





7.

Determine the MAXIMUM initial standard size nontime delay fuses permitted for branchcircuit, shortcircuit and groundfault protection for a 5 hp, 240volt, singlephase, ac continuousduty motor with no code letter indicated on the nameplate.


















As shown in Table 430.248 the fullload current rating of the motor is 28 amperes. This value is to be used when sizing the overcurrent protective devices from Table 430.52, which shows nontimedelay fuses are to be sized at not more than 300% of the FLC of the motor. 28 amperes x 300% = 84 amperes. According to Section 240.6(A), this is not a standard size fuse rating therefore, Section 430.52(C)(1),EX.1, permits you to go up to the next standard size fuses having a rating of 90 amperes.
FLC of motor = 28 amperes [Tbl. 430.248]
28 amperes x 300% = 84 amperes [Tbl. 430.52]
Incorrect answer, please choose another answer.






12.

Under which, if any, of the following conditions is the neutral NOT to be counted as a current carrying conductor?
I) When it is only carrying the unbalanced current.
II) When it is the neutral of a 3phase wyeconnected system where the major portion of the load consists of nonlinear loads.


















Section 310.15(B)(5)(a), indicates where a neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be counted as a currentcarrying conductor. An example of this would be a 120/240volt, singlephase, system supplying incandescent luminaires, generaluse receptacles and appliances of a residence. Section 310.15(B)(5)(c), mandates on a 3phase, 4wire circuit where the major portion of the load consists of nonlinear loads, the neutral conductor is considered a currentcarrying conductor. A good example of this type of system is a 208Y/120volt, 3phase system that supplies fluorescent luminaires in a school or office building.
Incorrect answer, please choose another answer.

















28.

Apply the general method of calculation and determine the demand load, in VA, for a onefamily dwelling having a total connected load of 96,400 VA for the general lighting, small appliance and laundry loads.


















When applying the general method of calculation for dwelling units the demand factors specified in Table 220.42 are to be applied. The concept for applying demand factors is that the entire connected load is not likely to be used at one time. To size the serviceentrance conductors for a dwelling unit with a connected load of 96,400 VA apply the values as specified in Table 220.42 as shown:
1st 3,000 VA @ 100% = 3,000 VA
remainder (96,400 VA  3,000 VA) = 93,400 VA @ 35% = 32,690 VA
Demand load = 35,690 VA
Incorrect answer, please choose another answer.








35.

What is the MINIMUM general lighting demand load, in VA, of an industrial commercial (loft) building having dimensions of 100 ft. x 300 ft.?


















As shown in Table 220.12 when calculating the general lighting load of a industrial commercial (loft) building, a unit load of 2 VA per sq. ft. is to be applied. Also as per Section 230.42(A)(1), because this load is considered a continuous load, you are to multiply this value by 125%. First, find the total square feet of the building.
100 ft. x 300 ft. = 30,000 sq. ft. x 2 VA = 60,000 VA
x 125% demand = 75,000 VA
Incorrect answer, please choose another answer.

























59.

Determine the ampacity of a size 3 AWG THHN copper conductor given the following related information:
Three (3) currentcarrying conductors are in the raceway.
The ambient temperature is 35 deg. C.
The terminations are rated at 60 deg. C.


















Section 110.14(C) states the temperature rating associated with the ampacity of a conductor shall be selected so as not to exceed the temperature of any connected terminal. But, for conductors with a temperature rating with a higher value than that of the terminations, the higher value can be used for ampacity adjustment and correction. Prior to derating, Table 310.15(B)(16) shows size 3 AWG THHN copper conductors to have an ampacity of 115 amperes, which is to be multiplied by 0.96, due to the ambient temperature [selected from Table 310.15(B)(2)(a), where the temperature factors are shown]. Therefore the allowable ampacity is reduced to 110.4 amperes. However, since the terminations are rated for 60C, the ampacity of the conductor is to be selected from the 60C column, which shows the allowable ampacity to be 85 amperes.
3 AWG THHN ampacity before derating = 115 amperes (90 deg. C Col.)
115 amperes x .96 (temperature correction) = 110.4 amperes.
3 AWG (60 deg. C Col.) = 85 amperes is to be used.
Incorrect answer, please choose another answer.










68.

All 125volt, singlephase, 15 and 20ampere receptacles installed in residential garages shall be _____.
I) protected by a listed arcfault circuit interrupter
II) provided with groundfault circuit interrupter protection


















According to Section 210.8(A)(2), all singlephase, 125volt, 15 or 20ampere receptacles located in residential garages are required to be provided with groundfault protection for personnel. This requirement improves safety for persons using portable handheld portable tools and similar tools that might be connected to these receptacles. GFCI protection is also required because auto repair work and maintenance is performed and workshop electrical tools are often used in the garage. There are no exceptions to this rule, regardless of where in the garage the receptacle is located or the equipment served. Most safety experts agree that GFCIs are directly responsible for saving numerous lives and preventing countless injuries. Section 210.12(A) lists the areas of a dwelling unit where all 120volt, single phase, 15 and 20ampere branchcircuits supplying outlets and devices are mandated to have AFCI protection. Residential garages are not are not included in this list. However, AFCI devices may also be capable of performing other functions such as overcurrent protection, groundfault protection and surge suppression. Therefore, AFCI devices are certainly permitted to be installed in residential garages, but not required.
Incorrect answer, please choose another answer.










77.

Where a 3phase rooftop mounted air conditioning unit is supplied by three (3) size 6 AWG THWN copper conductors installed in a 1 in. EMT, determine the allowable ampacity of the conductors, where given the following related information: ambient temperature is 110F the EMT is located within three (3) inches of the roof surface the EMT is exposed to direct sunlight


















Due to the close proximity of the conduit to the rooftop, first find the temperature adder, 40 F, as shown in Table 310.15(B)(3)(c). Then, add this value to the ambient temperature.
110F (ambient temperature) +40F (adder) 150F (to be used for derating)
Next, locate the temperature correction factor to be applied, 0.33, from Table 310.15(B)(2)(a). Finally, multiply 0.33 x 65 amperes, the allowable ampacity of size 6 AWG THWN copper conductors before derating, as indicated in Table 310.15(B)(16).
Size 6 AWG THWN ampacity (before derating) = 65 amperes 65 amperes x .33 (temperature correction) = 21.45 amperes
Incorrect answer, please choose another answer.




80.

Determine the MINIMUM number of 20ampere, 120volt general lighting branch circuits required for a 12 unit apartment building that is 12,000 sq. ft. in size. The apartment building is not designed for permanent residents and does not contain cooking facilities.


















Section 220.12 requires a unit load of not less than that specified in Table 220.12 is to be used for determining the minimum general lighting load.
Multiply this value by the sq. ft. of the occupancy. 12,000 sq. ft. x 2 VA = 24,000 VA (building lighting VA)
Next, find the VA of one circuit by multiplying volts x amperes. 120 volts x 20 amperes = 2,400 VA (one circuit VA)
Finally, divide the load, 24,000 VA, by one circuit, 2,400 VA. 24,000 VA (blg.) 2,400 VA (one circuit) = 10 general lighting circuits
Incorrect answer, please choose another answer.









88.

When a circuit breaker is used as the disconnecting means for a 20 hp, 480volt, 3phase, continuousduty motor, the circuit breaker shall have an ampere rating of at LEAST _____.


















First, locate the fullload current of the motor, 27 amperes, from Table 430.250 next, multiply by 115%, as per Section 430.110(A).
FLC of motor = 27 amperes x 115% = 31.05 amperes
Incorrect answer, please choose another answer.



90.

Storage batteries for legally required standby systems shall have the capacity to maintain not less than 87 percent of the system voltage of the circuits supplying the legally required standby power for a period of at LEAST _____ upon loss of the normal power.


















As covered in Section 700.12, an approved supply system for legally required standby purposes, in addition to the normal services to the building, may be a storage battery backup system. Section 700.12(A) requires a storage battery to have a suitable rating and capacity to supply and maintain not less than 87 percent of system voltage the total load of the circuits supplying legally required standby power for a period of not less than 1 hours. Legally required systems are intended to provide electric power to municipal first responders such as fire stations, sheriff offices, highway departments, police stations and similar operations to aid in fire fighting, emergency situations, rescue operations and control of health hazards. When normal power is lost, legally required systems must be able to supply standby power in 60 seconds or less.
Incorrect answer, please choose another answer.




93.

A 120/240volt, singlephase feeder of a retail department store is to supply a noncontinuous load of 18,000 VA and a continuous load of 13,356 VA. What MINIMUM size 75 deg. C copper conductors are required for this installation?


















In compliance with Section 215.2(A)(1)(a), feeder conductors are to have an ampacity of 100% of the continuous loads, plus 125% of the continuous loads. Therefore:
18,000 VA x 100% = 18,000 VA 13,356 VA x 125% = 16,695 VA 34,695 VA total
To find the load, apply the singlephase current formula:
I = 34,695 VA = 144 amperes 240 volts
As shown in Table 310.15(B)(16), size 1/0 AWG conductors with an allowable ampacity of 150 amperes should be selected.
Incorrect answer, please choose another answer.








