South Carolina Electrical Contractor License Practice Exam

Description

Sample Questions

Question 1. A South Carolina electrical contractor is sizing a feeder for a commercial tenant panelboard. The load calculation shows 64 amperes of continuous load and 80 amperes of noncontinuous load. The panelboard and overcurrent device are not listed for operation at 100 percent of their rating. Under NEC 215.2(A)(1), what is the minimum feeder conductor ampacity, before any temperature or adjustment factors are applied?

A. 144 amperes
B. 175 amperes
C. 180 amperes
D. 160 amperes

Correct Answer: D. 160 amperes

Answer Explanation:
Feeder conductors must have an allowable ampacity of at least the noncontinuous load plus 125 percent of the continuous load. The continuous portion is 64 × 1.25, which equals 80 amperes. Adding the 80-ampere noncontinuous load gives 160 amperes. Only the continuous portion receives the multiplier. This figure is the minimum calculated ampacity, so the contractor must still apply temperature and adjustment factors, check equipment terminal ratings, and select a standard overcurrent device size under 240.6. A 175-ampere device would likely be chosen, but that is a device rating selection and not the calculated conductor requirement.

Why the Other Options Are Incorrect:
A. 144 amperes — Incorrect. This simply adds the two loads and skips the 125 percent factor, which leaves the conductors undersized for sustained heating.
B. 175 amperes — Incorrect. This is the next standard overcurrent device size above 160 amperes, not the calculated conductor ampacity.
C. 180 amperes — Incorrect. This applies 125 percent to the entire 144 amperes, which penalizes noncontinuous load that the rule does not multiply.

Quick Study Guide:
Treat continuous and noncontinuous loads separately. Multiply only the continuous portion by 1.25, then add the noncontinuous portion. That result is a minimum calculated ampacity, not a device size. Afterward, check terminal temperature limits, apply correction and adjustment factors, and choose the next standard overcurrent rating under 240.6.

Question 2. An electrician is grounding a new commercial service. The service-entrance conductors are 250 kcmil copper. The grounding electrode conductor will run to a metal underground water pipe that is in direct contact with the earth for more than 10 feet. Under NEC Table 250.66, what is the minimum size copper grounding electrode conductor?

A. 4 AWG
B. 2 AWG
C. 6 AWG
D. 1/0 AWG

Correct Answer: B. 2 AWG

Answer Explanation:
Table 250.66 sizes the grounding electrode conductor from the largest ungrounded service-entrance conductor. Copper service conductors larger than 3/0 AWG, up to 350 kcmil, require a 2 AWG copper grounding electrode conductor, and 250 kcmil falls in that range. The reduced limits in 250.66, such as 6 AWG for a rod, pipe, or plate electrode and 4 AWG for a concrete-encased electrode, apply only when the conductor serves solely that electrode type. A metal water pipe does not qualify for a reduction. The installer must also supplement the water pipe with another electrode and use listed connection fittings.

Why the Other Options Are Incorrect:
A. 4 AWG — Incorrect. This size serves 2/0 through 3/0 AWG copper service conductors, or the limit for a concrete-encased electrode, and it is too small here.
C. 6 AWG — Incorrect. This is the maximum required when the conductor connects solely to a driven rod, pipe, or plate electrode, not a water pipe.
D. 1/0 AWG — Incorrect. This applies to service conductors larger than 350 kcmil through 600 kcmil and would be oversized.

Quick Study Guide:
Size the grounding electrode conductor from the largest ungrounded service conductor using Table 250.66, not from the overcurrent device. Learn the reductions: 6 AWG copper for rod, pipe, or plate electrodes only, and 4 AWG copper for concrete-encased electrodes. Metal water piping needs a supplemental electrode, and parallel service conductors use their combined equivalent size.

Question 3. A contractor is installing a continuous-duty, three-phase, 230-volt motor. The nameplate lists a full-load current of 20.0 amperes and a service factor of 1.15. The contractor is selecting a separate overload device. Under the general rule in NEC 430.32(A)(1), what is the maximum overload trip current for the initial selection, before any allowance for higher settings?

A. 25 amperes
B. 23 amperes
C. 28 amperes
D. 20 amperes

Correct Answer: A. 25 amperes

Answer Explanation:
Motor overload protection is sized from the nameplate full-load current, not the Table 430.250 value used for conductor sizing. For continuous-duty motors marked with a service factor of 1.15 or greater, or with a temperature rise of 40°C or less, the overload device may be set at no more than 125 percent of nameplate current. Here, 20.0 × 1.25 equals 25 amperes. If that setting cannot start the motor or carry the load, 430.32(C) allows higher settings up to a stated ceiling. The contractor should confirm the nameplate markings first, because the percentage depends on them.

Why the Other Options Are Incorrect:
B. 23 amperes — Incorrect. This is 115 percent, which applies to motors without a 1.15 service factor or 40°C rise marking. This motor qualifies for the higher percentage.
C. 28 amperes — Incorrect. This is 140 percent, the ceiling for other motors under 430.32(C), and it applies only if the initial setting fails to start the motor.
D. 20 amperes — Incorrect. This is 100 percent of nameplate and is not the maximum. It understates the permitted setting and may cause nuisance tripping.

Quick Study Guide:
Use nameplate current for overloads and table current for conductors. For motors with a 1.15 service factor or a 40°C rise or less, the general maximum is 125 percent. Other motors use 115 percent. If the motor will not start, 430.32(C) allows higher settings up to 130 or 140 percent.

Question 4. A contractor is installing a 240-volt, single-phase circuit that supplies a 40-ampere load located 150 feet from the panelboard, measured one way. The conductors are 6 AWG copper, which has a circular mil area of 26,240. Use a resistivity constant of 12.9 ohm-circular mils per foot. What is the approximate voltage drop as a percentage of source voltage?

A. 1.2 percent
B. 2.1 percent
C. 2.5 percent
D. 4.9 percent

Correct Answer: C. 2.5 percent

Answer Explanation:
For a single-phase circuit, voltage drop equals 2 × K × I × L divided by the circular mil area, where L is the one-way length. That gives 2 × 12.9 × 40 × 150 divided by 26,240, or about 5.9 volts. Dividing by the 240-volt source gives 2.46 percent, which rounds to 2.5 percent. The factor of 2 accounts for current traveling out and back on two conductors. The informational notes in 210.19 and 215.2 suggest limiting branch-circuit drop to 3 percent and total feeder and branch drop to 5 percent. They are recommendations, not mandatory rules, unless a design specification requires them.

Why the Other Options Are Incorrect:
A. 1.2 percent — Incorrect. This leaves out the factor of 2 and counts only one conductor’s length, which understates the drop by half.
B. 2.1 percent — Incorrect. This uses the three-phase multiplier of 1.732 instead of 2, which does not fit a single-phase circuit.
D. 4.9 percent — Incorrect. This effectively doubles the length a second time, producing twice the correct drop.

Quick Study Guide:
Single-phase voltage drop uses 2KIL divided by circular mils, with L measured one way. Three-phase circuits use 1.732 instead of 2. Divide the result by source voltage for a percentage. The 3 percent and 5 percent figures in the NEC are informational recommendations unless a specification or other code section makes them mandatory.

Question 5. An electrician is pulling six current-carrying 10 AWG THHN copper conductors through a single EMT run in a commercial building. Ambient temperature is 86°F (30°C), and all equipment terminals are rated for 75°C. The 90°C table ampacity for 10 AWG copper is 40 amperes, and the 75°C table ampacity is 35 amperes. The foreman asks what the maximum allowable ampacity of these conductors is. Which answer is correct?

A. 35 amperes
B. 32 amperes
C. 40 amperes
D. 28 amperes

Correct Answer: B. 32 amperes

Answer Explanation:
When more than three current-carrying conductors share a raceway, the adjustment factors in NEC 310.15(C)(1) apply. Four to six conductors require an 80 percent adjustment. The adjustment may start from the 90°C ampacity of the THHN insulation, so 40 × 0.80 equals 32 amperes. That result is less than the 35-ampere limit from the 75°C terminal rating, so the lower figure of 32 amperes governs. No ambient temperature correction is needed at 30°C. The contractor should also remember that 240.4(D) limits overcurrent protection for 10 AWG copper to 30 amperes in most cases.

Why the Other Options Are Incorrect:
A. 35 amperes — Incorrect. This is the 75°C table value and ignores the 80 percent adjustment for six current-carrying conductors.
C. 40 amperes — Incorrect. This is the unadjusted 90°C table value, which overlooks both the adjustment factor and the terminal limit.
D. 28 amperes — Incorrect. This applies 70 percent, the factor for seven to nine conductors, which does not fit six conductors.

Quick Study Guide:
For adjusted ampacity, start with the insulation’s table value, apply the correction and adjustment factors, then compare the result to the terminal temperature limit. The lower number governs. Remember the conductor-count factors: 80 percent for four to six, 70 percent for seven to nine, and 50 percent for ten to twenty. Also check 240.4(D) for small-conductor limits.