


4.

For other than dwelling units, all 125volt, singlephase, 15 and 20ampere receptacles located _____ shall be provided with GFCI protection for personnel.


















You should have selected, in all these locations, because Sections 210.8(B)(1),(2)&(3) requires all 15 and 20ampere, 125volt receptacles located in nondwellingtype kitchens, bathrooms, and on rooftops to be GFCI protected. This rule is in place to protect persons from accidental shock hazards and possible electrocution when using cordandplug connected appliances and equipment.
Incorrect answer, please choose another answer.








11.

What is the MINIMUM number of 120volt, 15ampere, general lighting branch circuits required for a dwelling unit having a calculated general lighting load of 9,600 VA?


















First find the VA of one circuit: 120 volts x 15 amperes = 1,800 VA
Next, divide the load of 9,600 VA by 1,800 VA (one circuit):
9,600 VA (load) = 5.3 = 6 circuits
1,800 VA (1 circuit)
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16.

Given: A busway has an ampere rating of 1,100 amperes. What is the MAXIMUM standard size overcurrent protection devices that may be used to protect the busway?


















In compliance with Section 240.4(C), the ampacity of the conductors shall be equal
to or greater than the overcurrent device, when over 800 amperes. Therefore, the
maximum standard size overcurrent devices that may be used to protect a busway
with an ampere rating of 1,100 amperes is 1,000 ampere rated overcurrent
protective devices. The standard ratings of overcurrent protective devices are listed
in Section 240.6(A).
Incorrect answer, please choose another answer.























38.

What is the MAXIMUM balanced demand load, in VA, permitted to be connected to a new service of a commercial building where given the following conditions?
I. The service is 208Y/120volts, 3phase, with a 600 ampere rated main circuit breaker.
II. The maximum load must not exceed 80 percent of the ampere rating of the main circuit breaker.


















To solve this problem, apply the 3phase power formula: P = I x E x 1.732, and multiply by 80%:
P = 600 amperes x 208 volts x 1.732 x 80% = 172,923 VA
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48.

Given: A onefamily dwelling to be built will have 4,000 sq. ft. of livable space, a 600 sq. ft. garage, a 400 sq. ft. open porch, a 2,000 sq. ft. unfinished basement (adaptable for future use), three (3) smallappliance branchcircuits and a branch circuit for the laundry room. Determine the demand load, in VA, on the ungrounded serviceentrance conductors for the general lighting and receptacle loads using the standard method of calculation for a onefamily dwelling.


















To solve this problem take into consideration the following related information
regarding calculations for dwelling units:
1) According to Section 220.12, a unit load of 3 VA as specified in Table 220.12 shall constitute the minimum lighting load. The calculation floor area shall not include garages and open porches.
2) As per Section 220.52(A), each small appliance circuit shall be calculated at 1,500 VA each.
3) Section 220.52(B) specifies a load of not less than 1,500 VA is to be included for the laundry branch circuit.
4) As permitted by Section 220.42, the demand factors specified in Table 220.42 shall apply to that portion of the total branch circuit load calculated for general illumination.
Therefore, to calculate the demand load do the math as follows.
4,000 sq. ft. + 2,000 sq. ft. = 6,000 sq. ft. x 3 VA = 18,000 VA
three small appliance circuits @ 1,500 VA each = 4,500 VA
one laundry circuit @ 1,500 VA = 1,500 VA
Total connected load = 24,000 VA
1st 3,000 VA @ 100% 3,000 VA
24,000 VA  3,000 VA = 21,000 VA (remainder) @ 35% = 7,350 VA
Total demand load = 10,350 VA
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49.

Determine the MINIMUM size Type SOW flexible cord that may be used to supply a 30 hp, 3phase, 480volt, continuousduty, ac motor from the motor controller to the motor terminations. Assume voltagedrop and elevated ambient temperature are not considerations.


















In compliance with Section 430.6, where flexible cord is used to supply motors, the
size of the conductors shall be selected in accordance with Section 400.5. For
general motor applications, Section 430.6(A)(1) indicates the fullload current
values given in Table 430.250 are to be used where sizing the conductors for a 3
phase ac motor. Section 430.22 requires conductors that supply a single motor
used in a continuous duty application shall have an ampacity of not less than 125%
of the motor fullload current rating.
Locate the FLC of the motor, 40 amperes, as shown in Table 430.250, then as per Section 430.22 multiply by 125% to determine the required ampacity of the conductors to supply the motor.
FLC of 30 hp motor = 40 amperes x 125% = 50 amperes
Size 4 AWG SOW cord with an allowable ampacity of 60 amperes should
be selected from Column A of Table 400.5(A)(1).
Incorrect answer, please choose another answer.














62.

Where a mobile home park has 25 mobile home lots calculated at 15,000 VA each, determine the MINIMUM required ampacity required for the ungrounded serviceentrance conductors.


















As required by Section 550.30, the secondary distribution system for a mobile home
lot is to be singlephase, 120/240volts. Section 550.31(1), mandates each
mobile home lot is to be calculated at a minimum of 16,000 VA; this is the value to
be used for this calculation. A demand factor of 24% is permitted to be applied as
shown in Table 550.31. First determine the demand load:
25 lots x 16,000 VA (minimum) = 400,000VA
"‹ X .24 (demand factor) 96,000 VA (demand load)
Next determine the load in amperes:
I = power I = 96,000 VA = 400 amperes
volts 240 volts
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78.

A twelve (12) unit apartment building will have a 5,500 watt electric clothes dryer placed in each apartment. Apply the general method of calculation and determine the demand load, in VA, on the ungrounded serviceentrance conductors for the dryers.


















Use Table 220.54 and first find the demand factor, in percent, for 12 dryers:
47%  1% (minus 1% for each dryer exceeding 11) = 46% demand factor
12 dryers x 5,500 watts = 66,000 VA (connected load) x 46% = 30,360 VA demand
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92.

Where an apartment complex has a calculated connected lighting load of 205.4 kVA, what is the demand load, in kVA, on the ungrounded serviceentrance conductors where applying the standard (general) method of calculation? Given: Each dwelling unit in the complex has cooking facilities provided.


















The demand factors specified in Table 220.42 may be applied to calculate the
general lighting demand loads for dwelling units.
First, convert into VA: 205.4 kVA x 1,000 = 205,400 VA
Next, apply the values given in Table 220.42:
first 3,000 VA @ 100% = 3,000 VA
3,001 to 120,000 VA @ 35% = 117,000 VA @ 35% = 40,950 VA
remainder 205,400 VA  120,000 VA = 85,400 VA @ 25% = 21,350 VA
Demand = 65,300 VA
Finally, determine the kVA: 65,300 VA = 65.3 kVA
1,000
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93.

Cordandplug connected highpressure spray washing machines shall be provided with GFCI protection for personnel when operating at _____.
I. singlephase, 120volts, 30 amperes
II. 3phase, 208volts, 50 amperes


















The GFCI protection requirements for highpressure spray washers are addressed in Section 422.29. This requirement was initiated because at least ten documented electrocutions have been linked to nonindustrial high pressure spray washing
machines of the type purchased by, or rented to consumers. These cordandplug connected machines present shock hazards to users in wet conditions.
The same shock hazard exist for 3phase machines as for singlephase. However, at this date, there are no 3phase, 240volt listed GFCI devices available to be located in the attachment plug or in the cord of the appliance, or listed GFCI devices intended for use on circuits operating greater than 150volts to ground.
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95.

When supplying a 36,000 VA, 240volt, singlephase load in an area where the ambient temperature reaches 119F, determine the MINIMUM size 75C rated copper conductors required to supply the load.


















First, apply the singlephase current formula to determine the load in amperes:
I = power I = 36,000 VA = 150 amperes load
volts 240 volts
Next, due the elevated temperature, divide the load of 150 amperes by .75, the
temperature correction factor given in Table 310.15(B)(2)(a):
required ampacity = 150 amperes = 200 amperes
.75
Then, size 3/0 AWG 75C copper conductors with an allowable ampacity of 200
amperes should be selected from Table 310.15(B)(16).
Incorrect answer, please choose another answer.






