Journeyman Electrician Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

1. Givens: P = 4,800 W at 240 V; new supply = 208 V; the load is resistive, so R is constant.
2. Resistance: R = V² ÷ P = 240² ÷ 4,800 = 57,600 ÷ 4,800 = 12 Ω.
3. New power: P = V² ÷ R = 208² ÷ 12 = 43,264 ÷ 12 ≈ 3,605 W.
4. Check: (208 ÷ 240)² = 0.751, and 0.751 × 4,800 ≈ 3,605 W.

3,605 watts: Correct. The element's resistance stays fixed at 240² ÷ 4,800 = 12 ohms, so at 208 volts P = 208² ÷ 12 ≈ 3,605 watts; power falls with the square of the voltage.

4,160 watts: This scales the power in direct proportion to voltage (4,800 × 208 ÷ 240); for a fixed resistance, power varies with voltage squared.

4,800 watts: This assumes the rated wattage is delivered at any voltage; the rating applies only at 240 volts.

5,538 watts: This inverts the voltage ratio (4,800 × 240 ÷ 208), as if lower voltage raised the power.

For a fixed resistance, power is proportional to the square of the voltage: a 240-V element on 208 V gives about 75% of its rated watts.

Answer: B

3 feet: Three feet is Condition 1 (no live or grounded parts on the opposite side), or any condition at 0–150 volts to ground; a concrete wall counts as grounded, so Condition 1 does not apply.

3½ feet: Correct. Voltage to ground is 277 volts, so the 151–600-volt row of Table 110.26(A)(1) applies. Exposed live parts on one side with a grounded surface (concrete, brick or tile walls are considered grounded) on the other is Condition 2, which needs 3½ feet.

4 feet: Four feet is Condition 3 at 151–600 volts, which requires exposed live parts on both sides of the working space.

5 feet: Five feet is Condition 3 in the 601–1000-volt row; this switchboard is only 277 volts to ground.

Pick the depth by voltage to ground first, then by condition; a concrete or masonry wall opposite exposed live parts makes it Condition 2.

Answer: C

1. Givens: continuous load = 120 A; noncontinuous load = 40 A.
2. Rule: minimum ampacity = 125% of continuous + 100% of noncontinuous.
3. Arithmetic: 120 A × 1.25 = 150 A; 150 A + 40 A = 190 A.
4. Compare: 190 amperes is the minimum ampacity; 200 amperes would wrongly apply 125% to the noncontinuous load too.

160 amperes: This adds the loads at 100 percent and misses the 125 percent factor on the continuous load.

175 amperes: This rounds 160 amperes to the next standard breaker size instead of applying the 125 percent factor.

190 amperes: Correct. 215.2(A)(1) requires feeder conductor ampacity of at least 100 percent of the noncontinuous load plus 125 percent of the continuous load: 120 × 1.25 = 150, plus 40 = 190 amperes.

200 amperes: This applies 125 percent to the whole 160-ampere load; the noncontinuous part is taken at 100 percent.

Feeder conductors carry 125% of the continuous load plus 100% of the noncontinuous load before any adjustment or correction.

Answer: B

1. Givens: 40-ft wall space; doorways at both ends, so the wall space ends at each doorway.
2. Rule: no point along the floor line may be more than 6 ft from a receptacle, so each receptacle covers 6 ft on each side = 12 ft.
3. Arithmetic: 40 ft ÷ 12 ft = 3.33 receptacles.
4. Round up: 4 receptacles (for example at 6 ft, 18 ft, 30 ft and 38 ft from one end, covering every point).

3: This divides 40 by 12 and rounds down; three outlets leave a section of the wall more than 6 feet from a receptacle.

4: Correct. 210.52(A) requires that no point along the floor line of a wall space be more than 6 feet from a receptacle, so each outlet covers 12 feet of wall: 40 ÷ 12 = 3.33, rounded up to 4.

6: This places an outlet about every 6 feet and rounds down; 6 feet is the distance from any point to the nearest outlet, not the spacing between outlets.

7: This divides 40 by 6 and rounds up, treating 6 feet as the spacing between outlets instead of the reach of each outlet.

No point on a dwelling wall space may be more than 6 feet from a receptacle, so one outlet can cover up to 12 feet of wall; divide by 12 and round up.

Answer: B

Bonded to the enclosure with the bonding screw: That would make the feeder neutral and the EGC parallel paths, putting neutral current on the EGC and metal parts.

Isolated from the enclosure and EGC bar: Correct. Where the feeder includes an EGC, the grounded conductor at the separate building is not connected to the EGC or to the grounding electrode.

Connected only to a new ground rod: The building needs an electrode, but it connects to the EGC system, not to the neutral.

Combined with the EGC on one bar: Combining them is the same downstream neutral-to-ground bond the rule forbids.

At a building supplied by a feeder with an EGC, keep the neutral isolated; the EGC and the building’s grounding electrode connect to the enclosure.

Answer: B

8 AWG: Too small. 8 AWG is the Table 250.66 value for 2 AWG or smaller service conductors, and it is below the size the rule allows for rods.

6 AWG: Correct. The GEC portion that is the sole connection to rod, pipe or plate electrodes is not required to be larger than 6 AWG copper, whatever the service size.

4 AWG: This is the cap for the sole connection to a concrete-encased electrode, not to rods.

1/0 AWG: This is the full Table 250.66 value for 500 kcmil, but a GEC that serves only rods is not required to exceed 6 AWG copper.

A GEC that is the only connection to rods, pipes or plates need not exceed 6 AWG copper; to a concrete-encased electrode it need not exceed 4 AWG copper.

Answer: D

One rod is sufficient as installed: Wrong. A single rod may stand alone only if it is shown to have 25 ohms or less to earth; 40 ohms does not qualify.

Add rods until 25 ohms is measured: A common misconception. The NEC requires one supplemental electrode, not a resistance target; once the second electrode is installed, no further rods are required.

Add a second rod at least 3 feet away: Too close. Supplemental rod electrodes must be at least 6 feet apart.

Add a second rod at least 6 feet away: Correct. A single rod not shown to be 25 ohms or less must be supplemented by an additional electrode; when that is another rod, the rods must be at least 6 feet apart.

A single rod, pipe or plate electrode needs a supplemental electrode unless it measures 25 ohms or less; two rods must be at least 6 feet apart, and no resistance target applies after that.

Answer: C

1. Givens: tap length 10 ft; feeder OCPD = 600 A; tap ends in a 30-A fused disconnect; load = 24 A; field installation leaving the enclosure.
2. Rule: tap ampacity ≥ the calculated load, ≥ the rating of the device supplied, and ≥ 1/10 of the feeder OCPD rating.
3. Arithmetic: load = 24 A; device = 30 A; 600 A ÷ 10 = 60 A.
4. Compare: the largest value, 60 A, is the minimum tap conductor ampacity.

30 amperes: Thirty amperes satisfies the rule for the rating of the device supplied but misses the one-tenth-of-feeder-protection rule for field taps leaving the enclosure.

40 amperes: This adds 125 percent to the 30-ampere device rating; no such factor applies, and the one-tenth rule governs.

60 amperes: Correct. 240.21(B) requires a 10-foot tap to be at least the load and the rating of the device it supplies, and, for field taps leaving the enclosure, at least one-tenth of the feeder protection: 600 ÷ 10 = 60 amperes, the largest of the three.

200 amperes: This applies the one-third rule from the 25-foot tap rule instead of the one-tenth rule for 10-foot taps.

A 10-foot feeder tap must be at least the load, the rating of the device it supplies, and (for field taps leaving the enclosure) one-tenth of the feeder overcurrent device rating; the largest value governs.

Answer: C

3 AWG: Reads the copper side of Table 310.16; 3 AWG copper is 100 amperes at 75 °C, but the stem specifies aluminum.

2 AWG: Uses the 90 °C ampacity of XHHW-2 (100 amperes). The 75 °C terminals cap the usable ampacity at 90 amperes for 2 AWG aluminum.

1 AWG: Correct. Because the terminals are marked 75 °C, 110.14(C)(1)(a) permits the 75 °C column even though the circuit is 100 amperes or less. In the aluminum 75 °C column of Table 310.16, 1 AWG is rated 100 amperes.

1/0 AWG: Uses the 60 °C aluminum column. The 60 °C limit applies only when terminals are not marked for 75 °C.

Terminals marked 75 °C let a circuit of 100 amperes or less use the 75 °C column; always read the aluminum side when the stem says aluminum.

Answer: C

1. Givens: six 12 AWG insulated conductors, three 12 AWG EGCs, internal clamps, one duplex receptacle.
2. Conductors: 6. EGCs (all together): 1. Clamps (all together): 1. Device yoke: 2. Total = 10.
3. Volume per 12 AWG conductor: 2.25 in³.
4. 10 × 2.25 in³ = 22.50 in³.

18.00 cubic inches: Counts only the six insulated conductors and the grounding conductors, leaving out both the device (2) and the clamp (1) allowances.

20.25 cubic inches: Leaves out one allowance, either the clamps or one of the two device allowances; the total must be 10 volume units.

22.50 cubic inches: Correct. Under 314.16(B): 6 conductors + 1 for all equipment grounding conductors + 1 for the clamps + 2 for the device yoke = 10 allowances; 10 × 2.25 cubic inches (12 AWG) = 22.50 cubic inches.

27.00 cubic inches: Counts each of the three equipment grounding conductors separately; all of them together count as one.

Box fill: each conductor 1, all EGCs together 1, all clamps together 1, each device yoke 2, all at the volume of the largest conductor involved.

Answer: C

1. 1. Givens: EVSE maximum load 48 A at 240 V.
2. 2. Rule: EVSE is a continuous load; branch-circuit overcurrent protection is at least 125% of the maximum load (Article 625).
3. 3. Arithmetic: 48 A × 1.25 = 60 A.
4. 4. Standard rating: 60 A is a standard rating in 240.6(A), so 60 A is the minimum.

48 amperes: Uses the load current without the 125% continuous-duty factor; 48 amperes is also not a standard rating.

50 amperes: The next standard rating above 48 amperes, but it still omits the 125% factor.

60 amperes: Correct. EVSE loads are continuous, so Article 625 requires the overcurrent device to be rated at least 125% of the maximum load: 48 × 1.25 = 60 amperes, a standard rating in 240.6(A).

70 amperes: Larger than necessary; 60 amperes already meets the 125% minimum exactly.

Treat EVSE as a continuous load: size the overcurrent device at 125% of the maximum load, then pick the standard rating at or above it.

Answer: A

3 feet: Correct. Section 680.26(B)(2) defines the perimeter surfaces to be bonded as those extending 3 feet horizontally beyond the inside walls of the pool.

5 feet: A distance used elsewhere in Article 680 (for example the spa emergency switch and switching devices), not the perimeter bonding extent.

6 feet: Six feet is the minimum distance of pool-area receptacles from the inside walls, not the extent of perimeter bonding.

10 feet: Bonding the entire deck is allowed but not required; the code minimum is much smaller.

Perimeter surface bonding covers the paved or unpaved surface 3 feet horizontally from the inside walls of the pool.

Answer: B

6 inches: Six inches applies to recessed luminaires (and to surface-mounted fluorescent types), not to a surface-mounted LED luminaire with an enclosed light source.

12 inches: Correct. 410.16(C)(1) permits surface-mounted incandescent or LED luminaires with a completely enclosed light source on the wall above the door or on the ceiling with at least 12 inches to the nearest point of the storage space.

24 inches: Twenty-four inches is the horizontal depth that defines the storage space itself, not the luminaire clearance.

72 inches: Seventy-two inches (6 feet) is the height used to define the storage space, not the luminaire clearance.

In closets, surface-mounted incandescent or LED luminaires with enclosed sources need 12 in from storage space; recessed and surface fluorescent types need 6 in; open-lamp incandescent luminaires are not permitted.

Answer: C

1. Table 430.250: 15 hp at 460 V = 21 A (nameplate 19.5 A is not used here).
2. Table 430.52: inverse-time breaker = 250% of FLC.
3. 2.5 × 21 A = 52.5 A.
4. 240.6(A): 52.5 A is not standard; the next higher standard rating is 60 A (430.52(C)(1) Exception No. 1).
5. Answer: 60 amperes maximum.

40 amperes: This uses 175%, the dual-element fuse percentage, instead of 250% for an inverse-time breaker.

50 amperes: This starts from the 19.5-ampere nameplate (48.75 A). Short-circuit protection is based on the table FLC.

60 amperes: Correct. Table 430.52 allows 250% of the Table 430.250 FLC: 2.5 × 21 A = 52.5 A; 52.5 A is not a standard rating, so 430.52(C)(1) Exception No. 1 permits the next higher standard size, 60 A.

70 amperes: This uses 300%, the nontime-delay fuse percentage (63 A, rounded up), which does not apply to a breaker.

Motor short-circuit and ground-fault protection uses the table FLC times the Table 430.52 percentage, rounded up to the next standard rating when the result is not standard.

Answer: C

1. Givens: 25 hp, 460 V, three-phase; copper THWN; 75 °C terminations.
2. Table 430.250: FLC = 34 A at 460 V.
3. 430.22: minimum ampacity = 1.25 × 34 A = 42.5 A.
4. Table 310.16, 75 °C copper column: 10 AWG = 35 A (too small); 8 AWG = 50 A.
5. Answer: 8 AWG copper.

4 AWG: This uses the 230-volt full-load current (68 A × 1.25 = 85 A) instead of the 460-volt value.

6 AWG: This comes from the 60 °C column (8 AWG is only 40 amperes there), but the terminations are rated 75 °C.

8 AWG: Correct. 430.22 requires 125% of the table FLC: 1.25 × 34 A = 42.5 A, and 8 AWG copper at 75 °C (50 amperes in Table 310.16) is the smallest size that is large enough.

10 AWG: This carries the 34-ampere full-load current (35 amperes at 75 °C) but misses the 125% factor of 430.22.

Single-motor branch conductors must have an ampacity of at least 125% of the table full-load current, read from the termination-temperature column.

Answer: B

1. Givens: 45 kVA = 45,000 VA, three-phase, 480 V primary, primary-only protection.
2. Primary current: 45,000 ÷ (480 × 1.732) = 45,000 ÷ 831.4 = 54.1 A.
3. Table 450.3(B), primary only, 9 A or more: 125% → 54.1 × 1.25 = 67.7 A.
4. Note 1: next higher standard rating permitted → 70 A (240.6(A)).

60 amperes: Rounds 67.7 amperes down; for primary currents of 9 amperes or more, the next higher standard rating is permitted.

70 amperes: Correct. Primary current = 45,000 ÷ (480 × 1.732) = 54.1 amperes. Table 450.3(B) allows 125% for primary-only protection (54.1 × 1.25 = 67.7 amperes), and Note 1 permits the next higher standard rating in 240.6(A), 70 amperes.

90 amperes: Applies 167%, which is the primary-only limit for currents under 9 amperes, not 54.1 amperes.

150 amperes: Applies 250%, allowed only when secondary protection is also provided, and then rounds up as well.

Primary-only protection (1000 V or less, 9 A or more): 125% of primary current, next standard size up permitted.

Answer: B

1. Givens: 8 current-carrying THHN copper conductors, 30 °C (no correction), 24 A load, 25 A breaker, 75 °C terminations.
2. Adjustment for 7–9 conductors: 70%.
3. 12 AWG: 30 A (90 °C) × 0.70 = 21 A — less than 24 A, too small.
4. 10 AWG: 40 A (90 °C) × 0.70 = 28 A — at least 24 A and 25 A.
5. Termination check: 10 AWG at 75 °C = 35 A ≥ 24 A. Answer: 10 AWG.

8 AWG: This applies the 70% factor to the 60 °C column (10 AWG = 30 × 0.70 = 21 A); derating may start from the 90 °C rating of THHN.

10 AWG: Correct. Eight conductors adjust to 70% (310.15(C)(1)): 10 AWG THHN is 40 × 0.70 = 28 A, which covers the 24-ampere load and 25-ampere breaker, and its 75 °C value (35 A) also exceeds the load, satisfying 110.14(C).

12 AWG: This skips the adjustment: 12 AWG is 30 A at 90 °C, but 30 × 0.70 = 21 A, which is less than the 24-ampere load.

14 AWG: This compares the 24-ampere load with the unadjusted 90 °C value (25 A) and ignores both the 70% adjustment and the 15-ampere limit on 14 AWG.

Adjust the 90 °C ampacity for the number of current-carrying conductors, then confirm the size also works in the termination-temperature column.