1.

Photovoltaic systems with dc source circuits, dc output circuits, or both, operating at a PV system MAXIMUM system voltage of at LEAST_____ or greater, is required to be protected by a listed (dc) arcfault circuit interrupter, PV type, or other system components listed to provide equivalent protection.


















The requirement indicated in Section 690.11 is limited to PV systems with a
maximum voltage of 80 volts or greater. PV systems can be subjected to extreme
environmental conditions such as, wind, rain, snow, ice, and elevated temperature
extremes. These systems can deteriorate over time and eventually develop
insulation failures or internal PV module conductor fault conditions. The AFCI device
is provided to interrupt arcing faults resulting from a failure in the intended
continuity of a conductor, connection, module, or other system component in the dc
PV circuits.
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39.

Where given the following related information, determine the MAXIMUM standard size overcurrent protection required for the primary and secondary side of a transformer, when both primary and secondary overcurrent protection is to be provided.
* 150 kVA rating
* Primary  480volts, 3phase, 3wire
* Secondary  208Y/120volts, 3phase, 4wire


















As per Section 450.3(B), to determine the overcurrent protection required for a
transformer of less than 600 volts, the protection shall be provided in accordance
with Table 450.3(B). First determine the fullload current rating of the transformer
by applying the 3phase current formula, then multiply by the values shown in Table 450.3(B). (Primary)
I = VA I = 150,000 = 150,000 = 180 amps x 250% = 450 amperes
E x 1.732 480 x 1.732 831.36
(Secondary)
I = VA I = 150,000 = 150,000 = 416 amps x 125% = 520 amperes
E x 1.732 208 x 1.732 36.25
Since this fullload current value does not correspond to a standard size overcurrent
device, and it is permitted to go up to the next standard size, a 600 ampere rated
overcurrent protective device is permitted.
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41.

Apply no exceptions and determine the MAXIMUM standard size timedelay fuses permitted to be used for branchcircuit, shortcircuit, and groundfault protection for a 15 hp, 480volt, threephase, continuousduty, wound rotor motor.


















In compliance with the motor rules specified in Section 430.6(A)(1), for 3phase,
general motor applications, the ampere rating of the overcurrent protective devices
are to be based on the motor fullload running current value given in Table 430.250. The table shows the motor to have a fullload current rating of 21 amperes. The actual rating or setting of the overcurrent protection devices are to
be sized in accordance with Table 430.52.
FLC of motor = 21 amperes x 150% = 31.50 amperes
In this question exceptions are not to be applied, therefore you are required to go
down to the next standard size fuses, indicated in Section 240.6(A), having a rating
of 30 amperes.
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44.

A 6,000 sq. ft. bank building is to be constructed and the number of generalpurpose receptacles to be installed in the building is yet to be determined; the building will also have an electric sign. Determine the MINIMUM number of 120volt, 20ampere branch circuits required for the lighting, receptacles and the sign.


















To solve this problem, take into consideration the following:
1) As per Article 100, lighting circuits for a commercial building such as a bank
are considered continuous loads.
2) In compliance with Section 210.19(A)(1), branchcircuits must have an allowable
ampacity of not less than 125% of the continuous loads to be served.
3) Generaluse receptacles are not considered continuous loads.
4) Electric signs are to be on a separate dedicated circuit. Section 600.5(A)
5) Minimum lighting loads are to be calculated in accordance with the values
given on Table 220.12.
6) As per Section 220.14(K)(2), when the number of receptacles to be installed is
unknown, an additional one (1) VA per sq. ft. is to be added to the general
lighting load.
Therefore, do the math as shown to determine the minimum number of 120volt,
20ampere branch circuits required for this building.
6,000 sq. ft. x 3.5 VA = 21,000 VA x 125% = 26,250 VA (lighting)
6,000 sq. ft. x 1.0 VA = 6,000 VA (receptacles)
32,250 VA (receptacles & lighting)
120 volts x 20 amperes = 2,400 VA of one circuit
32,250 VA (load) 2,400 VA (one circuit) = 13.4 circuits = 14 circuits
14 circuits (lighting & receptacles) + 1 sign circuit = 15 circuits total
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46.

Where a 200 kW, 3phase, 480Y/277volt, 4wire backup generator is installed at a commercial building and the design and operation of the generator does not prevent overloading, determine the MINIMUM size THWN copper conductors required from the generator terminals to the first distribution device(s) containing overcurrent protection.


















To determine the allowable ampacity of the conductors required from the terminals
of the generator to the first overcurrent device, first apply the 3phase current
formula to find the fullload current rating of the generator:
I = kW x 1,000 I = 200 x 1,000 = 200,000 = 240.56 amperes (FLC)
volts x 1.732 480 x 1.732 831.36
Next, as per Section 445.13, multiply by 115%:
241 amperes x 115% = 277 amperes (required ampacity of conductors)
Finally, size 300 kcmil THWN conductors with an allowable ampacity of 285 amperes
should be selected from Table 310.15(B)(16).
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48.

A dwelling unit to be built will have 1,400 sq. ft. of livable space on the main floor, a 1,400 sq. ft. basement (unfinished but adaptable for future use), a 200 sq. ft. open porch and a 600 sq. ft. garage. Determine the MINIMUM number of 15 ampere, 120volt general lighting branchcircuits required for the dwelling.


















To determine the minimum number of 15ampere, 120volt general lighting circuits
required for the dwelling, refer to Section 220.12 and note that open porches and
garages are not to be included in the calculation. A unit load not less than specified
in Table 220.12 shall constitute the minimum lighting load.
First, determine the usable square feet of the building and multiply by 3 VA as
shown in Table 220.12:
1,400 sq. ft. + 1,400 sq. ft. = 2,800 sq. ft. x 3 VA = 8,400 VA
Next, find the VA of one circuit by multiplying 120 volts by 15 amperes:
120 volts x 15 amperes = 1,800 VA of one circuit
Then, divide the lighting load (8,400 VA) by the VA of one circuit (1,800 VA):
8,400 VA (load) = 4.6 = 5 circuits total
1800 VA (one circuit)
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51.

Thermal insulation is not permitted to be installed above a recessed luminaire or WITHIN _____ of the recessed luminaire's enclosure, wiring compartment, ballast, transformer, LED driver or power supply, unless the luminaire is identified as Type IC for insulation contact.


















A listed Type IC recessed luminaire is provided with thermal protection to deactivate
the lamp should the luminaire be mislamped so that it overheats. Unless a
recessed luminaire is listed as Type IC, Section 410.116(B) requires thermal
insulation to not be installed above a recessed luminaire or within 3 in. of the
recessed luminaire's enclosure, wiring compartment, ballast, transformer, LED driver
or power supply.
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60.

Determine the MAXIMUM standard size timedelay fuses permitted for branchcircuit, shortcircuit and groundfault protection for a 50 hp, 3phase, 480volt, inductiontype motor with a fullload ampere rating of 61 amperes marked on the nameplate. Apply no exceptions.


















Section 430.6(A)(1) indicates to size the overcurrent protection for a 3phase 50
hp, 480volt, motor used in a general application, the fullload running current
value, 65 amperes, shown on Table 430.250 shall be used. For 3phase motors,
Table 430.52 requires the rating of timedelay fuses to be no more than 175% of
the FLC of the motor.
FLC of motor  65 amperes x 175% = 113.75 amperes
Since exceptions were not applicable on this question, you should have selected
fuses with a standard rating of 110 amperes from Section 240.6(A).
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72.

A retail store is to have eighty (80) feet of accent track lighting installed where the track lighting will be in use continuously during business hours. For the purpose of sizing branch circuits, feeders and service entrance conductors, determine the connected load in VA.


















For track lighting in nondwelling occupancies, to comply with Section 220.43(B) a
load of 150 VA is to be included for every two (2) feet of lighting track. To solve this
problem , multiply 150 VA per linear foot by 80 ft. of track, by 125% (continuous
load), then divide by 2.
150 VA x 80 ft. x 1.25 = 15,000 = 7,500 VA
2 2
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86.

A 120/240volt, singlephase feeder of a retail shopping mall is to supply a noncontinuous load of 20,000 VA and a continuous load 16,000 VA. Consider the overcurrent device protecting the feeder conductors is not listed for operation at 100 percent of its rating and determine the MINIMUM size 75C copper conductors required for this installation.


















As per Section 215.2(A)(1)(a), feeder conductors are required to have an ampacity
to carry 100% of the noncontinuous load, plus 125% of the continuous load:
20,000 VA x 100% = 20,000 VA
16,000 VA x 125% = 20,000 VA
Total = 40,000 VA
Apply the singlephase current formula to find the load:
I = power I = 40,000 VA = 166.6 amperes
volts 240 volts
In accordance with Table 310.15(B)(16), size 2/0 AWG 75 rated copper conductors
with an allowable ampacity of 175 amperes should be selected.
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