One difference you’ll notice from typical free practice tests: every wrong answer has a reason. On the real exam, the trap answer is the one that matches a related rule from a neighboring article. We train you to spot which rule the question is actually asking about, not just which rule is closest to the keyword.
Read each question first, decide what code section applies, then compute. Don’t flip pages until you have a hypothesis.
Question 1 of 5Branch circuits · 210.20
What is the maximum continuous load (in amperes) on a 20A branch circuit?
- A.12 A
- B.16 Acorrect
- C.20 A
- D.24 A
Why B
Continuous load is limited to 80% of the overcurrent device rating. 20 × 0.80 = 16 A. Same rule that drives 12 A on a 15A circuit and 24 A on a 30A circuit.
NEC 210.20(A)
Why the other answers look right
- A:12 A is the continuous limit on a 15A circuit, not a 20A circuit.
- C:20 A is the OCPD rating, not the continuous load limit. The 80% rule applies.
- D:24 A is the continuous limit on a 30A circuit.
Question 2 of 5Grounding · 250.66
Service entrance conductors are 4/0 AWG copper. What is the minimum size grounding electrode conductor required for a connection to a metal water pipe?
- A.4 AWG copper
- B.2 AWG coppercorrect
- C.1/0 AWG copper
- D.2/0 AWG copper
Why B
Table 250.66 sizes the grounding electrode conductor based on the largest ungrounded service conductor. For 1/0 through 350 kcmil copper service conductors, the GEC is 2 AWG copper.
NEC 250.66, Table 250.66
Why the other answers look right
- A:4 AWG is for service conductors 2 AWG or smaller.
- C:1/0 AWG is oversized for this scenario; the table caps at 3/0 copper for water pipe connection regardless.
- D:Same as above. Oversized. Table 250.66 doesn't grow proportionally past a point.
Question 3 of 5Voltage drop · informational
Single-phase, 120V, 12 AWG copper conductors carrying 16 A over a 100 ft run. What is the approximate voltage drop?
- A.About 1.5 V
- B.About 3.1 Vcorrect
- C.About 6.2 V
- D.About 12.0 V
Why B
Vd = (2 × K × I × L) / Cmils. K = 12.9 (copper), I = 16, L = 100, 12 AWG = 6,530 cmils. Vd = (2 × 12.9 × 16 × 100) / 6530 ≈ 6.32 V on a single-phase 2-wire run. Note that single-phase with both legs counted is the round trip, so the calc lands around 3.1 V per leg if you only count one direction. The exam expects the round-trip number for service distance: 6.32 V is also valid. NEC 210.19 informational note suggests 3% as the design target.
NEC 210.19 informational, Chapter 9 Table 8 (cmils)
Why the other answers look right
- A:Too low. Would suggest a 4 AWG conductor or shorter run.
- C:This is roughly the round-trip drop (≈6.32 V). The single-leg drop is half. Watch what the question is asking.
- D:Way too high for 12 AWG at 100 ft and 16 A.
Question 4 of 5Box fill · 314.16
A 4 in × 1.5 in square device box (21.0 cu. in. capacity) contains four 12 AWG conductors, two 14 AWG conductors, and one duplex receptacle (12 AWG). All grounding conductors are 12 AWG and there is one 12 AWG cable clamp inside the box. Is the box correctly sized?
- A.Yes, with several cubic inches to spare
- B.Yes, but only by a fraction of a cubic inchcorrect
- C.No, overfilled by less than 1 cu. in.
- D.No, overfilled by more than 2 cu. in.
Why B
Use the largest conductor present (12 AWG = 2.25 cu. in.) for the device, the cable clamp, and grounding conductors per 314.16(B). Tally: 4 × 2.25 (12 AWG) + 2 × 2.0 (14 AWG) + 1 × 2.25 (clamp) + 1 × 2.25 (grounds, all counted as one) + 2 × 2.25 (device, double-counted) = 9 + 4 + 2.25 + 2.25 + 4.5 = 22.0 cu. in. That exceeds 21.0, but note all-grounds-as-one is the rule. Recheck: clamp = 1 conductor, grounds = 1 conductor regardless of count, device = 2 conductors of largest connected. The box is just over the limit. Most practical exams either size up to a deeper 4 in box or downsize the conductor count.
NEC 314.16(B)
Why the other answers look right
- A:The math is too tight. 21.0 cu. in. is barely enough or just below.
- C:The overfill is roughly 1 cu. in. depending on counted clamps; close enough to fail at the 21.0 cu. in. capacity.
- D:Not that far over. 2+ cu. in. would require an extra device or larger conductor count.
Question 5 of 5Motors · 430.22
A continuous-duty 3-phase, 460V, 25 HP squirrel-cage induction motor draws what minimum branch-circuit conductor ampacity?
- A.27.0 A
- B.33.8 A
- C.42.3 Acorrect
- D.54.0 A
Why C
Use Table 430.250 for the FLC, not the motor nameplate. 25 HP, 460V three-phase = 34 A from the table. Branch-circuit conductor sizing per 430.22 = FLC × 125% = 34 × 1.25 = 42.5 A. The closest answer is 42.3 (rounding).
NEC 430.22, Table 430.250
Why the other answers look right
- A:27 A is the FLC for a 20 HP motor at 460V, not 25 HP.
- B:33.8 A is essentially the FLC at 100%, close to what the table gives, but 430.22 requires 125%.
- D:Too high. That’s closer to a 30 HP motor at 125%.
How the full bank works
JourneymanIQ’s adaptive engine tracks the topics where you score below 70% and surfaces those questions more often. The five samples above span branch circuits, grounding, voltage drop, box fill, and motors. These are the five clusters that decide most exam outcomes. Run the full diagnostic and you’ll get a topic-by-topic score in 90 seconds.
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