conductor equivalents and the per-equivalent volume. Multiplying 6 by 2.25 cubic inches gives 13.5 cubic inches; no unlisted clamp or yoke should be invented.

2. A raceway has 1.00 square inch of internal area and the supplied allowable fill is 40 percent. What conductor area may be installed?

A2.50 square inches
B0.25 square inch
C1.40 square inches
0.40 square inch

Why D is correct

Allowed conductor area equals raceway area times the supplied fill percentage. One square inch times 0.40 gives 0.40 square inch; the calculation does not divide by the percentage.

3. A box provides 20.0 cubic inches for 10 AWG conductors, using a supplied 2.50-cubic-inch allowance per equivalent. How many conductor equivalents fit?

A6 equivalents
B10 equivalents
8 equivalents
D12 equivalents

Why C is correct

Divide available volume by the supplied allowance: 20.0 ÷ 2.50 = eight conductor equivalents. Keep units consistent and use only the factors stated; no optional method or unstated demand factor is assumed.

4. Conductors occupy 0.60 square inch at a supplied 40-percent fill limit. What minimum internal raceway area is required?

A0.24 square inch
B1.00 square inch
C2.40 square inches
1.50 square inches

Why D is correct

Divide conductor area by the permitted fraction: 0.60 ÷ 0.40 = 1.50 square inches of raceway internal area.

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