# NEC Practice Test (Code)

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#### A cord-and-plug-attached appliance is linked to a 25 ampere branch circuit. What is the maximum ampere rating that the appliance can handle?

Correct! Wrong!

Cord and plug connected equipment shall not exceed 80 percent of branch circuit rating 25 * 0.8 = 20 amperes

#### On a 15 amp branch circuit operating at 120 volts, what is the maximum number of receptacles that can be installed?

Correct! Wrong!

P = 120 * 15 = 1800 Number of receptacles = 1800 / 180 = 10 receptacles

#### The branch conductor's ampacity, which supplies power to the motor with an FLC of 25 amperes, is

Correct! Wrong!

Conductor ampacity = 125% of FLC = 1.25 * 25 = 31.25 amperes

#### A 2/0 THWN conductor is placed in a dry area at a temperature of 450 degrees Celsius. What is the conductor's corrected ampacity?

Correct! Wrong!

Ampacity of 2/0 THWN conductor in dry location = 195 A ( NEC Table 310.15 (B)(16)) Correction Ampacity at ambient temperature 450 C = 0.87 (NEC Table 310.15(B)(2)) Corrected Ampacity of conductor = 0.87 * 195 = 170 Amperes

#### A 1/0 THWN in a wet location has a rated ampacity of 150 amperes. If the same conductor is used at a temperature of 400 degrees Celsius. What is the corrected ampacity?

Correct! Wrong!

Temperature correction factor of 1/0 THWN Conductor for 40-750 C = 0.88 The corrected ampacity = 0.88 * 150 = 132 amperes

#### The entry points of underground ducts or conduits into the building must be sealed.

Correct! Wrong!

These must be sealed to prevent the entry of gases or moisture in to the building.

#### If the raceway contains four 1/0 THWN conductors, what is the ampacity of a 1/0 THWN conductor in a dry location at a temperature of 1040 F?

Correct! Wrong!

The conductor ampacity of 1/0 THWN in dry location ( 900C) = 170 amperes ( NEC Table 310.15(B)(16)) Ambient temperature correction factor at 1080F = 0.91 ( NEC Table 310.15(B)(2)) Raceway conductor bundle adjustment factor for 4 conductors = 0.8 ( NEC Table 310.15(B)(3)(a)) corrected ampacity = 170 * 0.8*0.91 = 124 amperes

#### What conductor size is needed for a single-family housing unit with an estimated load of 220 amps and a maximum unbalanced neutral load of 110 amps?

Correct! Wrong!

Service conductor for 225 Amperes load = 3/0 AWG neutral conductor for unbalanced load 110 = 3 AWG

#### Which of the following is the minimum size of a single supply-side bonding jumper for three metal raceways with 300 kcmil service conductors each?

Correct! Wrong!

NEC Table 250.102(C)(1) For over 3/0 AWG and 350 kcmil the required bonding jumper size is 2 AWG

#### The connected load of a single family dwelling unit is 4000 VA. The calculated demand load is which of the following?

Correct! Wrong!

First 3000 VA the demand factor is 100% = 3000 * 100 % = 3000 VA remaining (4000-3000) = 1000 ( demand factor = 35%) = 1000 * 0.35 = 350 VA total demand load = 3000 + 350 = 3350 VA

#### What is the adjustment ampacity of a 3/0 THWN conductor in a wet location with an ambient temperature of 1080F? There are four 3/0 THWN conductors in the raceway.

Correct! Wrong!

3/0 THWN conductor ampacity in wet location (750C) = 175 Amperes ( NEC 310.15(B)(16)) Ambient temperature correction factor at 1080F = 0.85 ( NEC 310.15(B)(2)) raceway conductor bundle adjustment factor for 4 conductors= 0.8 ( NEC 310.15(B)(3)(a)) The conductor ampacity = 175 * 0.85 * 0.8 = 119 Amperes

#### The voltmeter at one end of the feeder reads 240 volts. What is the volt-minimum meter's reading voltage at the outlet of lighting loads?

Correct! Wrong!

The maximum voltage drop in feeder is 3 % the voltmeter reading = 240 – 0.03 * 240 = 240 – 7.2 = 232.8 volts

#### The capacity of the conductor is 150 amperes. If five conductors are placed in a flexible cord, what is the maximum conductor ampacity allowed for each conductor?

Correct! Wrong!

For 5 conductors in flexible cord allowed ampacity percentage = 80 % of rated amapcity Allowed amapcity of each conductor with 150 Ampere rating = 150 * 0.8 = 120 amperes.

#### For a 4200 W, 240 V electric water heater connected in a building, what is the maximum size over current device required?

Correct! Wrong!

Protection rating for an electric water heater should be less than 150% of the rating of device. Protection size = 4200 / 240 * 1.5 Protections size = 26.25 A The nearest next size over current device is 30 A.

#### In a 3/4 EMT trade size, how many 10 AWG RHH conductors may be installed?

Correct! Wrong!

Total 6 10 AWG RHH conductors can be installed in trade size ¾ EMT.

#### What size service conductors are needed for a 190 A calculated load in a dwelling? The service disconnect breaker has a 200-amp rating.

Correct! Wrong!

Answer : 2/0 AWG Note : Service conductors that supply individual dwelling unit or single dwelling unit can have ampacity rating of 83% of service disconnect. The conductor ampacity for 200 A service disconnect = 0.83 * 200 = 166 amperes Refer : NEC Table 310.15(B)(16) 2/0 AWG with ampacity 175 amperes ( 750 C) is suitable as service conductor

#### A set of 450 kcmil feeder conductors is protected by a 400 breaker. With a 25 A over current device, a feeder fed from a 450 kcmil feeder disconnects. What is the smallest conductor size for these taps?

Correct! Wrong!

Answer : 8 AWG The over current device 25 A is less than 10 percent of the rating of the feeder breaker (400A.) As per code it cannot be less than 40 A. Refer : NEC Table 310.15(B)(16) 8 AWG at 40A 60 C is the minimum size required.

#### For a 4in * 4 in wireway, what is the maximum permitted conductor fill?

Correct! Wrong!

Wireway Area = 4 in. * 4 in. = 16 sq. in. Refer : NEC 376.22(A) The conductor fill area should be maximum 20 % of the total wireway area. Wireway conductor fill = 16 * 0.2 = 3.2 sq. in.

#### For a 4000 W, 240 electric water heater, what size branch circuit conductor is required?

Correct! Wrong!

Branch circuit conductor size should 125% of continuous load [ NEC 422.10(A)] Electric water heater current = 4000 / 240 = 16.6 Amperes Branch conductor size = 16.6 * 1.25 = 20.75 amperes As per branch conductor next size up rule 10 AWG with rating of 30 A at 60 C is suitable [ Refer NEC Table 310.15(B)(16)]

#### What is the general lighting load of a hospital with an area of 80,000 sq ft after the demand factor is applied?

Correct! Wrong!

As per NEC Table 220.12 , The unit general lighting load for hospital is 2 VA per sq ft. So total lighting load of hospital = 80,000 * 2 = 1,60,000 VA As per NEC Table 220.42 The lighting load demand factors are as follows First 50000 or less 40% i.e 50,000 * 0.4 = 20000 Reaming is 20% 1,10,000 * 20 = 22000 So total lighting load after demand factor is applied = 20000 + 22000 = 42000 VA

#### For a dwelling unit with a calculated load of 240 A, what size feeder conductors are required? The feeder conductors carry the whole load of the dwelling unit, and the service disconnect is rated for 250 A?

Correct! Wrong!

Answer: 4/0 AWG Note: Feeder conductors that supply entire load of dwelling unit can have the rating of 83% of service disconnect of dwelling unit. The service conductor ampacity for 250 A service disconnect = 0.83 * 250 = 207.5 amperes Refer : NEC Table 310.15(B)(16) 4/0 AWG with ampacity 250 amperes ( 75 C) is suitable as service conductor

#### For a constant load of 22 amperes at 750C, what size conductor should be used to protect against overcurrent?

Correct! Wrong!

The size of branch circuit conductor to protect against over current must be more than 125 % of continuous load. 22 * 1.25 = 27.5 amperes. Refer : Table 310.15(B)(16). No 10 THWN with 35 amperes at 75 0C is suitable

#### What is the needed conductor size for GEC if a service is fed by three parallel sets of 400 kcmil conductors?

Correct! Wrong!

The total underground conductor size = 3 * 400 = 1200 kcmil .Refer NEC Table 250.166 .The required GEC is 3/0 AWG for 1200 kcmil size

#### In a 4 in. * 4 in. wireway, how many 500 kcmil THWN conductors can be installed?

Correct! Wrong!

Wireway area = 4 * 4 = 16 sq. in. Refer NEC [376.22(A)] The conductor fill area should be maximum 20% of wireway area. WireWay conductor fill = 16 * 0.2 = 3.2 sq. in. Area of 500 kcmil THHN = 0.7073 sq in. [Chapter 9, Table 5] Number of 500 kcmil THWN conductors = 3.2 / 0.7073 = 4.5 ( 4 conductors)

#### A 45 KVA three phase transformer with a 115/230 delta configuration. What is the transformer secondary's high-leg voltage?

Correct! Wrong!

A delta connected transformer high leg voltage = ( Line voltage /2 ) * 1/732 High leg voltage = 230/2 * 1.732 High leg voltage = 199 Volts

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