300 questions
A single-family dwelling has 2,000 sq ft of habitable floor area. Using the standard calculation method at 3 VA per sq ft, what is the general lighting and general-use receptacle load before any demand factors?
- a.3,000 VA
- b.6,000 VA✓
- c.4,000 VA
- d.8,000 VA
General lighting is figured at 3 VA per sq ft: 2,000 sq ft x 3 VA = 6,000 VA. This value is calculated before applying the general-lighting demand factors.
In the standard dwelling calculation, what is the minimum load to include for two required small-appliance branch circuits plus one laundry branch circuit, before any demand factor?
- a.1,500 VA
- b.4,500 VA✓
- c.6,000 VA
- d.3,000 VA
Each small-appliance and laundry circuit is figured at 1,500 VA. Two small-appliance circuits (2 x 1,500 = 3,000) plus one laundry circuit (1,500) equals 4,500 VA.
A dwelling has 7,200 VA general lighting, 3,000 VA small-appliance, and 1,500 VA laundry. Applying the standard demand factor (first 3,000 VA at 100%, remainder at 35%), what is the net general lighting load?
- a.4,095 VA
- b.11,700 VA
- c.7,200 VA
- d.6,045 VA✓
Sum = 7,200 + 3,000 + 1,500 = 11,700 VA. First 3,000 at 100% = 3,000; remaining 8,700 at 35% = 3,045. Net = 3,000 + 3,045 = 6,045 VA.
Using the demand table for household electric ranges, what is the demand load for one 12-kW electric range in a dwelling (Column C, one appliance)?
- a.9,600 VA
- b.12,000 VA
- c.6,000 VA
- d.8,000 VA✓
For a single household range not over 12 kW, the Column C demand is 8 kW = 8,000 VA. The 12-kW nameplate is reduced to an 8-kW calculated demand for one range.
What is the demand load for two 12-kW household electric ranges served from one feeder (range demand table, Column C, two appliances)?
- a.11,000 VA✓
- b.12,000 VA
- c.19,200 VA
- d.16,000 VA
Two ranges not over 12 kW have a Column C demand of 11 kW = 11,000 VA. This is less than the 24,000 VA connected total because of diversity.
A single household range is rated 16 kW. Using the 5% increase per kW over 12 kW rule on the 8-kW base, what is the demand load?
- a.9,600 VA✓
- b.16,000 VA
- c.10,400 VA
- d.8,000 VA
16 kW is 4 kW over 12 kW, so add 5% x 4 = 20% to the 8-kW base: 8,000 x 1.20 = 9,600 VA.
A dwelling has one 4-kW electric clothes dryer. What value must be used as the dryer demand load in the service calculation?
- a.5,500 VA
- b.4,500 VA
- c.4,000 VA
- d.5,000 VA✓
The dryer load is the nameplate rating or 5,000 VA, whichever is larger. Since 4,000 VA is below 5,000 VA, 5,000 VA must be used.2023 NEC 220.54
A dwelling has four fastened-in-place appliances (dishwasher, disposal, water heater, compactor) totaling 6,000 VA. What is the demand load after applying the fastened-appliance demand factor?
- a.3,000 VA
- b.4,500 VA✓
- c.6,000 VA
- d.7,500 VA
Where four or more fastened-in-place appliances are on the same feeder, a 75% demand factor may be applied: 6,000 x 0.75 = 4,500 VA.2023 NEC 220.53
A dwelling's total calculated load is 24,000 VA on a 240-V, single-phase service. What is the calculated service current?
- a.167 A
- b.100 A✓
- c.120 A
- d.83 A
Current equals volt-amperes divided by voltage: 24,000 VA / 240 V = 100 A.
Using the 75 C column, what is the smallest copper conductor rated to carry a 100-A service load?
- a.#2 AWG
- b.#4 AWG
- c.#3 AWG✓
- d.#1 AWG
At 75 C, #4 Cu is rated 85 A and #3 Cu is rated 100 A. The #3 AWG copper is the smallest conductor that meets the 100-A requirement.2023 NEC Table 310.16
For a 200-A dwelling service using the 83% conductor allowance, what is the smallest copper ungrounded conductor (75 C) permitted for the main power feeder?
- a.#1/0 AWG
- b.#3/0 AWG
- c.#2/0 AWG✓
- d.#4/0 AWG
200 A x 0.83 = 166 A required ampacity. At 75 C, #2/0 Cu is rated 175 A, which satisfies 166 A, so #2/0 is the smallest permitted.2023 NEC 310.12
An electric range has a Column C demand of 8,000 VA. Using the 70% neutral demand for ranges, what is the range contribution to the feeder neutral load?
- a.8,000 VA
- b.5,600 VA✓
- c.4,000 VA
- d.7,000 VA
The feeder neutral for a household range may be figured at 70% of the range demand: 8,000 x 0.70 = 5,600 VA.
From the motor full-load current table, what is the full-load current of a 25-HP, 480-V, three-phase induction motor?
- a.28 A
- b.42 A
- c.34 A✓
- d.40 A
The table value for a 25-HP, 460-480-V three-phase motor is 34 A. Table full-load current (FLC), not nameplate, is used for conductor and protection sizing.2023 NEC Table 430.250
A single continuous-duty motor has a table full-load current of 34 A. What minimum conductor ampacity is required for its branch circuit?
- a.50 A
- b.34 A
- c.40 A
- d.42.5 A✓
Motor branch-circuit conductors must be at least 125% of the FLC: 34 A x 1.25 = 42.5 A minimum ampacity.2023 NEC 430.22
A motor with a service factor of 1.15 has a nameplate full-load current of 34 A. What is the maximum overload protection setting at 125%?
- a.42.5 A✓
- b.45.9 A
- c.34 A
- d.39.1 A
For a motor with service factor 1.15 or greater, overload is set at up to 125% of nameplate FLA: 34 x 1.25 = 42.5 A.2023 NEC 430.32
For a motor with a 34-A full-load current, what is the maximum inverse-time circuit breaker for branch-circuit short-circuit and ground-fault protection (250%, then next higher standard size)?
- a.70 A
- b.100 A
- c.90 A✓
- d.80 A
Inverse-time breaker: 250% of FLC = 34 x 2.50 = 85 A. Where this does not correspond to a standard size, the next higher standard rating (90 A) is permitted.2023 NEC 430.52
A feeder supplies three motors with full-load currents of 34 A, 28 A, and 22 A. What minimum conductor ampacity is required for the feeder?
- a.110 A
- b.92.5 A✓
- c.105 A
- d.84 A
Feeder ampacity = 125% of the largest motor FLC plus the sum of the others: (34 x 1.25) + 28 + 22 = 42.5 + 50 = 92.5 A.2023 NEC 430.24
What is the full-load primary current of a 75-kVA, 480-V, three-phase transformer?
- a.72 A
- b.104 A
- c.90 A✓
- d.156 A
Three-phase current = VA / (1.732 x V): 75,000 / (1.732 x 480) = 90.2 A, about 90 A.
What is the full-load secondary current of a 75-kVA, three-phase transformer with a 208-V secondary?
- a.208 A✓
- b.180 A
- c.156 A
- d.90 A
Secondary current = 75,000 / (1.732 x 208) = 208.2 A, about 208 A.
A 75-kVA transformer has a 90-A primary current (over 9 A). With primary-only protection at 125% max, what is the primary overcurrent device rating (next higher standard permitted)?
- a.150 A
- b.100 A
- c.125 A✓
- d.110 A
125% of 90 A = 112.5 A. Where this does not match a standard rating, the next higher standard device (125 A) is permitted for primary-only protection.2023 NEC 450.3(B)
A 120-V branch circuit carries 16 A over a one-way run of 80 ft using #12 Cu (circular mils = 6,530, K = 12.9). What is the approximate voltage drop?
- a.5.1 V✓
- b.7.4 V
- c.3.0 V
- d.2.5 V
Single-phase VD = (2 x K x I x L) / CM = (2 x 12.9 x 16 x 80) / 6,530 = 33,024 / 6,530 = 5.06 V.
Using the previous circuit (5.06-V drop on a 120-V circuit), what is the approximate percent voltage drop?
- a.3.0%
- b.6.0%
- c.4.2%✓
- d.2.1%
Percent VD = 5.06 / 120 = 0.042 = 4.2%, which exceeds the 3% recommended for branch circuits.
A 208-V three-phase feeder carries 40 A over 150 ft using #6 Cu (CM = 26,240, K = 12.9). What is the approximate voltage drop?
- a.3.0 V
- b.7.2 V
- c.9.0 V
- d.5.1 V✓
Three-phase VD = (1.732 x K x I x L) / CM = (1.732 x 12.9 x 40 x 150) / 26,240 = 134,050 / 26,240 = 5.11 V.
For a 112.5-kVA, 208-V, three-phase transformer with 2% impedance, what is the approximate available fault current at the secondary terminals (infinite primary)?
- a.15,617 A✓
- b.20,000 A
- c.6,250 A
- d.10,000 A
Secondary FLA = 112,500 / (1.732 x 208) = 312.3 A. Isc = FLA / %Z = 312.3 / 0.02 = 15,617 A.
A 45-kVA, 208-V, three-phase transformer has 2% impedance. What is the approximate available fault current at its secondary (infinite primary)?
- a.12,500 A
- b.3,120 A
- c.6,245 A✓
- d.9,000 A
Secondary FLA = 45,000 / (1.732 x 208) = 124.9 A. Isc = 124.9 / 0.02 = 6,245 A.
Six current-carrying THHN conductors share one raceway. If the 90 C ampacity of #6 Cu is 75 A, what is the adjusted ampacity after the fill adjustment factor for 4-6 conductors?
- a.55 A
- b.60 A✓
- c.52.5 A
- d.75 A
For 4 to 6 current-carrying conductors, apply an 80% adjustment factor: 75 A x 0.80 = 60 A.2023 NEC Table 310.15(C)(1)
A #3 Cu THHN conductor has a 90 C ampacity of 110 A. In a 50 C ambient (correction factor 0.82), what is the corrected ampacity?
- a.90 A✓
- b.82 A
- c.100 A
- d.110 A
Corrected ampacity = 110 A x 0.82 = 90.2 A, about 90 A, before comparing to the termination temperature limit.2023 NEC Table 310.15(B)(1)
A #2 Cu THHN (90 C ampacity 130 A) runs with 6 current-carrying conductors (0.80 factor) in a 40 C ambient (0.91 factor). What is the adjusted-and-corrected ampacity?
- a.130 A
- b.95 A✓
- c.104 A
- d.115 A
Apply both factors to the 90 C ampacity: 130 x 0.80 x 0.91 = 94.6 A, about 95 A.
An office has 45 general-use receptacle outlets. At 180 VA per outlet, what is the receptacle load before demand factors?
- a.5,400 VA
- b.4,500 VA
- c.9,000 VA
- d.8,100 VA✓
Each general-use receptacle outlet is figured at 180 VA: 45 x 180 = 8,100 VA.2023 NEC 220.14(I)
A commercial building has a 30,000-VA receptacle load. Applying the demand (first 10 kVA at 100%, remainder at 50%), what is the demand load?
- a.25,000 VA
- b.15,000 VA
- c.30,000 VA
- d.20,000 VA✓
First 10,000 VA at 100% = 10,000; remaining 20,000 at 50% = 10,000. Total demand = 20,000 VA.
What is the maximum number of service disconnecting means permitted for a single set of service-entrance conductors (each in a separate enclosure)?
- a.Eight
- b.Six✓
- c.One
- d.Four
A service is limited to a maximum of six disconnecting means. In the 2023 NEC each of the up-to-six disconnects must be in a separate enclosure or be part of listed equipment.2023 NEC 230.71
A service uses #2/0 Cu ungrounded conductors. What is the minimum copper grounding electrode conductor?
- a.#4 AWG✓
- b.#2 AWG
- c.#8 AWG
- d.#6 AWG
For service conductors over 1/0 through 3/0 Cu, the table requires a #4 Cu grounding electrode conductor.2023 NEC Table 250.66
Where the grounding electrode conductor connects only to a made ground rod, what is the largest copper conductor size ever required?
- a.#4 AWG
- b.#6 AWG✓
- c.#8 AWG
- d.#2 AWG
The portion of the GEC connecting solely to a rod, pipe, or plate electrode need not be larger than #6 Cu.2023 NEC 250.66(A)
A branch circuit is protected by a 100-A overcurrent device. What is the minimum copper equipment grounding conductor?
- a.#4 AWG
- b.#8 AWG✓
- c.#6 AWG
- d.#10 AWG
The equipment grounding conductor table requires #8 Cu for an overcurrent device rating up to 100 A.2023 NEC Table 250.122
A feeder is protected by a 400-A overcurrent device. What is the minimum copper equipment grounding conductor?
- a.#6 AWG
- b.#4 AWG
- c.#1 AWG
- d.#3 AWG✓
For a 400-A overcurrent device the table requires #3 Cu for the equipment grounding conductor.2023 NEC Table 250.122
A service has 500-kcmil Cu ungrounded conductors per phase. What is the minimum copper grounded (neutral) service conductor for bonding purposes?
- a.1/0 AWG✓
- b.#1 AWG
- c.#2 AWG
- d.#4 AWG
The grounded service conductor must be at least the size in Table 250.102(C)(1); for 500-kcmil Cu phase conductors that value is 1/0 Cu.2023 NEC Table 250.102(C)(1)
Service phase conductors total 1,500 kcmil Cu per phase (over 1,100 kcmil). Using the 12.5% rule, what is the minimum copper main bonding jumper?
- a.4/0 AWG✓
- b.250 kcmil
- c.2/0 AWG
- d.3/0 AWG
Above 1,100 kcmil, the main bonding jumper is at least 12.5% of the phase area: 0.125 x 1,500 = 187.5 kcmil, so 4/0 Cu (211.6 kcmil) is the smallest that qualifies.2023 NEC 250.28(D)
A panelboard is rated 225 A. What is the maximum rating of the overcurrent device that may protect it?
- a.225 A✓
- b.200 A
- c.400 A
- d.250 A
Each panelboard must be protected by an overcurrent device rated no more than the panelboard rating, so 225 A is the maximum.2023 NEC 408.36
Which of the following is the next standard overcurrent device rating above 95 A?
- a.90 A
- b.125 A
- c.110 A
- d.100 A✓
Standard ratings include 90, 100, 110, and 125 A. The next standard size above 95 A is 100 A.2023 NEC 240.6
Absent a specific exception, what is the maximum overcurrent protection for a #12 Cu conductor?
- a.20 A✓
- b.25 A
- c.30 A
- d.15 A
Under the small-conductor rule, #12 Cu is limited to a maximum overcurrent device of 20 A.2023 NEC 240.4(D)
Absent a specific exception, what is the maximum overcurrent protection for a #10 Cu conductor?
- a.25 A
- b.40 A
- c.30 A✓
- d.20 A
The small-conductor rule limits #10 Cu to a maximum 30-A overcurrent device.2023 NEC 240.4(D)
A conductor has an allowable ampacity of 115 A, which is not a standard rating. Under the next-standard-size rule (600 A or less), what is the largest permitted overcurrent device?
- a.125 A✓
- b.100 A
- c.110 A
- d.150 A
When ampacity does not match a standard rating and conditions are met, the next higher standard device (125 A) is permitted for 800-A or smaller circuits.2023 NEC 240.4(B)
What is the smallest conductor size generally permitted to be installed in parallel?
- a.1/0 AWG✓
- b.#1 AWG
- c.2/0 AWG
- d.#2 AWG
Conductors are generally permitted in parallel only in sizes 1/0 AWG and larger.2023 NEC 310.10(G)
Two paralleled sets of 500-kcmil Cu (75 C, 380 A each) supply one feeder. What is the combined ampacity before adjustment?
- a.620 A
- b.760 A✓
- c.380 A
- d.500 A
Paralleled conductor ampacities add: 380 A x 2 = 760 A total for the feeder.2023 NEC Table 310.16
What is the minimum service disconnecting-means rating for a one-family dwelling?
- a.200 A
- b.125 A
- c.60 A
- d.100 A✓
The service disconnecting means for a one-family dwelling must be rated at least 100 A, 3-wire.2023 NEC 230.79
What is the minimum vertical clearance for an overhead service drop, not over 300 V to ground, above a residential driveway that is not subject to truck traffic?
- a.15 ft
- b.12 ft✓
- c.10 ft
- d.18 ft
Service-drop conductors up to 300 V to ground require 12 ft of clearance over residential property and driveways not subject to truck traffic.2023 NEC 230.24(B)
At a separately derived system such as a transformer, what conductor connects the grounded conductor to the system grounding electrode?
- a.Equipment bonding jumper
- b.Main bonding jumper
- c.Grounding electrode conductor✓
- d.System bonding jumper
Per 2023 NEC 250.30(A)(5), the grounding electrode conductor connects the grounded conductor of a separately derived system to the grounding electrode. Trap: the system bonding jumper (250.30(A)(1)) bonds the grounded conductor to the equipment/enclosure, not to the electrode.2023 NEC 250.30(A)(5)
An emergency system must be capable of supplying power to its loads within what maximum time after normal supply is lost?
- a.120 seconds
- b.10 seconds✓
- c.60 seconds
- d.30 seconds
Emergency systems must restore power to life-safety loads within 10 seconds of loss of the normal source.2023 NEC 700.12
A legally required standby system must supply its loads within what maximum time after loss of normal power?
- a.30 seconds
- b.10 seconds
- c.60 seconds✓
- d.120 seconds
Legally required standby systems must be able to supply their loads within 60 seconds of a normal-source failure.2023 NEC 701.12
Generator output conductors must have an ampacity of at least what percent of the nameplate current?
- a.100%
- b.125%
- c.110%
- d.115%✓
Generator conductors from the terminals to the first overcurrent device must be at least 115% of the nameplate current rating.2023 NEC 445.13
这门考试有多难?
电工技师(Master)执照由各州主办、以 NEC 为依据,因此格式因州而异。例如在德州,考试分两个开卷部分(2023 NEC)——75 道 NEC 知识题与 33 道计算题(共 108 题)——各须 70% 方可通过,费用 78 美元。电工年薪中位数约 62,350 美元(BLS,2024 年 5 月)。
- 推荐学习时间
- 多数人 80-150 小时——技师考试比熟练工增加更重的计算与规范应用深度。
- 官方公布的通过率
- 19.08% (来源未说明统计的是哪些考次)(n = 3,946);22.21% (来源未说明统计的是哪些考次)(n = 3,472) —— Texas TDLR,FY 2025。请把它们看作两场考试而非一场:TDLR 并没有名为「Master Electrician」的行,只有「Master Calculations」(第一个数字)和「Master NEC」(第二个数字)。任何引用单一「德州 master 通过率 19%」的说法,都悄悄丢掉了一半考试。执照按州发放,因此这只适用于德州。来源: Texas TDLR — Electrician Exam Statistics, Fiscal Year 2025
- 重点学习方向
- 服务、服务设备与单独派生系统是最重的板块之一(德州约 18%),外加负载计算部分。
费用与薪资为近似值,会随时间变动。上方的通过率引自旁边链接的来源,并限于该来源覆盖的期间——凡是我们尚未核实来源的,都会直接说明并且不给数字。